Calorimetry is the study of heat transfer and changes of state resulting from chemical reactions, phase transitions, or physical changes. The tool used to measure heat change is the calorimeter. Two popular types of calorimeters are the coffee cup calorimeter and bomb calorimeter.

These problems demonstrate how to calculate heat transfer and enthalpy change using calorimeter data. While working these problems, review the sections on coffee cup and bomb calorimetry and the laws of thermochemistry.

### Coffee Cup Calorimetry Problem

The following acid-base reaction is performed in a coffee cup calorimeter:

- H
^{+}(aq) + OH^{-}(aq) → H_{2}O(l)

The temperature of 110 g of water rises from 25.0 C to 26.2 C when 0.10 mol of H^{+} is reacted with 0.10 mol of OH^{-}.

- Calculate q
_{water} - Calculate ΔH for the reaction
- Calculate ΔH if 1.00 mol OH
^{-}reacts with 1.00 mol H^{+}

### Solution

Use this equation:

- q = (specific heat) x m x Δt

Where q is heat flow, m is mass in grams, and Δt is the temperature change. Plugging in the values given in the problem, you get:

- q
_{water}= 4.18 (J / g·C;) x 110 g x (26.6 C - 25.0 C) - q
_{water}= 550 J - ΔH = -(q
_{water}) = - 550 J

You know that when 0.010 mol of H^{+} or OH^{-} reacts, ΔH is - 550 J:

- 0.010 mol H
^{+}~ -550 J

Therefore, for 1.00 mol of H^{+} (or OH^{-}):

- ΔH = 1.00 mol H
^{+}x (-550 J / 0.010 mol H^{+}) - ΔH = -5.5 x 10
^{4}J - ΔH = -55 kJ

### Answer

- 550 J (Be sure to have two significant figures.)
- -550 J
- -55 kJ

### Bomb Calorimetry Problem

When a 1.000 g sample of the rocket fuel hydrazine, N_{2}H_{4}, is burned in a bomb calorimeter, which contains 1,200 g of water, the temperature rises from 24.62 C to 28.16 C.

If the C for the bomb is 840 J/C, calculate:

- q
_{reaction}for combustion of a 1-gram sample - q
_{reaction}for combustion of one mole of hydrazine in the bomb calorimeter

### Solution

For a bomb calorimeter, use this equation:

- q
_{reaction}= -(qwater + qbomb) - q
_{reaction}= -(4.18 J / g·C x mwater x Δt + C x Δt) - q
_{reaction}= -(4.18 J / g·C x mwater + C)Δt

Where q is heat flow, m is mass in grams, and Δt is the temperature change. Plugging in the values given in the problem:

- q
_{reaction}= -(4.18 J / g·C x 1200 g + 840 J/C)(3.54 C) - q
_{reaction}= -20,700 J or -20.7 kJ

You now know that 20.7 kJ of heat is evolved for every gram of hydrazine that is burned. Using the periodic table to get atomic weights, calculate that one mole of hydrazine, N_{2}H_{4}, weight 32.0 g. Therefore, for the combustion of one mole of hydrazine:

- q
_{reaction}= 32.0 x -20.7 kJ/g - q
_{reaction}= -662 kJ

### Answers

- -20.7 kJ
- -662 kJ