What is the amount of heat required (in calories) to convert 10 g of ice at - 10 ℃ into steam at 100 ℃ ? Given that latent heat of vaporization of water is 540 cal g - 1 , latent heat of fusion of ice is 80 cal g - 1 , the specific heat capacity of water and ice are 1 cal g - 1 ° C - 1 and 0 . 5 cal g - 1 ° C - 1 respectively.

Step-by-step solution

10 g of ice at - 10 ℃ to ice at 0 ℃ Q 1 = c m , ∆ θ = 0.5 × 10 × 10 = 50 cal 10 g of ice 0 ℃ to water at 0 ℃ Q 2 = m L = 10 × 80 = 800 cal 10 g of water at 0 ℃ to water at 100 ℃ Q 3 = c m , ∆ θ = 1 × 10 × 100 = 1000 cal 10 g water at 100 ℃ to steam at 100 ℃ Q 4 = m L = 10 × 540 = 5400 cal Total heat required, Q = Q 1 + Q 2 + Q 3 + Q 4 = 50+800+1000+5400 = 7250 cal