Question

If there is no Heat loss to the surroundings, the heat released by the condensation of m₁ g of steam at 100°C into water at 100°C can be used to convert m₂ g of ice at 0°C into water at 0°C.

(i) Find:

(a) The heat lost by steam in terms of m₁

(b) The heat gained by ice in terms of m₂

(ii) Form a heat equation find the ratio of m₂ : m₁

Specific latent heat of vaporization of steam = 2268 kJ/kg

Specific latent heat of fusion of ice = 336 kJ/kg

Specific heat capacity of water = 4200 J/kg°C

Answer 1

(i)

(a) Heat lost by steam = mL + mst

= m₁ × 10^-3 × 2268 + m₁ × 10^-3 × 4.2 × 100 

(b) Heat gained by ice in converting to water at 0°C

mL = m₂ × 10^-3 × 336 kJ 

(ii) Heat lost by steam = Heat gained by ice

(m₁ × 10^-3 × 2268 + m₁ × 10^-3 × 4.2 × 100) = m₂ × 10^-3 × 336 

Or m₁ × 10^-3 (2268 + 420) = m₂ × 10^-3 × 336

∴ `"m"_2/"m"_1 = (10^-3 xx 2688)/(10^-3 xx 336) = 8/1`

∴ The ratio  m₂ : m₁ = 8 : 1