Question

Numerical Problem.

How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? (Specific latent heat of fusion of water = 3,34,000 J/kg, Specific heat capacity of water = 4200 JKg–1K–1).

Answer 1

Total heat = Heat required to convert 2Kg of ice into water at 0°C + Heat required to convert 2Kg of water at 0°C to 2Kg of water at 20°C
Heat = m (hfw) + mc∆T
Here, m(mass of ice) = 2Kg
hfw (specific latent heat of water) = 3,34,000J/Kg
C (specific heat capacity of water) = 4200JKg^{– 1}K^{– 1}
AT (Temperature difference) = 20°C
Therefore, Heat required = (2 × 334000) + (2 × 4200) (20 – 0)
= 668000 + 8400 (20)
= 668000+ 168000
Heat required = 8,36,000 J