A student adds 75.0 g of hot water at 80.0 0C into a calorimeter containing 100.0 g cold water at 20.0 0C. The final temperature is 42.5 0C. The heat capacity of water is 4.186 J/gK. What is the heat capacity of the calorimeter?

The heat capacity of the calorimeter is 104.65 J/K

Explanation:

Heat lost by the hot water = Heat gained by cold water + Heat gained by the calorimeter

Heat lost by hot water = mCΔT

m = 75 g, C = 4.186 J/g.K, ΔT = (80 - 42.5) = 37.5 K

Heat lost by hot water = 75 × 4.186 × 37.5 = 11773.125 J

Heat gained by cold water = mCΔT

m = 100 g, C = 4.186 J/g.K, ΔT = (42.5 - 20) = 22.5 K

Heat gained by cold water = 100 × 4.186 × 22.5 = 9418.5 J

Heat gained by the calorimeter = Heat lost by hot water - Heat gained by cold water

Heat gained by the calorimeter = 11773.125 - 9418.5 = 2354.625 J

Heat gained by the calorimeter = Heat capacity of the calorimeter × ΔT

ΔT = (42.5 - 20) = 22.5 K

Heat capacity of the calorimeter = (Heat gained by the calorimeter)/(ΔT) = 2351.25/22.5 = 104.65 J/K

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