Question

Calculate the amount of heat transfer required to change 88 kg of ice (solid state) which is at 00C to

water (liquid state) of 100C. The specific heat capacity of ice is 2090 J/kg0C and the specific latent heat

of fusion is 335 kJ/kg.

Answer 1

q₁ = m x ∆H_fus = 88 kg x 335 kJ/kg = 29 480 kJ

q₂ = m x c x ∆T = 88 x 2090 J/kg•°C x (10 - 0)°C = 1 839 kJ

Q = q₁ + q₂ = 29 480 + 1 839 = 31 319 kJ