how to find the Calorimeter constant

Question

Temp of cold water 2.0 C

Temp of hot water 92.0

Volume of cold water 100.0 mL

Volume of hot water 97.0 mL

Final temp after mixing 45.0 C

Mass of cold water 100.0

Mass of hot water 97.0

Calorimeter constant J/C (two decimal places)

Answer 1

The equation q = mC∆T relates heat (q) to mass (m), specific heat (C) and the change in temperature (∆T).

When mixing the hot and cold water together in a calorimeter, the heat lost by the hot water MUST equal the heat gained by the cold water PLUS the heat gained by the calorimeter (the calorimeter constant).

heat lost by hot water = q = mC∆T

q = (97.0 g)(4.184 J/gº)(92.0º - 45.0º) = (97.0 g)(4.184 J/gº)(47.0)

q = 19,075 J of heat lost by the hot water

heat gained by the cold water = q = mC∆T

q = (100.0 g)(4.184 J/gº)(45.0º- 2.0º) = (100.0 g)(4.184 J/gº)(43.0º)

q = 17,991 J of heat gained by the cold water

The difference between the heat lost by the hot water, and that gained by the cold water, is the heat that went into the calorimeter, and is referred to as the calorimeter constant. We actually need to know the initial temperature of the calorimeter to do this properly, so I'll assume the cold water was in the calorimeter first, and thus the initial temp of the calorimeter would be 2.0º, and the final temp of the calorimeter would be 45.0º. Thus, the change in temperature of the calorimeter would be 43º. The C_cal (calorimeter constant) would thus be calculated as:

C_cal = (19,075 J - 17,991 J) / 43º

C_cal = 25.21 J/º