Find the Heat Capacity of a Calorimeter:

A calorimeter is made using two nested styrofoam cups and filled with 25.0 g of water at an initial temperature of 25.0°C. The initial temperature of the calorimeter is then also 25°C. A quantity of water is heated to 50.0°C and 25.0 g of this hot water is added to the calorimeter. What is the heat capacity of the calorimeter if the final temperature is 35°C?

First, let's take the point of view that the calorimeter is a perfect insulator and absorbs no heat from the hot water. It makes sense then that equal volumes of water at different temperatures will reach a temperature halfway between their initial values when they reach equilibrium. Specifically, 25.0 g of water at 25.0°C mixed with 25.0 g of water at 50.0°C will have an equilibrium temperature of 37.5°C. Even so, let's calculate this.

Ti = 25.0°C for the water in the calorimeter
Ti = 50.0°C for the hot water to be added
Tf is to be determined
The heat capacity of water is s = 4.18 J/(°C g)

The equation to use is: qhot + qcold = 0 (the first law of thermodynamics demands that no energy is created or destroyed so the sum of the heat lost by the hot water plus the heat gained by the cold water is zero.)

Generally, q = smΔT

(4.18 J/°C·g)(25.0 g)(Tf - 50°C) + (4.18 J/°C·g)(25.0 g)(Tf - 25°C) = 0

this simplifies to
(209 J/°C)Tf = 7837.5 J so
Tf = 37.5°C

Now let’s come back to reality. Some of the heat from the hot water will go toward warming up the calorimeter and not the cold water. The actual final temperature will therefore be less. So we re-write the equation with this in mind:

qhot + qcold + qcalorimeter = 0

In this case the final temperature is measured in the lab and turns out to be 35°C. Obviously, all components of the system have the same final temperature.

(4.18 J/°C·g)(25.0 g)(35 - 50°C) + (4.18 J/°C·g)(25.0 g)(35 - 25°C) + (s)(35 - 25°C) = 0
-1.5675 × 103 J + 1.3585 × 103 J + (s)(10°C) = 0
(s)(10°C) = 522.5 J
s = 52.3 J/°C

Note that there is no unit for amount of material for the heat capacity of a calorimeter (no g and no mol). This is because the heat capacity of the calorimeter is a constant for that piece of equipment and the mass or moles of calorimeter will not vary.

#### Heat Capacity of a Metal

A calorimeter is made using two nested styrofoam cups and filled with 175.0 g of water at an initial temperature of 25.0°C. A quantity of water is heated to 100.0°C and 25.0 g of aluminum is heated in the boiling water for five minutes. Once the aluminum has the same temperature as the boiling water it is removed with a pair of tongs and placed into the water in the calorimeter. What is the heat capacity of the piece of aluminum if the final temperature is 27.2°C?

For a calorimeter that we assume is perfect (it neither absorbs nor releases any heat) we can set up an equation showing that the heat lost by the metal plus the heat gained by the water equals zero:

qAl + qH2O = 0
qAl = (25.0 g)(s)(27.2°C – 100.0°C)
qH2O = (175.0 g)(4.184 J/g·°C)(27.2°C – 25.0°C)

Setting qH2O equal to –qAl and solving for s gives:
s = 0.897 J/g·°C

This example was written so that the answer gives the accepted reference value for the specific heat of aluminum metal. Your work in the lab will involve a similar measurement but may not give a result that exactly matches the reference value. Differences may be due to heat lost in heating the cups of the calorimeter or due to the fact that a little hot water may have still been clinging to the metal when it was dropped in the cups. The thermometer itself may be a source of heat or absorb some heat and play a role in determining the final temperature. Finally, inaccuracies in thermometer readings cannot be ruled out. Each of these will have a different effect on the outcome of your measurement, making it higher or lower.

Find the Molar Heat of Solution for Potassium Nitrate:

Now the calorimeter can be put to use to measure heat in chemical and physical changes. Take for example the formation of a solution of potassium nitrate (KNO3). Dissolve 1.00 g of KNO3 in 25.0 g of water at 25.0°C. When this is done in the calorimeter above the final temperature is 22.8°C

qKNO3 + qwater + qcalorimeter = 0
qKNO3 + (4.18 J/°C·g)(25.0 g)(22.8 - 25.0°C) + (52.3 J/°C)(22.8 - 25.0°C) = 0
qKNO3 = 345 J
This amount of heat is for 1.00 g of KNO3 so the molar heat of solution is:
``` 345 J      101.1 g      1 kJ
-------- * --------- * ------- = 34.9 kJ/mol
1.00 g      1 mol      1000 J
```
Last updated: Sep 11, 2014        Home