Question

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.

Answer 1

Given: water = 250 gm = 30 °C

mass of vessel = 50 gm

Final temperature = 5 °C

By the principle of calorimeter.

Heat given = Heat taken,

Let mass of ice = m gm

250 × 4.2 × (30 - 5) + 50 × 0.4 × 25 = m × 336 + m × 4.2 × 5

26250 + 500 = 336 m + 21 m

26750 = 357 m

m = 74.9 gm.