How to Calculate a Calorimeter Constant

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Example #1: When 40.0 mL of water at 60.0 °C is added to 40.0 mL at 25.0 °C water already in a calorimeter, the temperature rises 15.0 °C. What is the calorimeter constant?

Solution:

We need to find the difference between the heat lost by the hot water when it droped from 60.0 to 40.0 and the heat gained by the cold water when it was heated up to 40.0 from 25.0. Note that the values in the problem are in mL and the values in the solution are grams. The volume (mL) is converted to the mass (grams) by using the density of water (1.00 g/mL).

1) Hot water lost:

q = m Δt Cp

q = (40.0 g) (20.0 °C) (4.184 J g¯1 °C¯1)

q = 3347.2 J

2) Cold water got:

q = m Δt Cp

q = (40.0 g) (15.0 °C) (4.184 J g¯1 °C¯1)

q = 2510.4 J

3) The calorimeter got the rest:

3347.2 − 2510.4 = 836.8 J

4) Find the heat capacity of the calorimeter:

836.8 J / 15.0 °C = 55.8 J / °C

Example #2: Calculate the calorimeter constant if 25.0 g of water at 60.0 °C was added to 25.0 g of water at 25.0 °C with a resulting temperature of 35.0 °C?

Solution:

If the constant were zero, the final temperature of the water would be 42.5 °C. So the amount of heat used by the calorimeter to heat from 25 to 35 is:

(25.0) (4.184) (7.5) = 784.5 J.

Since the constant is Joules/degree, the constant is

784.5 J / 10.0 °C = 78.4 J/°C (to three sig figs).

Determine a Calorimeter Constant II

Example #3: A calorimeter is to be calibrated: 72.55 g of water at 71.6 °C added to a calorimeter containing 58.85 g of water at 22.4 °C. After stirring and waiting for the system to equilibrate, the final temperature reached 47.3 °C. Calculate the heat capacity of the calorimeter. (The specific heat capacity of water is 4.184 J g¯1 °C¯1).

Solution:

1) Energy lost by the hot water:

q = m Cp ΔT

q = (72.55 g) (4.184 J g¯1 °C¯1) (24.3 °C)

q = 7376.24 J

2) Energy gained by the cold water:

q = m Cp ΔT

q = (58.85 g) (4.184 J g¯1 °C¯1) (24.9 °C)

q = 5818.54 J

3) The calorimeter got the rest:

7376.24 − 5818.54 = 1557.7 J

4) Heat capacity of the calorimeter:

1557.7 J / 24.9 °C = 62.6 J/°C

Example #4: A student wishes to determine the heat capacity of a coffee-cup calorimeter. After mixing 100.0 g of water at 58.5 °C with 100.0 g of water, already in the calorimeter, at 22.8 °C, the final temperature of the water is 39.7 °C. Calculate the heat capacity of the calorimeter in J/°C. (Use 4.184 J g¯1 °C¯1 as the specific heat of water.)

Solution:

1) Heat given up by warm water:

q = (100.0 g) (18.8 °C) (4.184 J g¯1 °C¯1) = 7865.92 J

2) Heat absorbed by water in the calorimeter:

q = (100.0 g) (16.9 °C) (4.184 J/g °C) = 7070.96 J

3) The difference was absorbed by the calorimeter:

7865.92 − 7070.96 = 794.96 J

4) Calorimeter constant:

794.96 J / 16.9 °C = 47.0 J/°C