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?
hint : heat lost by liquid +heat lost by calorimeter= heat gained by ice
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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 calorimetry, we have
110(30−T)= 800 + 10T
⇒T=20.83°C
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