Question

250 g of water at 30℃ is contained 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℃. Given: specific latent heat of fusion of ice = 336 × 10³J kg^-1, specific heat capacity of copper = 400 J kg^-1 K^-1, specific heat capacity of water = 4200 J kg^-1 K^-1.

Answer 1

The mass of ice at 0°C is denoted as m. The water in the calorimeter is maintained at 30°C, and this ice melts it. More heat is gained by this water, which is melted ice at 0°C. The mixture's temperature is 5°C.

According to the Principle of mixtures

Heat energy gained by ice = Heat energy lost by hot water and vessels

mL + mC_W (5 - 0) = [250 × C_W + 50 × C_V] (20 + 5)

m[L + 4.2 × 5] = [250 × 4.2 + 50 × 0.400] × 25

m [336 + 21.0] = `[25 × 42 + 50 × 400/1000]` × 25

m [357] = (1050 + 20) × 25

m [357] = 1070 × 25

m = `(1070 xx 25)/357`

= `26750/357`

= 74.93 g

Mass of ice m = 74.93 g

∴ Temperature will remain 0°C.