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}.**

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.