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Heat of Solution

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Heat of Solution

Acid should always be added to water because a large amount of heat can be released

How do you make solutions safely?

When preparing dilutions of concentrated sulfuric acid, the directions usually call for adding the acid slowly to water with a lot of stirring. When this acid is mixed with water, a great deal of heat is released in the dissolving process. If water were added to acid, the water would quickly heat and splatter, causing harm to the person making the solution.

Heat of Solution

Enthalpy changes also occur when a solute undergoes the physical process of dissolving into a solvent. Hot packs and cold packs (see Figure below ) use this property. Many hot packs use calcium chloride, which releases heat when it dissolves according to the equation below.

\text{CaCl}_2(s) \rightarrow \text{Ca}^{2+}(aq)+2\text{Cl}^-(aq)+82.8 \text{ kJ}

The molar heat of solution   (\Delta H_{\text{soln}}) of a substance is the heat absorbed or released when one mole of the substance is dissolved in water. For calcium chloride, \Delta H_{\text{soln}}=-82.8 \text{ kJ/mol} .

Instant hot and cold packs generate or absorb heat when chemicals are dissolved in water

Chemical hot packs and cold packs work because of the heats of solution of the chemicals inside them. When the bag is squeezed, an inner pouch bursts, allowing the chemical to dissolve in water. Heat is released in the hot pack and absorbed in the cold pack.

Many cold packs use ammonium nitrate, which absorbs heat from the surroundings when it dissolves.

\text{NH}_4\text{NO}_3(s)+25.7 \text{ kJ} \rightarrow \text{NH}_4^+(aq)+\text{NO}_3^-(aq)

Cold packs are typically used to treat muscle strains and sore joints. The cold pack is activated and applied to the affected area. As the ammonium nitrate dissolves, it absorbs heat from the body and helps to limit swelling. For ammonium nitrate, \Delta H_{\text{soln}}= 25.7 \text{ kJ/mol} .

Sample Problem: Heat of Solution

The molar heat of solution, \Delta H_{\text{soln}} , of NaOH is -445.1 kJ/mol. In a certain experiment, 5.00 g of NaOH is completely dissolved in 1.000 L of 20.0°C water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.

Step 1: List the known quantities and plan the problem .


  • mass NaOH = 5.00 g
  • molar mass NaOH = 40.00 g/mol
  • \Delta H_{\text{soln}}(\text{NaOH})=-445.1 \text{ kJ/mol}
  • mass H 2 O = 1.000 kg = 1000. g (assumes density = 1.00 g/mL)
  • T_{\text{initial}}(\text{H}_2\text{O}) = 20.0^\circ \text{C}
  • c_{p} (\text{H}_2\text{O})=4.18 \text{ J/g}^\circ \text{C}


  • T_{\text{final}} \text{ of H}_2\text{O} = ? \ ^\circ \text{C}

This is a multiple-step problem: 1) the grams NaOH is converted to moles; 2) the moles is multiplied by the molar heat of solution; 3) the joules of heat released in the dissolving process is used with the specific heat equation and the total mass of the solution to calculate the \Delta T ; 4) the  T_{\text{final}} is determined from \Delta T .

Step 2: Solve .

& 5.00 \text{ g NaOH} \times \frac{1 \text{ mol NaOH}}{40.00 \text{ g NaOH}} \times \frac{-445.1 \text{ kJ}}{1 \text{ mol NaOH}} \times \frac{1000 \text{ J}}{1 \text{ kJ}}=-5.56 \times 10^4 \text{ J} \\& \Delta T= \frac{\Delta H}{c_p \times m}=\frac{-5.56 \times 10^4 \text{ J}}{4.18 \text{ J/g}^\circ \text{C} \times 1005 \text{ g}}=13.2^\circ \text{C} \\& T_{\text{final}}=20.0^\circ \text{C}+13.2^\circ \text{C}=33.2^\circ \text{C}

Step 3: Think about your result .

The dissolving process releases a large amount of heat, which causes the temperature of the solution to rise. Care must be taken when preparing concentrated solutions of sodium hydroxide because of the large amounts of heat released.


  • Molar heat of solution is defined.
  • Sample calculations using molar heat of solution are given.


Work on the problems at the link below:




  1. Does NaOH in solution warm or cool the water?
  2. How can you tell whether a material will produce an increase or decrease in heat when dissolved?
  3. The sample problem was done at 20°C. Would the temperature increase be the same if the sample was run at 72°C?

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