250 g of water at 30 °C is present in a copper vessel of mass 50 g.

Hi, I have a question and I hope anyone could answer it:

250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C.
Specific latent heat of fusion of ice = 336 × 103 J kg-1
Specific heat capacity of copper vessel = 400 J kg-1 °C-1
Specific heat capacity of water = 4200 J kg-1 °C-1.

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1. 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.