1.7: Mechanisms of Heat Transfer
By the end of this section, you will be able to:
- Explain some phenomena that involve conductive, convective, and radiative heat transfer
- Solve problems on the relationships between heat transfer, time, and rate of heat transfer
- Solve problems using the formulas for conduction and radiation
Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is a temperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through the walls of a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where no heat transfer occurs. Yet every heat transfer takes place by only three methods:
- Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature above absolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the pan is transferred by conduction .
- Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example.
- Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A less obvious example is thermal radiation from the human body.
In the illustration at the beginning of this chapter, the fire warms the snowshoers’ faces largely by radiation. Convection carries some heat to them, but most of the air flow from the fire is upward (creating the familiar shape of flames), carrying heat to the food being cooked and into the sky. The snowshoers wear clothes designed with low conductivity to prevent heat flow out of their bodies.
In this section, we examine these methods in some detail. Each method has unique and interesting characteristics, but all three have two things in common: They transfer heat solely because of a temperature difference, and the greater the temperature difference, the faster the heat transfer (Figure \(\PageIndex{1}\)).
Name an example from daily life (different from the text) for each mechanism of heat transfer.
Solution
Conduction: Heat transfers into your hands as you hold a hot cup of coffee. Convection: Heat transfers as the barista “steams” cold milk to make hot cocoa. Radiation: Heat transfers from the Sun to a jar of water with tea leaves in it to make “Sun tea.” A great many other answers are possible.
Conduction
As you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet feel colder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with the tiles than with the carpet, so the sensation of cold is more intense.
Some materials conduct thermal energy faster than others. Figure \(\PageIndex{2}\) shows a material that conducts heat slowly—it is a good thermal insulator, or poor heat conductor—used to reduce heat flow into and out of a house.
A molecular picture of heat conduction will help justify the equation that describes it. Figure \(\PageIndex{3}\) shows molecules in two bodies at different temperatures, \(T_h\) and \(T_c\) for “hot” and “cold.” The average kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, energy transfers from the high-energy to the low-energy molecule. In a metal, the picture would also include free valence electrons colliding with each other and with atoms, likewise transferring energy. The cumulative effect of all collisions is a net flux of heat from the hotter body to the colder body. Thus, the rate of heat transfer increases with increasing temperature difference \(\Delta T = T_h - T_c\). If the temperatures are the same, the net heat transfer rate is zero. Because the number of collisions increases with increasing area, heat conduction is proportional to the cross-sectional area—a second factor in the equation.
A third quantity that affects the conduction rate is the thickness of the material through which heat transfers. Figure \(\PageIndex{4}\) shows a slab of material with a higher temperature on the left than on the right. Heat transfers from the left to the right by a series of molecular collisions. The greater the distance between hot and cold, the more time the material takes to transfer the same amount of heat.
All four of these quantities appear in a simple equation deduced from and confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure \(\PageIndex{4}\), is given by
\[P = \dfrac{dQ}{dt} = \dfrac{kA(T_h - T_c)}{d} \nonumber \]
where P is the power or rate of heat transfer in watts or in kilocalories per second, A and d are its surface area and thickness, as shown in Figure \(\PageIndex{4}\), \(T_h - T_c\) is the temperature difference across the slab, and k is the thermal conductivity of the material. Table \(\PageIndex{1}\) gives representative values of thermal conductivity.
More generally, we can write
\[P = -kA\dfrac{dT}{dx}, \nonumber \]
where x is the coordinate in the direction of heat flow. Since in Figure \(\PageIndex{4}\), the power and area are constant, dT / dx is constant, and the temperature decreases linearly from \(T_h\) to \(T_c\).
| Substance | Thermal Conductivity \(k(W/m^oC)\) |
|---|---|
| Diamond | 2000 |
| Silver | 420 |
| Copper | 390 |
| Gold | 318 |
| Aluminum | 220 |
| Steel iron | 80 |
| Steel (stainless) | 14 |
| Ice | 2.2 |
| Glass (average) | 0.84 |
| Concrete brick | 0.84 |
| Water | 0.6 |
| Fatty tissue (without blood) | 0.2 |
| Asbestos | 0.16 |
| Plasterboard | 0.16 |
| Wood | 0.08–0.16 |
| Snow (dry) | 0.10 |
| Cork | 0.042 |
| Glass wool | 0.042 |
| Wool | 0.04 |
| Down feathers | 0.025 |
| Air | 0.023 |
| Polystyrene foam | 0.010 |
A polystyrene foam icebox has a total area of \(0.950 \, m^2\) and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at \(0^oC\). The inside of the box is kept cold by melting ice. How much ice melts in one day if the icebox is kept in the trunk of a car at \(35.0^oC\)?
Strategy
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.
Solution
First we identify the knowns.
\(k = 0.010 \, W/m \cdot \, ^oC\) for polystyrene foam; \(A = 0.950 \, m^2\); \(d = 2.50 \, cm = 0.0250 \, m\); \(T_c = 0^oC\); \(T_h = 35.0^oC\); \(t = 1 \, day = 24 \, hour = 86,400 \, s\).
Then we identify the unknowns. We need to solve for the mass of the ice, m . We also need to solve for the net heat transferred to melt the ice, Q . The rate of heat transfer by conduction is given by
\[P = \dfrac{dQ}{dT} = \dfrac{kA(T_h - T_c)}{d}. \nonumber \]
The heat used to melt the ice is \(Q = mL_f\). We insert the known values:
\[P = \dfrac{(0.010 \, W/m \cdot ^oC)(0.950 \, m^2)(35.0 ^oC - 0^oC)}{0.0250 \, m} = 13.3 \, W. \nonumber \]
Multiplying the rate of heat transfer by the time we obtain
\[Q = Pt = (13.3 \, W)(86,400 \, s) = 1.15 \times 10^6 \, J. \nonumber \]
We set this equal to the heat transferred to melt the ice, \(Q = mL_f\) and solve for the mass m :
\[m = \dfrac{Q}{L_f} = \dfrac{1.15 \times 10^6 \, J}{334 \times 10^3 \, J/kg} = 3.44 \, kg. \nonumber \]
Significance
The result of 3.44 kg, or about 7.6 lb, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Table \(\PageIndex{1}\) shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goosedown feathers. Like polystyrene foam, these all contain many small pockets of air, taking advantage of air’s poor thermal conductivity.
In developing insulation , the smaller the conductivity k and the larger the thickness d , the better. Thus, the ratio d/k , called the R factor , is large for a good insulator. The rate of conductive heat transfer is inversely proportional to R . R factors are most commonly quoted for household insulation, refrigerators, and the like. Unfortunately, in the United States, R is still in non-metric units of \(ft^2 \cdot ^oF \cdot h/Btu\), although the unit usually goes unstated [1 British thermal unit (Btu) is the amount of energy needed to change the temperature of 1.0 lb of water by \(1.0^oF\) which is 1055.1 J]. A couple of representative values are an R factor of 11 for 3.5-inch-thick fiberglass batts (pieces) of insulation and an R factor of 19 for 6.5-inch-thick fiberglass batts (Figure \(\PageIndex{5}\)). In the US, walls are usually insulated with 3.5-inch batts, whereas ceilings are usually insulated with 6.5-inch batts. In cold climates, thicker batts may be used.
Note that in Table \(\PageIndex{1}\), most of the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an electrical insulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors, but the handles of those used on the stove are made from good insulators (bad conductors).
A steel rod and an aluminum rod, each of diameter 1.00 cm and length 25.0 cm, are welded end to end. One end of the steel rod is placed in a large tank of boiling water at \(100^oC\), while the far end of the aluminum rod is placed in a large tank of water at \(20^oC\). The rods are insulated so that no heat escapes from their surfaces. What is the temperature at the joint, and what is the rate of heat conduction through this composite rod?
Strategy
The heat that enters the steel rod from the boiling water has no place to go but through the steel rod, then through the aluminum rod, to the cold water. Therefore, we can equate the rate of conduction through the steel to the rate of conduction through the aluminum.
We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since it simply adds when two rods are joined end to end. (We will use a similar method in the chapter on direct-current circuits.)
Solution
- Identify the knowns and convert them to SI units. The length of each rod is \(L_{A1} = L_{steel} = 0.25 \, m\), the cross-sectional area of each rod is \(A_{A1} = A_{steel} = 7.85 \times 10^{-5} m^2\), the thermal conductivity of aluminum is \(k_{A1} = 220 \, W/m \cdot ^oC\), the thermal conductivity of steel is \(k_{steel} = 80 \, W/m \cdot ^oC\) the temperature at the hot end is \(T = 100^oC\) and the temperature at the cold end is \(T = 20^oC\).
- Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through the aluminum rod in terms of the unknown temperature T at the joint: \[\begin{align*} P_{steel} &= \dfrac{k_{steel}A_{steel}\Delta T_{steel}}{L_{steel}} \\[4pt] &= \dfrac{(80 \, W/m \cdot^oC)(7.85 \times 10^{-5}m^2)(100 \, ^oC - T)}{0.25} \\[4pt] &= (0.0251 \, W/^oC)(100 ^oC - T);\end{align*} \]\[\begin{align*} P_{A1} &= \dfrac{k_{A1}A_{A1}\Delta T_{A1}}{L_{A1}} \\[4pt] &= \dfrac{(220 \, W/m \cdot ^oC)(7.85 \times 10^{-5} m^2)(T - 20^oC)}{0.25 \, m} \\[4pt] &= (0.0691 \, W/ ^oC)(T - 20^oC). \end{align*} \]
- Set the two rates equal and solve for the unknown temperature: \[(0.0691 \, W/ ^oC)(T - 20^oC) = (0.0251 \, W/^oC)(100^oC - T) \nonumber \]\[T = 41.3 ^oC. \nonumber \]
- Calculate either rate: \[P_{steel} = (0.0251 \, W/^oC)(100^oC - 41.3^oC) = 1.47 \, W. \nonumber \]
- If desired, check your answer by calculating the other rate.
Solution
- Recall that \(R = L/k\). Now \(P = A \Delta T/R\), or \(\Delta T = PR/A\).
- We know that \(\Delta T_{steel} + \Delta T_{A1} = 100^oC - 20^oC = 80^oC\). We also know that \(P_{steel} = P_{A1}\), and we denote that rate of heat flow by P . Combine the equations: \[\dfrac{PR_{steel}}{A} + \dfrac{PR_{A1}}{A} = 80^oC. \nonumber \] Thus, we can simply add R factors. Now, \(P = \frac{80^oC}{A(R_{steel} + R_{A1})}.\)
- Find the \(R_s\) from the known quantities: \[R_{steel} = 3.13 \times 10^{-3}m^2 \cdot \, ^oC/W \nonumber \] and \[R_{A1} = 1.14 \times 10^{-3}m^2 \cdot \, ^oC/W. \nonumber \]
- Substitute these values in to find \(P = 1.47 \, W\) as before.
- Determine \(\Delta T\) for the aluminum rod (or for the steel rod) and use it to find T at the joint. \[\Delta T_{A1} = \dfrac{PR_{A1}}{A} = \dfrac{(1.47 \, W)(1.14 \times 10^{-3} m^2 \cdot \, ^oC/W)}{7.85 \times 10^{-5}m^2} = 21.3 ^oC, \nonumber \] so \(T = 20^oC + 21.3 ^oC = 41.3^oC\), as in Solution 1.
- If desired, check by determining \(\Delta T\) for the other rod.
Significance
In practice, adding R values is common, as in calculating the R value of an insulated wall. In the analogous situation in electronics, the resistance corresponds to AR in this problem and is additive even when the areas are unequal, as is common in electronics. Our equation for heat conduction can be used only when the areas are equal; otherwise, we would have a problem in three-dimensional heat flow, which is beyond our scope.
How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?
- Answer
-
Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled \((A_{final} = (2d)^2 = 4d^2 = 4A_{initial})\). The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two:
\[P_{final} = \dfrac{kA_{final}(T_h - T_c)}{d_{final}} = \dfrac{k(4A_{final}(T_h - T_c))}{2d_{initial}} = 2 \dfrac{kA_{final}(T_h - T_c)}{d_{initial}} = 2 P_{initial}. \nonumber \]
Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short times. For example, the temperature on Earth would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere were only through conduction. Also, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons. The next module discusses the important heat-transfer mechanism in such situations.
Convection
In convection , thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forced convection , the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hot surroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is the cooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the engine and a fan blows air to cool the radiator.
In free or natural convection , the flow is driven by buoyant forces: hot fluid rises and cold fluid sinks because density decreases as temperature increases. The house in Figure \(\PageIndex{6}\) is kept warm by natural convection, as is the pot of water on the stove in Figure \(\PageIndex{7}\). Ocean currents and large-scale atmospheric circulation, which result from the buoyancy of warm air and water, transfer hot air from the tropics toward the poles and cold air from the poles toward the tropics. (Earth’s rotation interacts with those flows, causing the observed eastward flow of air in the temperate zones.)
Natural convection like that of Figures \(\PageIndex{6}\) and \(\PageIndex{7}\), but acting on rock in Earth’s mantle, drives plate tectonics that are the motions that have shaped Earth’s surface.
Convection is usually more complicated than conduction. Beyond noting that the convection rate is often approximately proportional to the temperature difference, we will not do any quantitative work comparable to the formula for conduction. However, we can describe convection qualitatively and relate convection rates to heat and time. However, air is a poor conductor. Therefore, convection dominates heat transfer by air, and the amount of available space for airflow determines whether air transfers heat rapidly or slowly. There is little heat transfer in a space filled with air with a small amount of other material that prevents flow. The space between the inside and outside walls of a typical American house, for example, is about 9 cm (3.5 in.)—large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. On the other hand, the gap between the two panes of a double-paned window is about 1 cm, which largely prevents convection and takes advantage of air’s low conductivity reduce heat loss. Fur, cloth, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection (Figure \(\PageIndex{8}\)).
Some interesting phenomena happen when convection is accompanied by a phase change. The combination allows us to cool off by sweating even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.
The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (For simplicity, we assume this evaporation occurs when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)
Strategy
Energy is needed for this phase change (\(Q = mL_v\)). Thus, the energy loss per unit time is \[\dfrac{Q}{t} = \dfrac{mL_v}{t} = 120 \, W = 120 \, J/s. \nonumber \]
We divide both sides of the equation by \(L_v\) to find that the mass evaporated per unit time is \[\dfrac{m}{t} = \dfrac{120 \, J/s}{L_v}. \nonumber \]
Solution
Insert the value of the latent heat from [link] , \(L_v = 2430 \, kJ/kg = 2430 \, J/g\). This yields \[\dfrac{m}{t} = \dfrac{120 \, J/s}{2430 \, J/g} = 0.0494 \, g/s = 2.96 \, g/min. \nonumber \]
Significance
Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz.) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.
Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, possibly far from the ocean, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere (Figure \(\PageIndex{9}\)). Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise to colder altitudes. More condensation occurs in these regions, which in turn drives the cloud even higher. This mechanism is an example of positive feedback, since the process reinforces and accelerates itself. It sometimes produces violent storms, with lightning and hail. The same mechanism drives hurricanes.
This time-lapse video shows convection currents in a thunderstorm, including “rolling” motion similar to that of boiling water.
Explain why using a fan in the summer feels refreshing.
- Answer
-
Using a fan increases the flow of air: Warm air near your body is replaced by cooler air from elsewhere. Convection increases the rate of heat transfer so that moving air “feels” cooler than still air.
Radiation
You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun warms us without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. In these examples, heat is transferred by radiation (Figure \(\PageIndex{10}\)). That is, the hot body emits electromagnetic waves that are absorbed by the skin. No medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.
The energy of electromagnetic radiation varies over a wide range, depending on the wavelength: A shorter wavelength (or higher frequency) corresponds to a higher energy. Because more heat is radiated at higher temperatures, higher temperatures produce more intensity at every wavelength but especially at shorter wavelengths. In visible light, wavelength determines color—red has the longest wavelength and violet the shortest—so a temperature change is accompanied by a color change. For example, an electric heating element on a stove glows from red to orange, while the higher-temperature steel in a blast furnace glows from yellow to white. Infrared radiation is the predominant form radiated by objects cooler than the electric element and the steel. The radiated energy as a function of wavelength depends on its intensity, which is represented in Figure \(\PageIndex{11}\) by the height of the distribution. ( Electromagnetic Waves explains more about the electromagnetic spectrum, and Photons and Matter Waves discusses why the decrease in wavelength corresponds to an increase in energy.)
The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white is the least effective.  The reverse is also true—black radiates better than gray. Thus, on a clear summer night, the asphalt is colder than the gray sidewalk, because black radiates the energy more rapidly than gray. A perfectly black object would be an ideal radiator and an ideal absorber , as it would capture all the radiation that falls on it. In contrast, a perfectly white object or a perfect mirror would reflect all radiation, and a perfectly transparent object would transmit it all (Figure \(\PageIndex{13}\)). Such objects would not emit any radiation. Mathematically, the color is represented by the emissivity e . A “blackbody” radiator would have an \(e = 1\), whereas a perfect reflector or transmitter would have \(e = 0\). For real examples, tungsten light bulb filaments have an e of about 0.5, and carbon black (a material used in printer toner) has an emissivity of about 0.95.
To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal equilibrium. We know from experience that they will stay in equilibrium (the result of a principle that will be discussed at length in Second Law of Thermodynamics ). For the black object’s temperature to stay constant, it must emit as much radiation as it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the silver object must radiate as little as it absorbs. Thus, one property, emissivity, controls both radiation and absorption.
Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you knock apart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area.
The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation : \[P = \sigma AeT^4, \nonumber \] where \(\sigma = 5.67 \times 10^{-8} \, J/s \cdot m^2 \cdot K^4\) is the Stefan-Boltzmann constant, a combination of fundamental constants of nature; A is the surface area of the object; and T is its temperature in kelvins.
The proportionality to the fourth power of the absolute temperature is a remarkably strong temperature dependence. It allows the detection of even small temperature variations. Images called thermographs can be used medically to detect regions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be used to detect heat leaks in homes (Figure \(\PageIndex{14}\)), optimize performance of blast furnaces, improve comfort levels in work environments, and even remotely map Earth’s temperature profile.
The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object’s absorption of radiation from its surroundings. Assuming that an object with a temperature \(T_1\) is surrounded by an environment with uniform temperature \(T_2\), the net rate of heat transfer by radiation is
\[P_{net} = \sigma e A(T_2^4 - T_1^4), \nonumber \]
where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray, or black: The balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When \(T_2 > T_1\), the quantity \(P_{net}\) is positive, that is, the net heat transfer is from hot to cold.
Before doing an example, we have a complication to discuss: different emissivities at different wavelengths. If the fraction of incident radiation an object reflects is the same at all visible wavelengths, the object is gray; if the fraction depends on the wavelength, the object has some other color. For instance, a red or reddish object reflects red light more strongly than other visible wavelengths. Because it absorbs less red, it radiates less red when hot. Differential reflection and absorption of wavelengths outside the visible range have no effect on what we see, but they may have physically important effects. Skin is a very good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared spectrum. Thus, in spite of the obvious variations in skin color, we are all nearly black in the infrared. This high infrared emissivity is why we can so easily feel radiation on our skin. It is also the basis for the effectiveness of night-vision scopes used by law enforcement and the military to detect human beings.
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient temperature is \(22.0^oC\)? The person has a normal skin temperature of \(33.0^oC\) and a surface area of \(1.50 \, m^2\). The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.
Strategy
We can solve this by using the equation for the rate of radiative heat transfer.
Solution
Insert the temperature values \(T_2 = 295 \, K\) and \(T_1 = 306 \, K\), so that
\[\dfrac{Q}{t} = \sigma e A (T_2^4 - T_1^4) \nonumber \]
\[= (5.67 \times 10^{-8} \, J/s \cdot m^2 \cdot K^4)(0.97)(1.50 \, m^2) [(295 \, K)^4 - (306 \, K)^4] \nonumber \]
\[= -99 \, J/s = -99 \, W. \nonumber \]
Significance
This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at rest may produce energy at the rate of 125 W and that conduction and convection are also transferring energy to the environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer to the environment by all mechanisms, because clothing slows down both conduction and convection, and has a lower emissivity (especially if it is light-colored) than skin.
The average temperature of Earth is the subject of much current discussion. Earth is in radiative contact with both the Sun and dark space, so we cannot use the equation for an environment at a uniform temperature. Earth receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Conversely, dark space is very cold, about 3 K, so that Earth radiates energy into the dark sky. The rate of heat transfer from soil and grasses can be so rapid that frost may occur on clear summer evenings, even in warm latitudes.
The average temperature of Earth is determined by its energy balance. To a first approximation, it is the temperature at which Earth radiates heat to space as fast as it receives energy from the Sun.
An important parameter in calculating the temperature of Earth is its emissivity ( e ). On average, it is about 0.65, but calculation of this value is complicated by the great day-to-day variation in the highly reflective cloud coverage. Because clouds have lower emissivity than either oceans or land masses, they reflect some of the radiation back to the surface, greatly reducing heat transfer into dark space, just as they greatly reduce heat transfer into the atmosphere during the day. There is negative feedback (in which a change produces an effect that opposes that change) between clouds and heat transfer; higher temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing the temperature.
The often-mentioned greenhouse effect is directly related to the variation of Earth’s emissivity with wavelength (Figure \(\PageIndex{15}\)). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for life on Earth and for making Venus unsuitable for human life. Most of the infrared radiation emitted from Earth is absorbed by carbon dioxide (\(CO_2\)) and water (\(H_2O\)) in the atmosphere and then re-radiated into outer space or back to Earth. Re-radiation back to Earth maintains its surface temperature about \(40^oC\) higher than it would be if there were no atmosphere. (The glass walls and roof of a greenhouse increase the temperature inside by blocking convective heat losses, not radiative losses.)
The greenhouse effect is central to the discussion of global warming due to emission of carbon dioxide and methane (and other greenhouse gases) into Earth’s atmosphere from industry, transportation, and farming. Changes in global climate could lead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest biodiversity, and rising sea levels.
You can explore a simulation of the greenhouse effect that takes the point of view that the atmosphere scatters (redirects) infrared radiation rather than absorbing it and reradiating it. You may want to run the simulation first with no greenhouse gases in the atmosphere and then look at how adding greenhouse gases affects the infrared radiation from the Earth and the Earth’s temperature.
- Examine the situation to determine what type of heat transfer is involved.
- Identify the type(s) of heat transfer—conduction, convection, or radiation.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
- Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
- Solve the appropriate equation for the quantity to be determined (the unknown).
- For conduction, use the equation \(P = \frac{kA\Delta T}{d}\). Table \(\PageIndex{1}\) lists thermal conductivities. For convection, determine the amount of matter moved and the equation \(Q = mc\Delta T\), along with \(Q = mL_f\) or \(Q = mL_v\) if a substance changes phase. For radiation, the equation \(P_{net} = \sigma e A (T_2^4 - T_1^4)\) gives the net heat transfer rate.
- Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units.
- Check the answer to see if it is reasonable. Does it make sense?
How much greater is the rate of heat radiation when a body is at the temperature \(40^oC\) than when it is at the temperature \(20^oC\)?
[Hide solution]
The radiated heat is proportional to the fourth power of the absolute temperature . Because \(T_1 = 293 \, K\) and \(T_2 = 313 \, K\), the rate of heat transfer increases by about 30% of the original rate.


Polarity of Chemical Bonds
Skills to Develop
- Distinguish polarity from polarizability
- Define polarizability
Polarity means separation, in this case of electrical charge. If a bonding pair of electrons are pulled more toward one atom and away from the other, this will cause the first atom to be partially negatively charged, and the second to be partially positively charged. This will create an electric dipole moment, such as the the dipole moment in water that makes water so good as a solvent.
Although covalent and ionic substances might seem really different, with Lewis theory we can think of them as being basically similar. At one extreme, we have a complete transfer of an electron, such as in CsF, making a positive and negative ion that are then attracted to each other. At the other extreme, we have a completely equally shared pair, such as in F 2 . In between, we have bonds with unequal sharing. Either way, usually all atoms will have a noble gas electron configuration , either by sharing or by losing electrons.
In addition to average polarity of a bond, we can have temporary polarity. The electrons move around, and sometimes it will happen that both of them move toward one atom. In general, the more electrons an atom has, the looser they are held. For example, iodine has 53 electrons, which is a lot! They can move around pretty easily. Even though I 2 has a non-polar bond on average, because the atoms are the same, it can easily become polar because the electrons are held loosely. In solution, I 2 can split a little bit into I + and I – .
To keep these ideas separate, polarity means the permanent average separation of charge, and polarizability means the ability to become polarized temporarily. In a non-polar polarizable molecule like I 2 , the average polarity is 0, but if we take many precise measurements of the instantaneous (very short time) polarity, many of them will be far from 0. In a polar, non-polarizable molecule like HF, all the instantaneous measurements will be very similar, but the average will not be 0. AgI is both polar and polarizable. Both of these have important effects on properties of materials, such as reactivity, solubility, and boiling point, that we will talk about later.


5.3: Multiple Regression Explanation, Assumptions, Interpretation, and Write Up
At the end of this section you should be able to answer the following questions:
- Explain the difference between Multiple Regression and Simple Regression.
- Explain the assumptions underlying Multiple Regression.
Multiple Regression is a step beyond simple regression. The main difference between simple and multiple regression is that multiple regression includes two or more independent variables – sometimes called predictor variables – in the model, rather than just one.
As such, the purpose of multiple regression is to determine the utility of a set of predictor variables for predicting an outcome, which is generally some important event or behaviour. This outcome can be designated as the outcome variable, the dependent variable, or the criterion variable. For example, you might hypothesise that the need to belong will predict motivations for Facebook use and that self-esteem and meaningful existence will uniquely predict motivations for Facebook use.
Before beginning your analysis, you should consider the following points:
- Regression analyses reveal relationships among variables (relationship between the criterion variable and the linear combination of a set of predictor variables) but do not imply a causal relationship.
- A regression solution – or set of predictor variables – is sensitive to combinations of variables. Whether a predictor is important in a solution depends on the other predictors in the set. If the predictor of interest is the only one that assesses some important facet of the outcome, it will appear important. If a predictor is only one of several predictors that assess the same important facet of the outcome, it will appear less important. For a good set of predictor variables – the smallest set of uncorrelated variables is best.
PowerPoint: Venn Diagrams
Please click on the link labeled “Venn Diagrams” to work through an example.
In these Venn Diagrams, you can see why it is best for the predictors to be strongly correlated with the dependent variable but uncorrelated with the other Independent Variables. This reduces the amount of shared variance between the independent variables. The illustration in Slide 2 shows logical relationships between predictors, for two different possible regression models in separate Venn diagrams. On the left, you can see three partially correlated independent variables on a single dependent variable. The three partially correlated independent variables are physical health, mental health, and spiritual health and the dependent variable is life satisfaction. On the right, you have three highly correlated independent variables (e.g., BMI, blood pressure, heart rate) on the dependent variable of life satisfaction. The model on the left would have some use in discovering the associations between those variables, however, the model on the right would not be useful, as all three of the independent variables are basically measuring the same thing and are mostly accounting for the same variability in the dependent variable.
There are two main types of regression with multiple independent variables:
- Standard or Single Step: Where all predictors enter the regression together.
- Sequential or Hierarchical: Where all predictors are entered in blocks. Each block represents one step.
We will now be exploring the single step multiple regression:
All predictors enter the regression equation at once. Each predictor is treated as if it had been analysed in the regression model after all other predictors had been analysed. These predictors are evaluated by the shared variance (i.e., level of prediction) shared between the dependant variable and the individual predictor variable.
Multiple Regression Assumptions
There are a number of assumptions that should be assessed before performing a multiple regression analysis:
- The dependant variable (the variable of interest) needs to be using a continuous scale.
- There are two or more independent variables. These can be measured using either continuous or categorical means.
- The three or more variables of interest should have a linear relationship, which you can check by using a scatterplot.
- The data should have homoscedasticity. In other words, the line of best fit is not dissimilar as the data points move across the line in a positive or negative direction. Homoscedasticity can be checked by producing standardised residual plots against the unstandardized predicted values.
- The data should not have two or more independent variables that are highly correlated. This is called multicollinearity which can be checked using Variance-inflation-factor or VIF values. High VIF indicates that the associated independent variable is highly collinear with the other variables in the model.
- There should be no spurious outliers.
- The residuals (errors) should be approximately normally distributed. This can be checked by a histogram (with a superimposed normal curve) and by plotting the of the standardised residuals using either a P-P Plot, or a Normal Q-Q Plot .
Multiple Regression Interpretation
For our example research question, we will be looking at the combined effect of three predictor variables – perceived life stress, location, and age – on the outcome variable of physical health?
PowerPoint: Standard Regression
Please open the output at the link labeled “Chapter Five – Standard Regression” to view the output.
Slide 1 contains the standard regression analysis output.
On Slide 2 you can see in the red circle, the test statistics are significant. The F-statistic examines the overall significance of the model, and shows if your predictors as a group provide a better fit to the data than no predictor variables, which they do in this example.
The R 2 values are shown in the green circle. The R 2 value shows the total amount of variance accounted for in the criterion by the predictors, and the adjusted R 2 is the estimated value of R 2 in the population.
Moving on to the individual variable effects on Slide 3, you can see the significance of the contribution of individual predictors in light blue. The unstandardized slope or the B value is shown in red, which represents the change caused by the variable (e.g., increasing 1 unit of perceived stress will raise physical illness by .40). Finally, you can see the standardised slope value in green, which are also known as beta values. These values are standardised ranging from +/-0 to 1, similar to an r value.
We should also briefly discuss dummy variables:
A dummy variable is a variable that is used to represent categorical information relating to the participants in a study. This could include gender, location, race, age groups, and you get the idea. Dummy variables are most often represented as dichotomous variables (they only have two values). When performing a regression, it is easier for interpretation if the values for the dummy variable is set to 0 or 1. 1 usually resents when a characteristic is present. For example, a question asking the participants “Do you have a drivers license” with a forced choice response of yes or no.
In this example on Slide 3 and circled in red, the variable is gender with male = 0, and female = 1. A positive Beta (B) means an association with 1, whereas a negative beta means an association with 0. In this case, being female was associated with greater levels of physical illness.
Multiple Regression Write Up
Here is an example of how to write up the results of a standard multiple regression analysis:
In order to test the research question, a multiple regression was conducted, with age, gender (0 = male, 1 = female), and perceived life stress as the predictors, with levels of physical illness as the dependent variable. Overall, the results showed the utility of the predictive model was significant, F (3,363) = 39.61, R 2 = .25, p < .001. All of the predictors explain a large amount of the variance between the variables (25%). The results showed that perceived stress and gender of participants were significant positive predictors of physical illness ( β =.47, t = 9.96, p < .001, and β =.15, t = 3.23, p = .001, respectively). The results showed that age ( β =-.02, t = -0.49 p = .63) was not a significant predictor of perceived stress.


13.6: Notes
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- Kristen J. Warner
- University of California Press
1 Gloria T. Hull, Patricia Bell Scott, and Barbara Smith, eds., But Some of Us Are Brave: All the Women Are White, All the Blacks Are Men: Black Women’s Studies (New York: Feminist Press, 1993).
2 John Thornton Caldwell, Production Culture: Industrial Reflexivity and Critical Practice in Film and Television (Durham, NC: Duke University Press, 2008).
3 Darnell Hunt, “The 2014 Hollywood Writers Report: Turning Missed Opportunities into Realized Ones” (July 2014), www.wga.org/uploadedFiles/who_we_are/HWR14.pdf.
4 Ibid.
5 It should be noted that white males directing decreased by 3 percent in contrast to the year before, while minority males increased by 3 percent, although the study cites that the increase can be attributed to a higher number of episodes directed by Tyler Perry, who solely directed episodes for his three television series.
6 Directors Guild of America, “Employers Make No Improvement in Diversity Hiring in Episodic Television: DGA Report,” September 17, 2014, www.dga.org/News/PressReleases/2014/140917-Episodic-Director-Diversity-Report.aspx.
7 Screen Actors Guild, “2007 and 2008 Casting Data Reports,” www.sagaftra.org/files/sag/documents/2007–2008_CastingDataReports.pdf.
8 Katie M. Kleinman, “Minorities in Prime-Time Television,” November 15, 1999, www.katiekleinman.com/portfolio/minoritiesmedia.php.
9 Jennifer Armstrong and Margeaux Watson, “Diversity: Why Is TV So White,” Entertainment Weekly , June 13, 2008, www.ew.com/ew/article/0,,20206185,00.html.
10 NAACP, “Out of Focus—Out of Sync Take 4: A Report on the Television Industry,” December 2008, http://action.naacp.org/page/-/NAACP...OS%20Take4.pdf .
11 Armstrong and Watson, “Diversity.”
12 Heartsick, comment on the Deadline Team, April 14, 2014, 10:53 a.m., “WGA Diversity Report: Women Writers See Gains in TV, Slide on Film Side,” Deadline Hollywood , April 14, 2014, http://deadline.com/2014/04/writers-...riters-714478/ .
13 DevelopmentHell Exec, Twitter post, May 30, 2014, 3:46 p.m., https://twitter.com/DevHellExec .
14 Based on the pattern of Mystery account titles, the assumption is that unless modifiers are in place that suggest otherwise, the owners of said accounts are white males.
15 Keith Negus, “The Production of Culture,” in Production of Culture/Cultures of Production (Thousand Oaks, CA: Sage, 1997), 362–363.
16 Kristen Warner, The Cultural Politics of Colorblind TV Casting (New York: Routledge, 2015).
17 Stuart Hall, “The Spectacle of the Other,” in Representation: Cultural Representations and Signifiying Practices , ed. Stuart Hall (Thousand Oaks, CA: Sage, 2003), 236.
18 For evidence of this trend, look at the press junkets for the films and note how often the actors stress that these are movies “for everybody.”
19 Claude Brodesser-Akner, “ Think Like a Man Is the Best-Testing Film in Hollywood—but Can It Win Over Black Men and White Audiences,” Vulture , April 10, 2012, www.vulture.com/2012/04/think-like-a-man-kevin-hart-will-packer.html.
20 Ibid.
21 Ibid.
22 Clyde R. Taylor, “Black Silence and the Politics of Representation,” in African-American Filmmaking and Race Cinema of the Silent Era: Oscar Micheaux and His Circle , ed. Pearl Bowser, Jane Gaines, and Charles Musser (Bloomington: Indiana University Press, 2001), 3.
23 Mark Harris, “The Shonda Rhimes Revolution: Finishing What The Sopranos Started,” Grantland , October 16, 2014, http://grantland.com/hollywood-prosp...s-scandal-abc/ .
24 Lacey Rose, “Shonda Rhimes Opens Up about ‘Angry Black Woman’ Flap, Messy Grey’s Anatomy Chapter and the Scandal Impact,” Hollywood Reporter , October 8, 2014, www.hollywoodreporter.com/news/shonda-rhimes-opens-up-angry-738715?utm_source=twitter.


4.5: Geometric Meaning of Scalar Multiplication
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Outcomes
- Understand scalar multiplication, geometrically.
Recall that the point \(P=\left( p_{1},p_{2},p_{3}\right)\) determines a vector \(\vec{p}\) from \(0\) to \(P\). The length of \(\vec{p}\), denoted \(\| \vec{p} \|\), is equal to \(\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}\) by Definition 4.4.1 .
Now suppose we have a vector \(\vec{u} = \left[ \begin{array}{lll} u_1 & u_2 & u_3 \end{array} \right]^T\) and we multiply \(\vec{u}\) by a scalar \(k\). By Definition 4.2.2 , \(k\vec{u} = \left[ \begin{array}{rrr} ku_{1} & ku_{2} & ku_{3} \end{array} \right]^T\). Then, by using Definition 4.4.1 , the length of this vector is given by \[\sqrt{\left( \left( k u_{1}\right) ^{2}+\left( k u_{2}\right) ^{2}+\left( k u_{3}\right) ^{2}\right) }=\left\vert k \right\vert \sqrt{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}\nonumber \] Thus the following holds. \[\| k \vec{u} \| =\left\vert k \right\vert \| \vec{u} \|\nonumber \] In other words, multiplication by a scalar magnifies or shrinks the length of the vector by a factor of \(\left\vert k \right\vert\). If \(\left\vert k \right\vert > 1\), the length of the resulting vector will be magnified. If \(\left\vert k \right\vert <1\), the length of the resulting vector will shrink. Remember that by the definition of the absolute value, \(\left\vert k \right\vert >0\).
What about the direction? Draw a picture of \(\vec{u}\) and \(k\vec{u}\) where \(k\) is negative. Notice that this causes the resulting vector to point in the opposite direction while if \(k >0\) it preserves the direction the vector points. Therefore the direction can either reverse, if \(k < 0\), or remain preserved, if \(k > 0\).
Consider the following example.
Example \(\PageIndex{1}\): Graphing Scalar Multiplication
Consider the vectors \(\vec{u}\) and \(\vec{v}\) drawn below.
Draw \(-\vec{u}\), \(2\vec{v}\), and \(-\frac{1}{2}\vec{v}\).
Solution
In order to find \(-\vec{u}\), we preserve the length of \(\vec{u}\) and simply reverse the direction. For \(2\vec{v}\), we double the length of \(\vec{v}\), while preserving the direction. Finally \(-\frac{1}{2}\vec{v}\) is found by taking half the length of \(\vec{v}\) and reversing the direction. These vectors are shown in the following diagram.
Now that we have studied both vector addition and scalar multiplication, we can combine the two actions. Recall Definition 9.2.2 of linear combinations of column matrices. We can apply this definition to vectors in \(\mathbb{R}^n\). A linear combination of vectors in \(\mathbb{R}^n\) is a sum of vectors multiplied by scalars.
In the following example, we examine the geometric meaning of this concept.


7: Waves
Dandelion. Cello. Read those two words, and your brain instantly conjures a stream of associations, the most prominent of which have to do with vibrations. Our mental category of “dandelion-ness” is strongly linked to the color of light waves that vibrate about half a million billion times a second: yellow. The velvety throb of a cello has as its most obvious characteristic a relatively low musical pitch --- the note you're spontaneously imagining right now might be one whose sound vibrations repeat at a rate of a hundred times a second.
Evolution seems to have designed our two most important senses around the assumption that our environment is made of waves, whereas up until now, we've mostly taken the view that Nature can be understood by breaking her down into smaller and smaller parts, ending up with particles as her most fundamental building blocks. Does that work for light and sound? Sound waves are disturbances in air, which is made of atoms, but light, on the other hand, isn't a vibration of atoms. Light, unlike sound, can travel through a vacuum: if you're reading this by sunlight, you're taking advantage of light that had to make it through millions of miles of vacuum to get to you. Waves, then, are not just a trick that vibrating atoms can do. Waves are one of the basic phenomena of the universe. At the end of this book, we'll even see that the things we've been calling particles, such as electrons, are really waves! 1
Contributors and Attributions
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.


6: Eigenvalue Problems
An eigenvalue problem is a matrix equation of the form
\[\mathbf{A} \vec{x} = \lambda \vec{x},\]
where \(\mathbf{A}\) is a known \(N\times N\) matrix. The problem is to find one (or more than one) non-zero vector \(\vec{x}\), which is called an eigenvector , and the associated \(\lambda \in \mathbb{C}\), which is called an eigenvalue . Eigenvalue problems are ubiquitous in practically all fields of physics. Most prominently, they are used to describe the "modes" of a physical system, such as the modes of a classical mechanical oscillator, or the energy states of an atom.
Before discussing numerical solutions to the eigenvalue problem, let us quickly review the relevant mathematical facts.
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- 6.2: Numerical Eigensolvers
- There exist numerical methods, called eigensolvers, which can compute eigenvalues (and eigenvectors) even for very large matrices, with hundreds of rows/columns, or larger. How could this be? The answer is that numerical eigensolvers are approximate, not exact. But even though their results are not exact, they are very precise—they can approach the exact eigenvalues to within the fundamental precision limits of floating-point arithmetic.