Question

To change 25 kg of ice -10°C to steam 100°C, how much heat is required? The specific heat of water is 4.184 kJ/kg. K. The latent heat of fusion for water at 0°C is approximately 334 kJ/kg (or 80 cal/g), and the latent heat of vaporization at 100°C is about 2,230 kJ/kg (533 cal/g).

Answer 1

Specific heat of ice = 2.04 kJ / ( kg . K ).

Hence, heat required to raise the temperature of 25 kg ice by [ 0 - ( - 10 ) ]^o C or 10^o C or 10 K is :

H₁ = 25 kg x 2.04 kJ / ( kg . K ) x 10 K = 510 kJ.

Latent heat of fusion for water = 334 kJ/kg.

Hence, heat required to melt the ice completely is :

H₂ = 25 kg x 334 kJ/kg = 8350 kJ.

Specific heat of water = 4.184 kJ / ( kg . K ).

Hence, heat required to raise the temperature of 25 kg water by ( 100 - 0)^o C or 100^o C or 100 K is :

H₃ = 25 kg x 4.184 kJ / ( kg . K ) x 100 K = 10460 kJ.

Latent heat of vaporization for water = 2230 kJ/kg.

Hence, heat required to vaporize completely is :

H₄ = 25 kg x 2230 kJ/kg = 55750 kJ.

Hence, total heat required to change 25 kg of ice -10°C to steam 100°C is :

H = H₁ + H₂ + H₃ + H₄ = ( 510 + 8350 + 10460 + 55750 ) kJ = 75070 kJ.