Question

The temperature of 170 g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.

Given: Specific heat capacity of water = 4200 J kg^-1 °C^-1 and specific latent heat of ice = 336000 J kg^-1.

Answer 1

Given mass of water = 170 g = 1.17 kg,

Initial temperature = 50°C

Fall in temperature =  Δt = (50 - 5) = 45°C = 45K

Heat lost by water = mc Δt

= 0.17 × 4200 × 45

= 3.213 × 10⁴ J

If m' kg ice is added, heat gained by it to melt to 0°C = m'L

= m'C Δt

= m' × 4200 × 5

= m' × 2.1 × 10⁴ J

Total heat gained by ice

= 3.36 × 10⁵ m' + 2.1 × 10⁴ m'

= 3.57 × 10⁵ m' J

By the principle of method of mixtures heat lost by water = heat gained by ice

⇒ 3.213 × 10⁴ = 3.57 × 10⁵ m'

⇒ m' = `(3.213 xx 10^4)/(3.57 xx 10^5)`

⇒ m' = 0.09 kg (90 g)