Question

Heat is added to 1.53 kg of ice at -5 °C. How many kilocalories are required to change the ice to steam at 130 °C?

Answer 1

Latent heat of fusion of ice L = 79.5 kcal/kg

specific heat of ice = 0.5 kcal/kg.^oC

specific heat of water = 1 kcal/kg.^oC

Latent heat of fission of steam L = 539 kcal/kg

specific heat of steam Cp = 0.48 kcal/kg.^oC

-5^oC conveted into 130^oC steam , it is 5 step process

m = 1.53 kg

Q = (1.53*0.5*(0 +5)) +(1.53*79.5) +(1.53*1*(100-0))+(1.53*539)+(1.53*0.48*(130-100))

Q = ( 3.825 kcal) + ( 121.635 kcal) + ( 153 kcal) + ( 824.67 kcal) + ( 22.032 kcal)

Q = 1125.162 kcal