Question

the temperature of 170g of water at 50°C is lowered to 5°C by adding certain amount of ice to it. Find the mass of ice added. Given: specific heat capacity of water = 4200J kg^-1°C^-1 and specific latent heat of ice = 336000J kg^-1​

Answer 1

Explanation:

Given mass of water = 170 g = 0.17 kg

T1 = 50°C, T2 = 5°C,

specific heat capacity of water = 4200 J kg –1°C–1

Sp. Latent heat of ice = 336000 J kg–1

Let mass of ice added be x then we have heat lost by water = Heat used by ice.

⇒ water at 50°C to 5°C = ice at °C to water at °C + water at 0°C to 5°C

⇒ we should add 90 g of ice to water.