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Question

250 g of water at 30℃ is contained in a copper vessel of mass 50g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5℃. Given: specific latent heat of fusion of ice = 336×103Jkg1, specific heat capacity of copper = 400Jkg1K1, specific heat capacity of water = 4200Jkg1K1 .

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Solution

Mass of copper vessel m1=50g.
Mass of water contained in copper vessel m2=250g.
Mass of ice required to bring down the temperature of vessel = m
Final temperature =5°C.
Amount of heat gained when 'm' g of ice at 0°C converts into water at 0°C = m×336J
Amount of heat gained when temperature of 'm' g of water at 0°C rises to 5°C=m×4.2×5
Total amount of heat gained = m×336+m×4.2×5
Amount of heat lost when 250 g of water at 30°C cools to 5°C=250×4.2×25=26250J
Amount of heat lost when 50 g of vessel at 30°C cools to 5°C=50×0.4×25=500J
Total amount of heat lost =26250+500=26750J
We know that amount of heat gained = amount of heat lost m×336+m×4.2×5=26750/357
m=26750/357=74.93g
Hence, mass of ice required is 74.93g.

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