Question

A calorimeter has mass 100 g and specific heat 0.1 kcal/ kg°C. It contains 250 gm of liquid at 30°C having specific heat of 0.4 kcal/kg°C. If we drop a piece of ice of mass 10 g at 0°C, What will be the temperature of the mixture?

Answer 1

Let the final temperature of the mixture be T.

Amount of heat required in converting 10 g ice to 0°C to water at 0°C = 10 × 80 = 800 cal

Total amount of heat required in converting 10 g water to 0°C to water at T°C = 10 × 1 × T = 10T

Total heat energy required to convert 10 g ice at 0°C to water at T°C = 800 + 10T

Amount of heat released to raise the temperature of calorimeter at 30°C to T°C = 100 × 0.1 × (30 − T) = 10(30 − T)

Amount of heat released to raise the temperature of 250g of water at 30°C to T°C = 250 × 0.4 × (30 − T) = 100(30 − T)

Total amount of heat released in the process = 110(30 − T)

Using the principle of the calorimeter, we have 

110(30 − T) = 800 + 10T

⇒ T = 20.83° C