Question

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

Answer 1

SubstanceMassS.H.C.Initial TempFinal temp = xHot solid56 g0.9 J/kg K220°CθR = (x−20°C)Cold water220 g4.2 J/kg K20°CθF = (220°C−x),Heat given out by solid=mC θF=56 × 0.9 × (220 − x)=11088 − 50.4 xHeat absorbed by cold water=mC θR=220 × 4.2 × (x − 20)=924x − 18480Heat given out by solid=Heat absorbed by water11088 − 50.4 x=924x − 18480∴ 974.4x=29568∴ x=30.34°C

Answer 2

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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