CHM1 12 Worked Problem2 Collection

Solution The hot water loses a quantity of heat q_hot;

q _hot = C m ?T = 4.18 J/g°C × 50.0 g × (27.30°C - 32.00°C) = -982 J

The cold water gains a quantity of heat q_cold;

q _cold = C m ?T = 4.18 J/g°C × 40.0 g × (27.30°C - 22.00°C) = 886 J

If no heat were lost to the surroundings,q_hot would equal - q_cold and their sum would be zero. But it appears that more heat was lost by the hot water than was gained by the cold. The missing heat was absorbed by the calorimeter and surroundings:

q _cal = q _hot + q _cold = 96 J.

The calorimeter heat capacity is this heat divided by the temperature change of the calorimeter:

C_cal = q/?T = 96 J / 5.3 °C = 18.1 J/°C