5.7: Kinked Worldline
kink in the worldline decreases aging along that worldline
Acceleration-proof clocks
The change in slope of the worldline from event to event in Figures 5.6.1 and 5.6.2 (bottom) means that the clock being carried along this worldline changes velocity: It accelerates. Different clocks behave differently when accelerated. Typically a clock can withstand a great acceleration only when it is small and compact. A pendulum clock is not an accurate timepiece when carried by car through stop-and-go traffic; a wristwatch is fine. A wristwatch is destroyed by being slammed against a wall; a radioactive nucleus is fine. Typically, the smaller the clock, the more acceleration it can withstand and still register properly, and the sharper can be the curves and kinks on its worldline. 1 In all figures like Figures 5.6.1 and 5.6.2 (bottom), we assume the ideal limit of small (acceleration-proof) clocks.
Simplify: Worldlines with straight segments
We are now free to analyze a motion in which particle and clock are subject to a great acceleration. In particular, consider the simple special case of the worldline of Figure 5.6.1. That worldline gradually changes slope as the particle speeds up and slows down. Now make the period of speeding up shorter and shorter (great driving force!); also make the period of slowing down shorter and shorter. 2 In this way come eventually to the limiting case in which episodes of acceleration and deceleration-curved portions of the worldline - are too short even to show up on the scale of the spacetime map (worldline \(O Q B\) in Figure \(\PageIndex{1}\)). In this simple limiting case the whole history of motion is specified by (1) initial event \(O\) , (2) final event \(B\) , and (3) turnaround event \(Q\) , halfway in time between \(O\) and \(B\) . In this case it is particularly easy to see how the lapse of proper time between \(O\) and \(B\) depends on the location of the halfway event - and thus to compare three worldlines, \(O P B, O Q B\) , and \(O R B\) .
Zero proper time for light
Path \(O P B\) is the worldline of a particle that does not move in space; it stays next to the reference-frame clock. Proper time from \(O\) to \(B\) by way of \(P\) is evidently equal to time as measured in the free-float frame of this reference clock:
\[\text { (total proper time along } O P B)=10 \text { meters of time } \nonumber\]
In contrast, on the way from \(O\) to \(B\) via \(R\), for each segment the space separation equals the time separation, so the proper time has the value zero: 3
\[\begin{align*}
(\text { proper time along leg } O R)^{2} &=(\text { time })^{2}-(\text { space })^{2} \\
&=(5 \text { meters })^{2}-(5 \text { meters })^{2} \\
&=0 \\
(\text { total proper time along } O R B) &=2 \times(\text { proper time along } O R) \\
&=0
\end{align*}\]
As far as we know, only three things can travel 5 meters of distance in 5 meters of time: light (photons), neutrinos, and gravitons (see Box 8-1). No material clock can travel at light speed. Therefore the worldline \(O R B\) is not actually attainable by a material particle. However, it can be approached arbitrarily closely. One can find a speed sufficiently close to light speed - and yet less than light speed - so that a trip with this speed first one way then the other will bring an ideal clock back to the reference clock with a lapse of proper time that is as short as one pleases. In the same way we can, in principle, go to the star Canopus and back in as short a round-trip rocket time as we choose (Section 4.8).
Reduced proper time along kinked worldline
As distinguished from the limiting case \(O R B\) , worldline \(O Q B\) demands an amount of proper time that is greater than zero but still less than the 10 meters of proper time along the direct worldline \(O P B\) : 4
\[\begin{array}{rlr} (\text { proper time along leg } O Q)^{2} & =(5 \text { meters })^{2}-(4 \text { meters })^{2} & \\ & =25 \text { (meters) }^{2}-16 \text { (meters) }^{2} & \\ & =9 \text { (meters) }^{2} & \\ & =(3 \text { meters })^{2}\end{array} \nonumber\]
so
\[(\text { proper time along leg } O Q)=3 \text { meters } \nonumber\]
and
\[\begin{aligned} (\text { total proper time along both legs } O Q B) &=2 \times(\text { proper time along } O Q) \\ &=6 \text { meters } \end{aligned}\]
This is less proper time than (proper time along \(O P B\) ) \(=10\) meters that characterized the "direct" worldline \(O P B\) . Our trip to Canopus and back described in Chapter 4 follows a worldline similar to \(O Q B\) .
In the spacetime map shown, time and space are measured in years. A table shows space and time locations of numbered events in this frame.
- One traveler moves along the solid straight worldline segments from event 1 to events 2,3, and 4 . Calculate the time increase on her clock between event 1 and event 2; between event 2 and event 3; between event 3 and event 4. Calculate total proper time-her aging-along worldline 1, 2, 3, 4.
- Another traveler, her twin brother, moves along the straight dotted worldline from event 1 directly to event 4. Calculate the time increase on his clock along the direct worldline 1, 4.
- Which twin (solid-line traveler or dotted-line traveler) is younger when they rejoin at event 4?
Solution
-
From the table next to the map, space separation between events 1 and 2 equals 0. Time separation equals 1 year. Therefore the interval is reckoned from (interval)
\(^{2}=1^{2}-0^{2}=1^{2}\)
. Thus the proper time lapse on a clock carried between events 1 and 2 equals 1 year.
Space separation between event 2 and event 3 equals \(1-(-0.5)=1.5\) light-years. Time separation equals 2 years. Therefore the square of the interval is \(2^{2}-(1.5)^{2}=4-2.25=1.75\) (years) \(^{2}\) and the advance of proper time equals the square root of this, or \(1.32\) years.
Between event 3 and event 4 space separation equals \(2.5\) light-years and time separation 3 years. The square of the interval has the value \(3^{2}-(2.5)^{2}=9-\) \(6.25=2.75\) (years) \(^{2}\) and proper time between these two events equals the square root of this, or \(1.66\) years. Total proper time - aging - along worldline 1, 2, 3, 4 equals the sum of proper times along individual segments: \(1+1.32+1.66=3.98\) years. - Space separation between events 1 and 4 equals 1 light-year. Time separation is 6 years. The squared interval between them equals \(6^{2}-1^{2}=36-1=35\) (years) \(^{2}\) . A traveler who moves along the direct worldline from event 1 to event 4 records a span of proper time equal to the square root of this value, or \(5.92\) years.
- The brother who moves along straight worldline 1,4 ages \(5.92\) years during the trip. The sister who moves along segmented worldline \(1,2,3,4\) ages less: \(3.98\) years. As always in Lorentz geometry, the direct worldline (shown dotted) is longer - that is, it has more elapsed proper time, greater aging - than the indirect worldline (shown solid).
1 Acceleration-proof clocks
2 Simplify: Worldlines with straight segments
3 Zero proper time for light
4 Reduced proper time along kinked worldline


1.4: Drawing the line- preventing sexually transmitted infections
At the end of this case, students will be able to:
- Explain the incidence of sexually transmitted infections (STIs) in the United States
- Describe groups of individuals at risk of developing complications related to specific STIs
- List the strategies for preventing and controlling STIs
- Recommend scheduling for immunizations to protect against STIs and other related infectious diseases
Introduction
Sexually transmitted infections (STIs) present a unique public health problem. Approximately 20 million new STIs are diagnosed in the US each year, and a large number of cases remain undiagnosed or unreported. 1 And while appropriate treatment options exist for many STIs including syphilis, gonorrhea, and chlamydia, 2 the number of new cases continues to increase each year. 1 These infections increase the risk of chronic health issues such as complications in reproductive and fetal health as well as increase risk of acquiring other STIs such as human immunodeficiency virus (HIV). 1
Certain groups have been identified as having a higher risk of acquiring particular STIs and/or developing serious long-term complications associated with STIs. 1 Approximately 50% of patients diagnosed with an STI are between the ages of 15 and 24 years of age. 3 Women of childbearing age are at high risk of long-term complications; the CDC estimates that approximately 20,000 women become infertile annually due to undiagnosed and/or untreated STIs. 4 Importantly, increasing rates of syphilis in women of childbearing age has led to an increase in congenital syphilis, which leads to significant morbidity and mortality in infants. 4 Another group with significant STI risk is men who have sex with men (MSM), and cases of reportable STIs among this population are also consistently increasing year to year. 1
Many behavioral and socioeconomic factors also influence the spread of STIs. 5 Hispanic, Black, and American Indian patients have higher rates of STIs compared to white patients as these groups also experience decreased access to care, poverty, and communities/sexual networks with higher rates of STIs. 5 According to Healthy People 2020 , STIs affect marginalized and indigent patients disproportionately due to decreased access to care and/or social networks with higher risk behaviors. 5 Patients with substance abuse disorders are also at a higher risk of acquiring an STI due to an increased likelihood of engaging in high-risk behaviors. 5 An important aspect of decreasing the societal burden of STIs is the likelihood of patients to seek treatment for these infectious diseases; however, the stigma associated with STIs including HIV may limit patients from accessing diagnosis and care. 5
Education, prevention measures, and prompt diagnosis and treatment are of utmost importance in controlling the STI epidemic in the US as rates of chlamydia, gonorrhea, and syphilis have consistently increased each year from 2013-2017. 1 Funding has also been cut from state resources including health departments; therefore, evaluating patients for sexual history and risky behaviors at any point of contact with the healthcare system is needed. 1 Prevention efforts should be coordinated between community, public health, and medical services. In addition, system-level obstacles should be reevaluated to allow for expedited partner therapy (EPT) for certain types of STIs as well as community-based test and treat programs. Pharmacists are easily accessible to many patients who otherwise may not seek medical care and are in a position to provide much needed patient education, counseling, and linkage to care for those patients who may benefit from STI evaluation and/or treatment.
Case
You are a pharmacist working in an ambulatory care clinic in New York City where you often counsel patients about prevention and treatment of STIs.
CC: “I have a crazy rash that covers most of my body. I am really worried about it because I don’t know where it came from.”
Patient: JB is a 20-year-old African American male who is a senior art major at New York University (NYU).  He does not have a primary care physician in the city and was referred to the clinic by a friend.
HPI: New onset rash that covers ~60% of his body, including the soles of his feet. No fever, chills, or systemic signs of infection. No complaints of pain or trouble urinating.
PMH: No significant history or surgeries
FH:
- Father: unknown
- Mother: hypertension and hyperlipidemia
- One younger sister with no significant medical history
SH:
- Drinks socially (7-8 vodka drinks) on weekend nights
- Denies cigarette smoking
- Occasional drug abuse when “partying with friends” in the city
- Sexually active with multiple male partners (reports condom use ~60% of the time), states that he is typically the receptive partner
SDH: American-born student at NYU with a part-time job at an art studio, full scholarship to NYU with on-campus housing and meals provided, raised by a single mother in rural, upstate New York with minimal access to healthcare service
Medications:
- Acetaminophen PRN for headaches
- Melatonin PRN for sleep
- Multivitamin daily
Allergies: NKDA
Vaccinations: No documentation available, patient states that he thinks he has received all routine childhood vaccines but is unsure
Vitals:
- BP 116/70 mmHg
- HR 70 bpm
Labs: None available at this time
Case Questions
- Is JB considered to be a patient at high risk for acquiring STIs? Why or why not?
- Without further laboratory data, which STI does JB most likely have? What is the appropriate therapy for JB at this time (include appropriate follow-up)? Without proper treatment, which additional STI is JB at high risk for?
- JB is extremely upset with his diagnosis and wants to know more about how to avoid STIs in the future. What non-pharmacologic recommendations can you provide JB with at this time?
- Which screening tests should be performed at least every year in MSMs who are sexually active?
- According to the CDC & the 2015 STD Treatment Guidelines, what are the five major strategies for preventing and controlling the spread of STIs?
- JB wants to know if there are any vaccinations available to protect patients against STIs. What information can you provide JB with at this time? What are the recommended age and dosing schedule for each of these vaccinations?
Author Commentary
STIs are on the rise despite available education, prevention strategies, and antibiotic treatment. For the fifth consecutive year (2013-2018), STI rates, including chlamydia, gonorrhea, and primary/secondary syphilis, have increased based on CDC reports. 1 Resources for testing and treating STIs are limited, especially among groups who are at highest risk for infection. Without appropriate diagnosis and treatment, patients are at risk for long-term health consequences as well as transmitting the infection to others, increasing the societal burden. Partner services are often limited due to lack of appropriate health department resources and/or state laws that prevent EPT. Pharmacists may be one resource that can bridge the gap between patients and health departments/clinics by counseling patients on the importance of being tested and treated for STIs.
Patient Approaches and Opportunities
Patients often do not understand that STIs can be transmitted by anal and oral sex; therefore, patient education at points of contact within the healthcare system is of utmost importance. Additionally, taking a thorough sexual history is necessary to assess the patient for STI risk factors and to recommend routine screening. Ensuring that the patient never feels “judged” by any healthcare worker is an imperative aspect of building a strong relationship. Pharmacists in community and/or ambulatory care settings have a unique opportunity to educate patients about STI transmission, the importance of partner screening, and available prevention measures such as vaccines and barrier contraceptives. In addition, pharmacists can link patients to the nearest health department or local clinics to be tested and treated for STIs. When discussing STIs and sexual health, it is imperative to keep in mind cultural differences and the health literacy of the individual patient.
Related chapters of interest:
- HIV and hepatitis C co-infection: a double-edged sword
- An ounce of prevention: pharmacy applications of the USPSTF guidelines
- Sex education: counseling patients from various cultural backgrounds
- PrEPare yourself: let’s talk about sex
External resources:
- Healthy People 2020 : https://www.healthypeople.gov/2020/topics-objectives/topic/sexually-transmitted-diseases
- Centers for Disease Control and Prevention- Sexually Transmitted Diseases (STDs): https://www.cdc.gov/std/tg2015/tg-2015-print.pdf
- 2021 Sexually Transmitted Disease Treatment Guidelines: https://www.cdc.gov/std/treatment-guidelines/default.htm


13.7: Chapter 11 Review Exercises
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Chapter Review Exercises
Distance and Midpoint Formulas; Circles
Exercise \(\PageIndex{1}\) Use the Distance Formula
In the following exercises, find the distance between the points. Round to the nearest tenth if needed.
- \((-5,1)\) and \((-1,4)\)
- \((-2,5)\) and \((1,5)\)
- \((8,2)\) and \((-7,-3)\)
- \((1,-4)\) and \((5,-5)\)
- Answer
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2. \(d=3\)
4. \(d=\sqrt{17}, d \approx 4.1\)
Exercise \(\PageIndex{2}\) Use the Midpoint Formula
In the following exercises, find the midpoint of the line segments whose endpoints are given.
- \((-2,-6)\) and \((-4,-2)\)
- \((3,7)\) and \((5,1)\)
- \((-8,-10)\) and \((9,5)\)
- \((-3,2)\) and \((6,-9)\)
- Answer
-
2. \((4,4)\)
4. \(\left(\frac{3}{2},-\frac{7}{2}\right)\)
Exercise \(\PageIndex{3}\) Write the Equation of a Circle in Standard Form
In the following exercises, write the standard form of the equation of the circle with the given information.
- radius is \(15\) and center is \((0,0)\)
- radius is \(\sqrt{7}\) and center is \((0,0)\)
- radius is \(9\) and center is \((-3,5)\)
- radius is \(7\) and center is \((-2,-5)\)
- center is \((3,6)\) and a point on the circle is \((3,-2)\)
- center is \((2,2)\) and a point on the circle is \((4,4)\)
- Answer
-
2. \(x^{2}+y^{2}=7\)
4. \((x+2)^{2}+(y+5)^{2}=49\)
6. \((x-2)^{2}+(y-2)^{2}=8\)
Exercise \(\PageIndex{4}\) Graph a Circle
In the following exercises,
- Find the center and radius, then
- Graph each circle.
- \(2 x^{2}+2 y^{2}=450\)
- \(3 x^{2}+3 y^{2}=432\)
- \((x+3)^{2}+(y-5)^{2}=81\)
- \((x+2)^{2}+(y+5)^{2}=49\)
- \(x^{2}+y^{2}-6 x-12 y-19=0\)
- \(x^{2}+y^{2}-4 y-60=0\)
- Answer
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2.
- radius: \(12,\) center: \((0,0)\)
4.
- radius: \(7,\) center: \((-2,-5)\)
6.
- radius: \(8,\) center: \((0,2)\)
Parabolas
Exercise \(\PageIndex{5}\) Graph Vertical Parabolas
In the following exercises, graph each equation by using its properties.
- \(y=x^{2}+4 x-3\)
- \(y=2 x^{2}+10 x+7\)
- \(y=-6 x^{2}+12 x-1\)
- \(y=-x^{2}+10 x\)
- Answer
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2.
4.
Exercise \(\PageIndex{6}\) Graph Vertical Parabolas
In the following exercises,
- Write the equation in standard form, then
- Use properties of the standard form to graph the equation.
- \(y=x^{2}+4 x+7\)
- \(y=2 x^{2}-4 x-2\)
- \(y=-3 x^{2}-18 x-29\)
- \(y=-x^{2}+12 x-35\)
- Answer
-
2.
- \(y=2(x-1)^{2}-4\)
4.
- \(y=-(x-6)^{2}+1\)
Exercise \(\PageIndex{7}\) Graph Horizontal Parabolas
In the following exercises, graph each equation by using its properties.
- \(x=2 y^{2}\)
- \(x=2 y^{2}+4 y+6\)
- \(x=-y^{2}+2 y-4\)
- \(x=-3 y^{2}\)
- Answer
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2.
4.
Exercise \(\PageIndex{8}\) Graph Horizontal Parabolas
In the following exercises,
- Write the equation in standard form, then
- Use properties of the standard form to graph the equation.
- \(x=4 y^{2}+8 y\)
- \(x=y^{2}+4 y+5\)
- \(x=-y^{2}-6 y-7\)
- \(x=-2 y^{2}+4 y\)
- Answer
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2.
- \(x=(y+2)^{2}+1\)
4.
- \(x=-2(y-1)^{2}+2\)
Exercise \(\PageIndex{9}\) Solve Applications with Parabolas
In the following exercises, create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in standard form.
1.
2.
- Answer
-
2. \(y=-\frac{1}{9} x^{2}+\frac{10}{3} x\)
Ellipses
Exercise \(\PageIndex{10}\) Graph an Ellipse with Center at the Origin
In the following exercises, graph each ellipse.
- \(\frac{x^{2}}{36}+\frac{y^{2}}{25}=1\)
- \(\frac{x^{2}}{4}+\frac{y^{2}}{81}=1\)
- \(49 x^{2}+64 y^{2}=3136\)
- \(9 x^{2}+y^{2}=9\)
- Answer
-
2.
4.
Exercise \(\PageIndex{11}\) Find the Equation of an Ellipse with Center at the Origin
In the following exercises, find the equation of the ellipse shown in the graph.
1.
2.
- Answer
-
2. \(\frac{x^{2}}{36}+\frac{y^{2}}{64}=1\)
Exercise \(\PageIndex{12}\) Graph an Ellipse with Center Not at the Origin
In the following exercises, graph each ellipse.
- \(\frac{(x-1)^{2}}{25}+\frac{(y-6)^{2}}{4}=1\)
- \(\frac{(x+4)^{2}}{16}+\frac{(y+1)^{2}}{9}=1\)
- \(\frac{(x-5)^{2}}{16}+\frac{(y+3)^{2}}{36}=1\)
- \(\frac{(x+3)^{2}}{9}+\frac{(y-2)^{2}}{25}=1\)
- Answer
-
2.
4.
Exercise \(\PageIndex{13}\) Graph an Ellipse with Center Not at the Origin
In the following exercises,
- Write the equation in standard form and
- Graph.
- \(x^{2}+y^{2}+12 x+40 y+120=0\)
- \(25 x^{2}+4 y^{2}-150 x-56 y+321=0\)
- \(25 x^{2}+4 y^{2}+150 x+125=0\)
- \(4 x^{2}+9 y^{2}-126 x+405=0\)
- Answer
-
2.
- \(\frac{(x-3)^{2}}{4}+\frac{(y-7)^{2}}{25}=1\)
4.
- \(\frac{x^{2}}{9}+\frac{(y-7)^{2}}{4}=1\)
Exercise \(\PageIndex{14}\) Solve Applications with Ellipses
In the following exercises, write the equation of the ellipse described.
- A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately \(10\) AU and the furthest is approximately \(90\) AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.
- Answer
-
1. Solve
Hyperbolas
Exercise \(\PageIndex{15}\) Graph a Hyperbola with Center at \((0,0)\)
In the following exercises, graph.
- \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\)
- \(\frac{y^{2}}{49}-\frac{x^{2}}{16}=1\)
- \(9 y^{2}-16 x^{2}=144\)
- \(16 x^{2}-4 y^{2}=64\)
- Answer
-
1.
3.
Exercise \(\PageIndex{16}\) Graph a Hyperbola with Center at \((h,k)\)
In the following exercises, graph.
- \(\frac{(x+1)^{2}}{4}-\frac{(y+1)^{2}}{9}=1\)
- \(\frac{(x-2)^{2}}{4}-\frac{(y-3)^{2}}{16}=1\)
- \(\frac{(y+2)^{2}}{9}-\frac{(x+1)^{2}}{9}=1\)
- \(\frac{(y-1)^{2}}{25}-\frac{(x-2)^{2}}{9}=1\)
- Answer
-
1.
3.
Exercise \(\PageIndex{17}\) Graph a Hyperbola with Center at \((h,k)\)
In the following exercises,
- Write the equation in standard form and
- Graph.
- \(4 x^{2}-16 y^{2}+8 x+96 y-204=0\)
- \(16 x^{2}-4 y^{2}-64 x-24 y-36=0\)
- \(4 y^{2}-16 x^{2}+32 x-8 y-76=0\)
- \(36 y^{2}-16 x^{2}-96 x+216 y-396=0\)
- Answer
-
1.
- \(\frac{(x+1)^{2}}{16}-\frac{(y-3)^{2}}{4}=1\)
3.
- \(\frac{(y-1)^{2}}{16}-\frac{(x-1)^{2}}{4}=1\)
Exercise \(\PageIndex{18}\) Identify the Graph of Each Equation as a Circle, Parabola, Ellipse, or Hyperbola
In the following exercises, identify the type of graph.
-
- \(16 y^{2}-9 x^{2}-36 x-96 y-36=0\)
- \(x^{2}+y^{2}-4 x+10 y-7=0\)
- \(y=x^{2}-2 x+3\)
- \(25 x^{2}+9 y^{2}=225\)
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- \(x^{2}+y^{2}+4 x-10 y+25=0\)
- \(y^{2}-x^{2}-4 y+2 x-6=0\)
- \(x=-y^{2}-2 y+3\)
- \(16 x^{2}+9 y^{2}=144\)
- Answer
-
1.
- Hyperbola
- Circle
- Parabola
- Ellipse
Solve Systems of Nonlinear Equations
Exercise \(\PageIndex{19}\) Solve a System of Nonlinear Equations Using Graphing
In the following exercises, solve the system of equations by using graphing.
- \(\left\{\begin{array}{l}{3 x^{2}-y=0} \\ {y=2 x-1}\end{array}\right.\)
- \(\left\{\begin{array}{l}{y=x^{2}-4} \\ {y=x-4}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}+y^{2}=169} \\ {x=12}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}+y^{2}=25} \\ {y=-5}\end{array}\right.\)
- Answer
-
1.
3.
Exercise \(\PageIndex{20}\) Solve a System of Nonlinear Equations Using Substitution
In the following exercises, solve the system of equations by using substitution.
- \(\left\{\begin{array}{l}{y=x^{2}+3} \\ {y=-2 x+2}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}+y^{2}=4} \\ {x-y=4}\end{array}\right.\)
- \(\left\{\begin{array}{l}{9 x^{2}+4 y^{2}=36} \\ {y-x=5}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}+4 y^{2}=4} \\ {2 x-y=1}\end{array}\right.\)
- Answer
-
1. \((-1,4)\)
3. No solution
Exercise \(\PageIndex{21}\) Solve a System of Nonlinear Equations Using Elimination
In the following exercises, solve the system of equations by using elimination.
- \(\left\{\begin{array}{l}{x^{2}+y^{2}=16} \\ {x^{2}-2 y-1=0}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}-y^{2}=5} \\ {-2 x^{2}-3 y^{2}=-30}\end{array}\right.\)
- \(\left\{\begin{array}{l}{4 x^{2}+9 y^{2}=36} \\ {3 y^{2}-4 x=12}\end{array}\right.\)
- \(\left\{\begin{array}{l}{x^{2}+y^{2}=14} \\ {x^{2}-y^{2}=16}\end{array}\right.\)
- Answer
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1. \((-\sqrt{7}, 3),(\sqrt{7}, 3)\)
3. \((-3,0),(0,-2),(0,2)\)
Exercise \(\PageIndex{22}\) Use a System of Nonlinear Equations to Solve Applications
In the following exercises, solve the problem using a system of equations.
- The sum of the squares of two numbers is \(25\). The difference of the numbers is \(1\). Find the numbers.
- The difference of the squares of two numbers is \(45\). The difference of the square of the first number and twice the square of the second number is \(9\). Find the numbers.
- The perimeter of a rectangle is \(58\) meters and its area is \(210\) square meters. Find the length and width of the rectangle.
- Colton purchased a larger microwave for his kitchen. The diagonal of the front of the microwave measures \(34\) inches. The front also has an area of \(480\) square inches. What are the length and width of the microwave?
- Answer
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1. \(-3\) and \(-4\) or \(4\) and \(3\)
3. If the length is \(14\) inches, the width is \(15\) inches. If the length is \(15\) inches, the width is \(14\) inches.
Practice Test
Exercise \(\PageIndex{23}\)
In the following exercises, find the distance between the points and the midpoint of the line segment with the given endpoints. Round to the nearest tenth as needed.
- \((-4,-3)\) and \((-10,-11)\)
- \((6,8)\) and \((-5,-3)\)
- Answer
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1. distance: \(10,\) midpoint: \((-7,-7)\)
Exercise \(\PageIndex{24}\)
In the following exercises, write the standard form of the equation of the circle with the given information.
- radius is \(11\) and center is \((0,0)\)
- radius is \(12\) and center is \((10,-2)\)
- center is \((-2,3)\) and a point on the circle is \((2,-3)\)
- Find the equation of the ellipse shown in the graph.
- Answer
-
1. \(x^{2}+y^{2}=121\)
3. \((x+2)^{2}+(y-3)^{2}=52\)
Exercise \(\PageIndex{25}\)
In the following exercises,
- Identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola, and
- Graph the equation.
- \(4 x^{2}+49 y^{2}=196\)
- \(y=3(x-2)^{2}-2\)
- \(3 x^{2}+3 y^{2}=27\)
- \(\frac{y^{2}}{100}-\frac{x^{2}}{36}=1\)
- \(\frac{x^{2}}{16}+\frac{y^{2}}{81}=1\)
- \(x=2 y^{2}+10 y+7\)
- \(64 x^{2}-9 y^{2}=576\)
- Answer
-
1.
- Ellipse
3.
- Circle
5.
- Ellipse
7.
- Hyperbola
Exercise \(\PageIndex{26}\)
In the following exercises,
- Identify the type of graph of each equation as a circle, parabola, ellipse, or hyperbola,
- Write the equation in standard form, and
- Graph the equation.
- \(25 x^{2}+64 y^{2}+200 x-256 y-944=0\)
- \(x^{2}+y^{2}+10 x+6 y+30=0\)
- \(x=-y^{2}+2 y-4\)
- \(9 x^{2}-25 y^{2}-36 x-50 y-214=0\)
- \(y=x^{2}+6 x+8\)
- Solve the nonlinear system of equations by graphing: \(\left\{\begin{array}{l}{3 y^{2}-x=0} \\ {y=-2 x-1}\end{array}\right.\).
- Solve the nonlinear system of equations using substitution: \(\left\{\begin{array}{l}{x^{2}+y^{2}=8} \\ {y=-x-4}\end{array}\right.\).
- Solve the nonlinear system of equations using elimination: \(\left\{\begin{array}{l}{x^{2}+9 y^{2}=9} \\ {2 x^{2}-9 y^{2}=18}\end{array}\right.\)
- Create the equation of the parabolic arch formed in the foundation of the bridge shown. Give the answer in \(y=a x^{2}+b x+c\) form.
10. A comet moves in an elliptical orbit around a sun. The closest the comet gets to the sun is approximately \(20\) AU and the furthest is approximately \(70\) AU. The sun is one of the foci of the elliptical orbit. Letting the ellipse center at the origin and labeling the axes in AU, the orbit will look like the figure below. Use the graph to write an equation for the elliptical orbit of the comet.
11. The sum of two numbers is \(22\) and the product is \(−240\). Find the numbers.
12. For her birthday, Olive’s grandparents bought her a new widescreen TV. Before opening it she wants to make sure it will fit her entertainment center. The TV is \(55\)”. The size of a TV is measured on the diagonal of the screen and a widescreen has a length that is larger than the width. The screen also has an area of \(1452\) square inches. Her entertainment center has an insert for the TV with a length of \(50\) inches and width of \(40\) inches. What are the length and width of the TV screen and will it fit Olive’s entertainment center?
- Answer
-
2.
- Circle
- \((x+5)^{2}+(y+3)^{2}=4\)
4.
- Hyperbola
- \(\frac{(x-2)^{2}}{25}-\frac{(y+1)^{2}}{9}=1\)
6. No solution
8. \((0,-3),(0,3)\)
10. \(\frac{x^{2}}{2025}+\frac{y^{2}}{1400}=1\)
12. The length is \(44\) inches and the width is \(33\) inches. The TV will fit Olive’s entertainment center.
Glossary
- system of nonlinear equations
- A system of nonlinear equations is a system where at least one of the equations is not linear.


6.4E: Coupling Specific Genes to Specific Organisms Using PCR
- Describe how polymerase chain reaction (PCR) allows for the amplification and mutation of DNA and enables researchers to study very small samples
Polymerase chain reaction (PCR) is a useful technique for scientists, because it allows for the amplification and mutation of DNA. Through PCR, small quantities of DNA can be replicated by orders of magnitude, not only essentially preserving the sample if successful, but allowing for study on a much larger scale.. Without PCR, the studies we perform would be limited by the amount of DNA we were able to isolate from samples. Through PCR, the original DNA is essentially limitless, allowing scientists to induce various mutations in different genes for further study.
Through site-directed mutagenesis or customized primers, individual mutations in DNA can be made. By changing the amino acids transcribed from DNA through individual mutations, the importance of those amino acids with respect to gene function can be analyzed. However, this process can be difficult, particularly when genes act in concert (with varying expression with respect to gene activity). The length of time it takes to run a successful PCR and perform other techniques before additional studies can be done (protein expression, isolation, and purification, for example), makes biochemical research time-consuming and difficult. However, PCR, coupled with other biochemical techniques, allows us to analyze the very core of organisms and the processes by which they function. Common PCR protocols in labs today include knockout genotyping, fluorescence genotyping and mutant genotyping. Researchers can use PCR as a method of searching for genes by using primers that flank the target sequence of the gene along with all other necessary components for PCR. If the gene is present, the primers will bind and amplify the DNA, giving a band of amplified DNA on the agarose gel that will be run. If the gene is not present, the primers will not anneal and no amplification will occur.
The ability to identify specific genes to specific organisms has increased the use of PCR and has allowed it to be more specific and eliminate the possibility of cross contaminants. The identification of specific genes to specific organisms has important medical diagnostic value.
PCR is a reliable method to detect the presence of unwanted genetic materials, such as infections and bacteria in the clinical setting. It can even allow identification of an infectious agent without culturing. For example, in diagnosis of diseases like AIDS, PCR can be used to detect the small percentage of cells that are infected with HIV by utilizing primers that are specific for genes specialized to the HIV virus. PCR can reveal the presence of HIV in people who have not mounted an immune response to this pathogen, which may otherwise be missed with an antibody assay). Additionally, PCR is used for identifying bacterial species, such as Mycobacterium tuberculosis in tissue specimens. With the use of PCR, as few as 10 bacilli per million human cells can be readily detected. The bacilli are identified by using Mycobacterium tuberculosis specific genes.
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Key Points
- PCR allows for identification of an infectious agent without the need for culturing.
- Researchers can use PCR as a method of searching for specific genes and/or mutations.
- PCR, coupled with other biochemical techniques, allows us to analyze the very core of organisms and the processes by which they function.
Key Terms
- polymerase chain reaction : A technique in molecular biology for creating multiple copies of DNA from a sample; used in genetic fingerprinting etc.


9.3: Calorimetry
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Learning Objectives
By the end of this section, you will be able to:
- Explain the technique of calorimetry
- Calculate and interpret heat and related properties using typical calorimetry data
One technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry . Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study. The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system).
A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure \(\PageIndex{1}\)). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.
Calorimetry measurements are important in understanding the heat transferred in reactions involving everything from microscopic proteins to massive machines. During her time at the National Bureau of Standards, research chemist Reatha Clark King performed calorimetric experiments to understand the precise heats of various flourine compounds. Her work was important to NASA in their quest for better rocket fuels.
Scientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment, which effectively limits the “surroundings” to the nonsystem components with the calorimeter (and the calorimeter itself). This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure \(\PageIndex{2}\)). These easy-to-use “coffee cup” calorimeters allow more heat exchange with the outside environment, and therefore produce less accurate energy values.
Commercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure \(\PageIndex{3}\)).
Before discussing the calorimetry of chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\PageIndex{4}\)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either its external environment. Under these ideal circumstances, the net heat change is zero:
\[q_{\text {substance }} M +q_{\text {substance }} W =0 \nonumber \]
This relationship can be rearranged to show that the heat gained by substance \(M\) is equal to the heat lost by substance \(W\):
\[q_{\text {substance M }}=-q_{\text {substance W }} \nonumber \]
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that \(q_{\text {substance M }}\) and \(q_{\text {substance W}}\) are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, \(q_{\text {substance M}}\) is a negative value and \(q_{\text {substance W}}\) is positive, since heat is transferred from M to W.
Example \(\PageIndex{1}\): Heat Transfer between Substances at Different Temperatures
A 360.0-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table 5.1), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the outside environment, then heat given off by rebar = −heat taken in by water , or:
\[q_{\text {rebar }}=-q_{\text {water }} \nonumber \]
Since we know how heat is related to other measurable quantities, we have:
\[(c \times m \times \Delta T)_{\text {rebar }}=-(c \times m \times \Delta T)_{\text {water }} \nonumber \]
Letting \(f\) = final and \(i\) = initial, in expanded form, this becomes:
\[c_{\text {rebar }} \times m_{\text {rebar }} \times\left(T_{ f , \text { rebar }}-T_{ i , \text { rebar }}\right)=-c_{\text {water }} \times m_{\text {water }} \times\left(T_{ f \text {,water }}-T_{ i , \text { water }}\right) \nonumber \]
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
\[\left(0.449 J / g { }^{\circ} C \right)(360.0 g )\left(42.7{ }^{\circ} C -T_{ i , \text { rebar }}\right)=-\left(4.184 J / g { }^{\circ} C \right)(425 g )\left(42.7{ }^{\circ} C -24.0^{\circ} C \right) \nonumber \]
\[T_{i, \text { rebar }}=\frac{\left(4.184 J / g^{\circ} C \right)(425 g )\left(42.7{ }^{\circ} C -24.0^{\circ} C \right)}{\left(0.449 J / g { }^{\circ} C \right)(360.0 g )}+42.7{ }^{\circ} C \nonumber \]
Solving this gives \(T_{i, \text { rebar }}\) = 248 °C, so the initial temperature of the rebar was 248 °C.
Exercise \(\PageIndex{1A}\)
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.
- Answer
-
The initial temperature of the copper was 335.6 °C.
Exercise \(\PageIndex{1B}\)
A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.
- Answer
-
The final temperature (reached by both copper and water) is 38.7 °C.
This method can also be used to determine other quantities, such as the specific heat of an unknown metal.
Example \(\PageIndex{2}\): Identifying a Metal by Measuring Specific Heat
A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.
Solution
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water , or:
\[q_{\text {metal }}=-q_{\text {water }} \nonumber \]
In expanded form, this is:
\[c_{\text {metal }} \times m_{\text {metal }} \times\left(T_{ f , \text { metal }}-T_{ i , \text { metal }}\right)=-c_{\text {water }} \times m_{\text {water }} \times\left(T_{ f \text {,water }}-T_{ i , \text { water }}\right) \nonumber \]
Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:
\[\left(c_{\text {metal }}\right)(59.7 g )\left(28.5^{\circ} C -100.0^{\circ} C \right)=-\left(4.18 J / g { }^{\circ} C \right)(60.0 g )\left(28.5^{\circ} C -22.0^{\circ} C \right) \nonumber \]
Solving this:
\[c_{\text {metal }}=\frac{-\left(4.184 J / g^{\circ} C \right)(60.0 g )\left(6.5^{\circ} C \right)}{(59.7 g )\left(-71.5^{\circ} C \right)}=0.38 J / g { }^{\circ} C \nonumber \]
Comparing this with values in Table 5.1, our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.
Exercise \(\PageIndex{2}\)
A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).
- Answer
-
c metal = 0.13 J/g °C
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.
When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the outside environment. Because energy is neither created nor destroyed during a chemical reaction, the heat produced or consumed in the reaction (the “system”), \( q_{\text {reaction }}), plus the heat absorbed or lost by the solution (the “surroundings”), \( q_{\text {solution }}\), must add up to zero:
\[q_{\text {reaction }} + q_{\text {solution }}=0 \nonumber \]
This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:
\[q_{\text {reaction }}=-q_{\text {solution }} \nonumber \]
This concept lies at the heart of all calorimetry problems and calculations.
Example \(\PageIndex{3}\): Heat Produced by an Exothermic Reaction
When 50.0 mL of 1.00 M HCl( aq ) and 50.0 mL of 1.00 M NaOH( aq ), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?
\[\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)} \nonumber \]
Solution
To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.
The heat given off by the reaction is equal to that taken in by the solution. Therefore:
\[q_{\text {reaction }}=-q_{\text {solution }} \nonumber \]
(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)
Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:
\[q_{\text {solution }}=(c \times m \times \Delta T)_{\text {solution }} \nonumber \]
To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 10 2 g (two significant figures). The specific heat of water is approximately 4.184 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:
\[q_{\text {solution }}=\left(4.184 J / g^{\circ} C \right)\left(1.0 \times 10^2 g \right)\left(28.9{ }^{\circ} C -22.0^{\circ} C \right)=2.9 \times 10^3 J \nonumber \]
Finally, since we are trying to find the heat of the reaction, we have:
\[q_{\text {reaction }}=-q_{\text {solution }}=-2.9 \times 10^3 J \nonumber \]
The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.
Exercise \(\PageIndex{3}\)
When 100 mL of 0.200 M NaCl( aq ) and 100 mL of 0.200 M AgNO 3 ( aq ), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?
- Answer
-
1.34 1.3 kJ; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water
Chemistry in Everyday Life: Thermochemistry of Hand Warmers
When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure \(\PageIndex{5}\)). A common reusable hand warmer contains a supersaturated solution of \(\ce{NaC2H3O2}\) (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable \(\ce{NaC2H3O2}\) quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).
The process
\[\ce{NaC2H3O2(aq) -> NaC2H3O2(s)} \nonumber \]
is exothermic, and the heat produced by this proces s is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the \(\ce{NaC2H3O2}\) redissolves and can be reused.
Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is
\[\ce{2 Fe (s) + 3/2 O2(g) -> Fe2O3(s)} \nonumber \]
Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.
Link to Learning
This link shows the precipitation reaction that occurs when the disk in a chemical hand warmer is flexed.
Example \(\PageIndex{4}\): Heat Flow in an Instant Ice Pack
When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an “instant ice pack” (Figure \(\PageIndex{6}\)). When 3.21 g of solid NH 4 NO 3 dissolves in 50.0 g of water at 24.9 °C in a calorimeter, the temperature decreases to 20.3 °C.
Calculate the value of q for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.
Solution
We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:
\[q_{ rxn }=-q_{ soln } \nonumber \]
with “rxn” and “soln” used as shorthand for “reaction” and “solution,” respectively.
Assuming also that the specific heat of the solution is the same as that for water, we have:
\[\begin{aligned}
q_{ rxn } &=-q_{ soln }=-(c \times m \times \Delta T)_{ soln } \\[4pt]
&=-\left[\left(4.184 J / g { }^{\circ} C \right) \times(53.2 g ) \times\left(20.3{ }^{\circ} C -24.9{ }^{\circ} C \right)\right] \\[4pt]
& =-\left[\left(4.184 J / g { }^{\circ} C \right) \times(53.2 g ) \times\left(-4.6{ }^{\circ} C \right)\right] \\[4pt]
& +1.0 \times 10^3 J =+1.0 kJ
\end{aligned} \nonumber \]
The positive sign for q indicates that the dissolution is an endothermic process.
Exercise \(\PageIndex{4}\)
When a 3.00-g sample of \(\ce{KCl}\) was added to 3.00 10 2 g of water in a coffee cup calorimeter, the temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the \(\ce{KCl}\)? What assumptions did you make?
- Answer
-
1.33 kJ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water
If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.
The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter , is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term “bomb” comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water (Figure \(\PageIndex{7}\)). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is absorbed by the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known q , such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.
Link to Learning
Click on this link to view how a bomb calorimeter is prepared for action. This site shows calorimetric calculations using sample data.
Example \(\PageIndex{5}\): Bomb Calorimetry
When 3.12 g of glucose, C 6 H 12 O 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases from 23.8 °C to 35.6 °C. The calorimeter contains 775 g of water, and the bomb itself has a heat capacity of 893 J/°C. How much heat was produced by the combustion of the glucose sample?
Solution
The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)
The heat produced by the reaction is absorbed by the water and the bomb:
\[\begin{aligned}
q_{ rxn } &=-\left(q_{\text {water }}+q_{\text {bomb }}\right) \\[4pt]
&=-\left[\left(4.184 J / g { }^{\circ} C \right) \times(775 g ) \times\left(35.6{ }^{\circ} C -23.8{ }^{\circ} C \right)+893 J /{ }^{\circ} C \times\left(35.6{ }^{\circ} C -23.8{ }^{\circ} C \right)\right] \\[4pt]
&=-(38,300 J +10,500 J ) \\[4pt]
&=-48,800 J =-48.8 kJ
\end{aligned} \nonumber \]
This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.
Exercise \(\PageIndex{5}\)
When 0.963 g of benzene, C 6 H 6 , is burned in a bomb calorimeter, the temperature of the calorimeter increases by 8.39 °C. The bomb has a heat capacity of 784 J/°C and is submerged in 925 mL of water. How much heat was produced by the combustion of the benzene sample?
- Answer
-
q rx = –39.0 kJ (the reaction produced 39.0 kJ of heat)
Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. 2 These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes.
For example Carla Prado's team at University of Alberta undertook whole-body calorimetry to understand the energy expenditures of women who had recently given birth. Studies like this help develop better recommendations and regimens for nutrition, exercise, and general wellbeing during this period of significant physiological change. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories (1 kcal), the amount of energy needed to heat 1 kg of water by 1 °C.
Chemistry in Everyday Life: Measuring Nutritional Calories
In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) = 1000 cal, or 1 kcal. (This is approximately the amount of energy needed to heat 1 kg of water by 1 °C.)
The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure \(\PageIndex{8}\)).
For the example shown in (b), the total energy per 228-g portion is calculated by:
\[(5 g \text { protein } \times 4 \text { Calories } / g )+(31 g \text { carb } \times 4 \text { Calories } / g )+(12 g \text { fat } \times 9 \text { Calories } / g )=252 \text { Calories } \nonumber \]
So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freeze-dried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.
Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.
Link to Learning
Click on this link to access the US Department of Agriculture (USDA) National Nutrient Database, containing nutritional information on over 8000 foods.
Footnotes
- 2 Francis D. Reardon et al. “The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.” Medical and Biological Engineering and Computing 8 (2006)721–28, http://link.springer.com/article/10....517-006-0086-5 .


10.11: Why It Matters- Social Media
Why use social media to present business messages?
While the content of business communications should be the central item of concern, there is no question that the type of media used to express the message is important. Social media, which has its origins all the way back to the 1970s ARPANET, or the original web, presents society with many conflicting and competing concepts. Some consider social media one of the greatest developments of the modern age, and people can be seen every day using their phones to connect with people across the world. However, others consider social media to be something sinister that saps real social connections and harms our ability to communicate more meaningfully.
Taking stock of where you personally stand on these issues can be an interesting exercise, and will most certainly affect your perspective on social media use in business. Most social media in business is used around two or three main functions, which are largely oriented around business marketing:
- To bring services or products to market; that is, to increase awareness.
- To engage customers, often with respect to service issues or problems.
- To encourage a “buzz” or interest around a company, its product(s) or its service(s).
Do these seem useful, correct, interesting and helpful? Most certainly, bringing new products or services to market requires a well developed marketing plan. While marketing is outside the scope of this course, it’s easy to see the importance of and relationship between business communication and marketing. The relationship is around increasing awareness: have you helped your customers—external or internal [1] —to make meaningful decisions?
The following video explores the importance of social media in business communications.
Note the emphasis on awareness, especially awareness gained via relatively inexpensive and commonly found means. We might position inexpensive awareness as our core concept for social media and business communication.
In this module, we will address which platforms are the most popular. We will also look at why they are popular for a given type of message or communication. Lastly, you will gain insight on which type of social media might be best for your organization.
- internal customer: another member of your own organization, especially a non-executive or non-manager
Contributors and Attributions
- Why It Matters: Social Media. Authored by : Freedom Learning Group. Provided by : Lumen Learning. License : CC BY: Attribution
- How Social Media and Electronic Communication Are Revolutionizing Business Communication, 2016. Authored by : Bovee and Thill Business Communication Videos. Located at : https://youtu.be/h1dz3htWqkg . License : All Rights Reserved . License Terms : Standard YouTube License


30.5: The Search for Extraterrestrial Intelligence
By the end of this section, you will be able to:
- Explain why spaceships from extraterrestrial civilizations are unlikely to have visited us
- List efforts by humankind to communicate with other civilizations via messages on spacecraft
- Understand the various SETI programs scientists are undertaking
Given all the developments discussed in this chapter, it seems likely that life could have developed on many planets around other stars. Even if that life is microbial, we saw that we may soon have ways to search for chemical biosignatures. This search is of fundamental importance for understanding biology, but it does not answer the question, “Are we alone?” that we raised at the beginning of this chapter. When we ask this question, many people think of other intelligent creatures, perhaps beings that have developed technology similar to our own. If any intelligent, technical civilizations have arisen, as has happened on Earth in the most recent blink of cosmic time, how could we make contact with them?
This problem is similar to making contact with people who live in a remote part of Earth. If students in the United States want to converse with students in Australia, for example, they have two choices. Either one group gets on an airplane and travels to meet the other, or they communicate by sending a message remotely. Given how expensive airline tickets are, most students would probably select the message route.
In the same way, if we want to get in touch with intelligent life around other stars, we can travel, or we can try to exchange messages. Because of the great distances involved, interstellar space travel would be very slow and prohibitively expensive. The fastest spacecraft the human species has built so far would take almost 80,000 years to get to the nearest star. While we could certainly design a faster craft, the more quickly we require it to travel, the greater the energy cost involved. To reach neighboring stars in less than a human life span, we would have to travel close to the speed of light. In that case, however, the expense would become truly astronomical.
Interstellar Travel
Bernard Oliver , an engineer with an abiding interest in life elsewhere, made a revealing calculation about the costs of rapid interstellar space travel. Since we do not know what sort of technology we (or other civilizations) might someday develop, Oliver considered a trip to the nearest star (and back again) in a spaceship with a “perfect engine”—one that would convert its fuel into energy with 100% efficiency. Even with a perfect engine, the energy cost of a single round-trip journey at 70% the speed of light turns out to be equivalent to several hundred thousand years’ worth of total U.S. electrical energy consumption. The cost of such travel is literally out of this world.
This is one reason astronomers are so skeptical about claims that UFOs are spaceships from extraterrestrial civilizations. Given the distance and energy expense involved, it seems unlikely that the dozens of UFOs (and even UFO abductions) claimed each year could be visitors from other stars so fascinated by Earth civilization that they are willing to expend fantastically large amounts of energy or time to reach us. Nor does it seem credible that these visitors have made this long and expensive journey and then systematically avoided contacting our governments or political and intellectual leaders.
Not every UFO report has been explained (in many cases, the observations are sketchy or contradictory). But investigation almost always converts them to IFOs (identified flying objects) or NFOs (not-at-all flying objects). While some are hoaxes, others are natural phenomena, such as bright planets, ball lightning, fireballs (bright meteors), or even flocks of birds that landed in an oil slick to make their bellies reflective. Still others are human craft, such as private planes with some lights missing, or secret military aircraft. It is also interesting that the group of people who most avidly look at the night sky, the amateur astronomers, have never reported UFO sightings. Further, not a single UFO has ever left behind any physical evidence that can be tested in a laboratory and shown to be of nonterrestrial origin.
Another common aspect of belief that aliens are visiting Earth comes from people who have difficulty accepting human accomplishments. There are many books and TV shows, for example, that assert that humans could not have built the great pyramids of Egypt, and therefore they must have been built by aliens. The huge statues (called Moai) on Easter Island are also sometimes claimed to have been built by aliens. Some people even think that the accomplishments of space exploration today are based on alien technology.
However, the evidence from archaeology and history is clear: ancient monuments were built by ancient people, whose brains and ingenuity were every bit as capable as ours are today, even if they didn’t have electronic textbooks like you do.
Messages on Spacecraft
While space travel by living creatures seems very difficult, robot probes can travel over long distances and over long periods of time. Five spacecraft—two Pioneers, two Voyagers, and New Horizons—are now leaving the solar system. At their coasting speeds, they will take hundreds of thousands or millions of years to get anywhere close to another star. On the other hand, they were the first products of human technology to go beyond our home system, so we wanted to put messages on board to show where they came from.
Each Pioneer carries a plaque with a pictorial message engraved on a gold-anodized aluminum plate (Figure 30.18). The Voyagers, launched in 1977, have audio and video records attached, which allowed the inclusion of over 100 photographs and a selection of music from around the world. Given the enormous space between stars in our section of the Galaxy, it is very unlikely that these messages will ever be received by anyone. They are more like a note in a bottle thrown into the sea by a shipwrecked sailor, with no realistic expectation of its being found soon but a slim hope that perhaps someday, somehow, someone will know of the sender’s fate.
An Excerpt from the Voyager Record:
“We cast this message into the cosmos. It is likely to survive a billion years into our future, when our civilization is profoundly altered. . . . If [another] civilization intercepts Voyager and can understand these recorded contents, here is our message:
This is a present from a small, distant world, a token of our sounds, our science, our images, our music, our thoughts, and our feelings. We are attempting to survive our time so we may live into yours. We hope, someday, having solved the problems we face, to join a community of galactic civilizations. This record represents our hope and our determination, and our goodwill in a vast and awesome universe.”
—Jimmy Carter, President of the United States of America, June 16, 1977
Communicating with the Stars
If direct visits among stars are unlikely, we must turn to the alternative for making contact: exchanging messages. Here the news is a lot better. We already use a messenger—light or, more generally, electromagnetic waves—that moves through space at the fastest speed in the universe. Traveling at 300,000 kilometers per second, light reaches the nearest star in only 4 years and does so at a tiny fraction of the cost of sending material objects. These advantages are so clear and obvious that we assume they will occur to any other species of intelligent beings that develop technology.
However, we have access to a wide spectrum of electromagnetic radiation, ranging from the longest-wavelength radio waves to the shortest-wavelength gamma rays. Which would be the best for interstellar communication? It would not be smart to select a wavelength that is easily absorbed by interstellar gas and dust, or one that is unlikely to penetrate the atmosphere of a planet like ours. Nor would we want to pick a wavelength that has lots of competition for attention in our neighborhood.
One final criterion makes the selection easier: we want the radiation to be inexpensive enough to produce in large quantities. When we consider all these requirements, radio waves turn out to be the best answer. Being the lowest-frequency (and lowest-energy) band of the spectrum, they are not very expensive to produce, and we already use them extensively for communications on Earth. They are not significantly absorbed by interstellar dust and gas. With some exceptions, they easily pass through Earth’s atmosphere and through the atmospheres of the other planets we are acquainted with.
The Cosmic Haystack
Having made the decision that radio is the most likely means of communication among intelligent civilizations, we still have many questions and a daunting task ahead of us. Shall we send a message, or try to receive one? Obviously, if every civilization decides to receive only, then no one will be sending, and everyone will be disappointed. On the other hand, it may be appropriate for us to begin by listening, since we are likely to be among the most primitive civilizations in the Galaxy who are interested in exchanging messages.
We do not make this statement to insult the human species (which, with certain exceptions, we are rather fond of). Instead, we base it on the fact that humans have had the ability to receive (or send) a radio message across interstellar distances for only a few decades. Compared to the ages of the stars and the Galaxy, this is a mere instant. If there are civilizations out there that are ahead of us in development by even a short time (in the cosmic sense), they are likely to have a technology head start of many, many years.
In other words, we, who have just started, may well be the “youngest” species in the Galaxy with this capability (see the discussion in Example 30.1). Just as the youngest members of a community are often told to be quiet and listen to their elders for a while before they say something foolish, so may we want to begin our exercise in extraterrestrial communication by listening.
Even restricting our activities to listening, however, leaves us with an array of challenging questions. For example, if an extraterrestrial civilization’s signal is too weak to be detected by our present-day radio telescopes, we will not detect them. In addition, it would be very expensive for an extraterrestrial civilization to broadcast on a huge number of channels. Most likely, they select one or a few channels for their particular message. Communicating on a narrow band of channels also helps distinguish an artificial message from the radio static that comes from natural cosmic processes. But the radio band contains an astronomically large number of possible channels. How can we know in advance which one they have selected, and how they have coded their message into the signal?
Table 30.1 summarizes these and other factors that scientists must grapple with when trying to tune in to radio messages from distant civilizations. Because their success depends on either guessing right about so many factors or searching through all the possibilities for each factor, some scientists have compared their quest to looking for a needle in a haystack. Thus, they like to say that the list of factors in Table 30.1 defines the cosmic haystack problem .
| Factors |
|---|
| From which direction (which star) is the message coming? |
| On what channels (or frequencies) is the message being broadcast? |
| How wide in frequency is the channel? |
| How strong is the signal (can our radio telescopes detect it)? |
| Is the signal continuous, or does it shut off at times (as, for example, a lighthouse beam does when it turns away from us)? |
| Does the signal drift (change) in frequency because of the changing relative motion of the source and the receiver? |
| How is the message encoded in the signal (how do we decipher it)? |
| Can we even recognize a message from a completely alien species? Might it take a form we don’t at all expect? |
Radio Searches
Although the cosmic haystack problem seems daunting, many other research problems in astronomy also require a large investment of time, equipment, and patient effort. And, of course, if we don’t search, we’re sure not to find anything.
The very first search was conducted by astronomer Frank Drake in 1960, using the 85-foot antenna at the National Radio Astronomy Observatory (Figure 30.19). Called Project Ozma, after the queen of the exotic Land of Oz in the children’s stories of L. Frank Baum, his experiment involved looking at about 7200 channels and two nearby stars over a period of 200 hours. Although he found nothing, Drake demonstrated that we had the technology to do such a search, and set the stage for the more sophisticated projects that followed.
Receivers are constantly improving, and the sensitivity of SETI programs— SETI stands for the search for extraterrestrial intelligence—is advancing rapidly. Equally important, modern electronics and software allow simultaneous searches on millions of frequencies (channels). If we can thus cover a broad frequency range, the cosmic haystack problem of guessing the right frequency largely goes away. One powerful telescope array (funded with an initial contribution from Microsoft founder Paul Allen) that is built for SETI searches is the Allen Telescope in Northern California. Other radio telescopes being used for such searches have included the giant Arecibo radio dish in Puerto Rico (which, sadly, collapsed due to storm damage in late 2020), the recently completed, and even larger, FAST dish in China, and the Green Bank Telescope in West Virginia, which is the largest steerable radio telescope in the world.
In 2015, Silicon Valley investor and philanthropist Yuri Milner donated 100 million dollars to the University of California, Berkeley, to enhance the SETI endeavor for ten years. The project, called B reakthrough: Listen , is enlisting more telescopes around the world in the search and is developing sophisticated artificial intelligence programs to scour the incoming signals for intelligent messages. The aim is to search more stars and more channels than any SETI project so far.
What kind of signals do we hope to pick up? We on Earth are inadvertently sending out a flood of radio signals, dominated by military radar systems. This is a kind of leakage signal, similar to the wasted light energy that is beamed upward by poorly designed streetlights and advertising signs. Could we detect a similar leakage of radio signals from another civilization? The answer is just barely, but only for the nearest stars. For the most part, therefore, current radio SETI searches are looking for beacons, assuming that civilizations might be intentionally drawing attention to themselves or perhaps sending a message to another world or outpost that lies in our direction. Our prospects for success depend on how often civilizations arise, how long they last, and how patient they are about broadcasting their locations to the cosmos.
1997 was quite a year for Jill Cornell Tarter (Figure 30.20), one of the world’s leading scientists in the SETI field. The SETI Institute announced that she would be the recipient of its first endowed chair (the equivalent of an endowed research professorship) named in honor of Bernard Oliver. The National Science Foundation approved a proposal by a group of scientists and educators she headed to develop an innovative hands-on high school curriculum based on the ideas of cosmic evolution (the topics of this chapter). And, at roughly the same time, she was being besieged with requests for media interviews as news reports identified her as the model for Ellie Arroway, the protagonist of Contact , Carl Sagan’s best-selling novel about SETI. The book had been made into a high-budget science fiction film, starring Jodie Foster, who had talked with Tarter before taking the role.
Tarter is quick to point out, “Carl Sagan wrote a book about a woman who does what I do, not about me.” Still, as the only woman in such a senior position in the small field of SETI, she was the center of a great deal of public attention. (However, colleagues and reporters pointed out that this was nothing compared to what would happen if her search for radio signals from other civilizations recorded a success.)
Being the only woman in a group is not a new situation to Tarter, who often found herself the only woman in her advanced science or math classes. Her father had encouraged her, both in her interest in science and her “tinkering.” As an undergraduate at Cornell University, she majored in engineering physics. That training became key to putting together and maintaining the complex systems that automatically scan for signals from other civilizations.
Switching to astrophysics for her graduate studies, she wrote a PhD thesis that, among other topics, considered the formation of failed stars—those whose mass was not sufficient to ignite the nuclear reactions that power more massive stars like our own Sun. Tarter coined the term “brown dwarf” for these small, dim objects, and it has remained the name astronomers use ever since.
It was while she was still in graduate school that Stuart Bowyer, one of her professors at the University of California, Berkeley, asked her if she wanted to be involved in a small experiment to siphon off a bit of radiation from a radio telescope as astronomers used it year in and year out and see if there was any hint of an intelligently coded radio message buried in the radio noise. Her engineering and computer programming skills became essential to the project, and soon she was hooked on the search for life elsewhere.
Thus began an illustrious career working full time searching for what she calls the technosignatures (signs of technology) from extraterrestrial civilizations. Tarter has received many awards, including being elected fellow of the American Association for the Advancement of Science in 2002, the Adler Planetarium Women in Space Science Award in 2003, and a 2009 TED Prize.
At the first scientific meeting devoted to SETI, Frank Drake wrote an equation on the blackboard that took the difficult question of estimating the number of civilizations in the Galaxy and broke it down into a series of smaller, more manageable questions. Ever since then, both astronomers and students have used this Drake equation as a means of approaching the most challenging question: How likely is it that we are alone? Since this is at present an unanswerable question, astronomer Jill Tarter has called the Drake equation a “way of organizing our ignorance.” (See Figure 30.21.)
The form of the Drake equation is very simple. To estimate the number of communicating civilizations that currently exist in the Galaxy (we will define these terms more carefully in a moment), we multiply the rate of formation of such civilizations (number per year) by their average lifetime (in years). In symbols,
To make this formula easier to use (and more interesting), however, Drake separated the rate of formation R total into a series of probabilities:
R star is the rate of formation of stars like the Sun in our Galaxy, which is, very roughly, around 10 stars per year. Each of the other terms is a fraction or probability (less than or equal to 1.0), and the product of all these probabilities is itself the total probability that each star will have an intelligent, technological, communicating civilization that we might want to talk to. We have:
- f p = the fraction of these stars with planets
- f e = the fraction of the planetary systems that include habitable planets
- f l = the fraction of habitable planets that actually support life
- f i = the fraction of inhabited planets that develop advanced intelligence
- f c = the fraction of these intelligent civilizations that develop science and the technology to build radio telescopes and transmitters
Each of these factors can be discussed and perhaps evaluated, but we must guess at many of the values. In particular, we don’t know how to calculate the probability of something that happened once on Earth but has not been observed elsewhere—and these include the development of life, of intelligent life, and of technological life (the last three factors in the equation). One important advance in estimating the terms of the Drake equation comes from the recent discovery of exoplanets. When the Drake equation was first written, no one had any idea whether planets and planetary systems were common. Now we know they are—another example of the Copernican principle.
Solution
Even if we don’t know the answers, we can make some guesses and calculate the resulting number N . Let’s start with the optimism implicit in the Copernican principle and set the last three terms equal to 1.0. If R is 10 stars/year and if we measure the average lifetime of a technological civilization in years, the units of years cancel. If we also assume that f p is 0.1, and f e is 1.0, the equation becomes
Now we see the importance of the term L, the lifetime of a communicating civilization (measured in years). We have had this capability (to communicate at the distances of the stars) for only a few decades.
Suppose we assume that this stage in our history lasts only one century.
- Answer
-
With our optimistic assumptions about the other factors, L = 100 years and N = 100 such civilizations in the entire Galaxy. In that case, there are so few other civilizations like ours that we are unlikely to detect any signals in a SETI search. But suppose the average lifetime is a million years; in that case, there are a million such civilizations in the Galaxy, and some of them may be within range for radio communication.
The most important conclusion from this calculation is that even if we are extremely optimistic about the probabilities, the only way we can expect success from SETI is if other civilizations are much older (and hence probably much more advanced) than ours.
Read Frank Drake’s own account of how he came up with his “equation.” And here is a recent interview with Frank Drake by one of the authors of this textbook.
SETI outside the Radio Realm
For the reasons discussed above, most SETI programs search for signals at radio wavelengths. But in science, if there are other approaches to answering an unsolved question, we don’t want to neglect them. So astronomers have been thinking about other ways we could pick up evidence for the existence of technologically advanced civilizations.
Recently, technology has allowed astronomers to expand the search into the domain of visible light. You might think that it would be hopeless to try to detect a flash of visible light from a planet given the brilliance of the star it orbits. This is why we usually cannot measure the reflected light of planets around other stars. The feeble light of the planet is simply swamped by the “big light” in the neighborhood. So another civilization would need a mighty strong beacon to compete with their star.
However, in recent years, human engineers have learned how to make flashes of light brighter than the Sun. The trick is to “turn on” the light for a very brief time, so that the costs are manageable. But ultra-bright, ultra-short laser pulses (operating for periods of a billionth of a second) can pack a lot of energy and can be coded to carry a message. We also have the technology to detect such short pulses—not with human senses, but with special detectors that can be “tuned” to hunt automatically for such short bursts of light from nearby stars.
Why would any civilization try to outshine its own star in this way? It turns out that the cost of sending an ultra-short laser pulse in the direction of a few promising stars can be less than the cost of sweeping a continuous radio message across the whole sky. Or perhaps they, too, have a special fondness for light messages because one of their senses evolved using light. Several programs are now experimenting with “ optical SETI ” searches, which can be done with only a modest telescope. (The term optical here means using visible light.)
If we let our imaginations expand, we might think of other for technosignatures —signs of technology from other civilizations. For example, what if a truly advanced civilization should decide to (or need to) renovate its planetary system to maximize the area for life? It could do so by breaking apart some planets or moons and building a ring of solid material that surrounds or encloses the star and intercepts some or all of its light. This huge artificial ring or sphere might glow very brightly at infrared wavelengths, as the starlight it receives is eventually converted to heat and re-radiated into space. That infrared radiation could be detected by our instruments, and searches for such infrared sources are also underway (Figure 30.22).
Should We Transmit in Addition to Listening?
Our planet has some leakage of radio waves into space, from FM radio, television, military radars, and communication between Earth and our orbiting spacecraft. However, such leakage radiation is still quite weak, and therefore difficult to detect at the distances of the stars, at least with the radio technology we have. So at the present time our attempts to communicate with other civilizations that may be out there mostly involve trying to receive messages, but not sending any ourselves.
Some scientists, however, think that it is inconsistent to search for beacons from other civilizations without announcing our presence in a similar way. (We discussed earlier the problem that if every other civilization confined itself to listening, no one would ever get in touch.) So, should we be making regular attempts at sending easily decoded messages into space? Some scientists warn that our civilization is too immature and defenseless to announce ourselves at this early point in our development. The decision whether to transmit or not turns out to be an interesting reflection of how we feel about ourselves and our place in the universe.
Discussions of transmission also raise the question of who should speak for planet Earth. Today, anyone and everyone can broadcast radio signals, and many businesses, religious groups, and governments do. It would be a modest step for the same organizations to use or build large radio telescopes and begin intentional transmissions that are much stronger than the signals that leak from Earth today. And if we intercept a signal from an alien civilization, then the issue arises whether to reply.
Who should make the decision about whether, when, and how humanity announces itself to the cosmos? Is there freedom of speech when it comes to sending radio messages to other civilizations? Do all the nations of Earth have to agree before we send a signal strong enough that it has a serious chance of being received at the distances of the stars? How our species reaches a decision about these kinds of questions may well be a test of whether or not there is intelligent life on Earth.
Conclusion
Whether or not we ultimately turn out to be the only intelligent species in our part of the Galaxy, our exploration of the cosmos will surely continue. An important part of that exploration will still be the search for biomarkers from inhabited planets that have not produced technological creatures that send out radio signals. After all, creatures like butterflies and dolphins may never build radio antennas, but we are happy to share our planet with them and would be delighted to find their counterparts on other worlds.
Whether or not life exists elsewhere is just one of the unsolved problems in astronomy that we have discussed in this book. A humble acknowledgment of how much we have left to learn about the universe is one of the fundamental hallmarks of science. This should not, however, prevent us from feeling exhilarated about how much we have already managed to discover, and feeling curious about what else we might find out in the years to come.
Our progress report on the ideas of astronomy ends here, but we hope that your interest in the universe does not. We hope you will keep up with developments in astronomy through media and online, or by going to an occasional public lecture by a local scientist. Who, after all, can even guess all the amazing things that future research projects will reveal about both the universe and our connection with it?


5.3C: Nails
Fingernails are made of keratin and they perform two major functions: protection and sensation.
LEARNING OBJECTIVE
Describe the structure of fingernails
Key Takeaways
Key Points
- The nail bed contains the blood vessels, nerves, and melanocytes or melanin-producing cells. As the nail is produced by the root, it streams down along the nail bed, which adds material to the undersurface of the nail and makes it thicker.
- The nail plate is the actual fingernail, composed of translucent keratin. The pink appearance of the nail comes from the blood vessels underneath the nail.
- The eponychium, or cuticle, is situated between the skin of the finger and the nail plate. It fuses these structures together and provides a waterproof barrier.
- Deformity or disease of the nails is referred to as onychosis. There are many diseases that can occur with the fingernails and toenails. The most common of these diseases are ingrown nails and fungal infections.
- Ingrown nails, also known as onychocryptosis, can affect either the fingers or the toes. In this condition, the nail cuts into one or both sides of the nail bed, resulting in inflammation and possibly infection.
Key Term
- keratin : A protein that makes up hair and nails.
Function of the Fingernail
The fingernail is an important structure made of keratin. The fingernail generally serves two purposes: it acts as a protective plate and enhances sensation of the fingertip. Nails can also help grasp small things.
The protection function of the fingernail is commonly known, but the sensation function is equally important. The fingertip has many nerve endings in it that allow it to receive volumes of information about the objects we touch. The nail acts as a counterforce to the fingertip, providing even more sensory input when an object is touched.
Nails grow from the nail bed continuously but they slow down their growth rate with age, poor nutrition, or poor circulation.
Anatomy of the Fingernail
The structure of the fingernail is divided into six specific parts:
- root
- nail bed
- nail plate
- eponychium (cuticle)
- perionychium
- hyponychium
Root and Nail Sinus
The nail sinus (sinus unguis) is where the nail root is—at the base of the nail underneath the skin. It originates from the actively growing tissue below, the matrix. The root of the fingernail is also known as the germinal matrix.
This portion of the nail is actually beneath the skin, behind the fingernail, and extends several millimeters into the finger. The fingernail root produces most of the volume of the nail and the nail bed. This portion of the nail does not have any melanocytes, or melanin-producing cells. The edge of the germinal matrix is seen as a white, crescent shaped structure called the lunula.
Nail Bed
The nail bed is a part of the nail matrix called the sterile matrix. It extends from the edge of the germinal matrix, or lunula, to the hyponychium. The nail bed contains the blood vessels, nerves, and melanocytes, or melanin-producing cells. As the nail is produced by the root, it streams down along the nail bed, which adds material to the undersurface of the nail and makes it thicker.
Nail Plate
The nail plate is the actual fingernail, composed of translucent keratin. The pink appearance of the nail comes from the blood vessels underneath the nail. The underneath surface of the nail plate has grooves along the length of the nail that help anchor it to the nail bed. The free margin or distal edge is the anterior margin of the nail plate corresponding to the abrasive or cutting edge of the nail.
Eponychium
The eponychium, or cuticle, is situated between the skin of the finger; the nail plate fuses these structures together and provides a waterproof barrier.
Perionychium
The perioncyhium is the skin that overlies the nail plate on its sides; it is also known as the paronychial edge. The perionychium is the site of hangnails, ingrown nails, and an infection of the skin called paronychia.
Hyponychium
The hyponychium is the area between the nail plate and the fingertip. It is the junction between the free edge of the nail and the skin of the fingertip, also providing a waterproof barrier.