### How to Calculate a Calorimeter Constant

Here are two videos which each discuss the solution to a 'determine the calorimeter constant' problem:

1) How to Calculate a Calorimeter Constant I

Two solved problems:

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 = 62.558 J/°C (round off as you see fit)

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 °C 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 °C) = 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