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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?
Solution
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 eater 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 calorimeter, we have
110(30 - T) = 800 + 10T
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