A heat transfer of 9.8 ✕ 105 J is required to convert a block of ice at - 12°C to water at 12°C. What was the mass of the block of ice?

Answer

0.238 Kg

Explanation

specific latent heat of fusion = 0.336 MJ/kg.

Specific heat capacity of water = 4200 J/kgK

Specific heat capacity of ice = 2100 J/kgK

Heat = mLf + mcΔθ + mcΔθ

(change of state) + (heat gained by ice) + (heat gained by water)

9.8 ✕ 10⁵ J = (336,000m) + (2100*12m) + (4200*12m)

98000 = m (336000 + 25200 + 50,400)

98000 = 411,600m

m = 98000/411600

= 0.238 Kg