A piece of ice of mass 50 g exists at a temperature of -200C. Determine the total heat required to convert it completely to steam at 1000C. (Specific heat capacity of ice = 0.5 cal/g-0C; specific latent heat of fusion for ice = 80 cal/g and specific latent heat of varporisation for water =540 cal/g).

For changing ice at -200C to steam at 1000C the  complete process is achieved in four different stages:(i) The rise in temperature of ice from -200C to 00C (i.e., melting point of ice).Heat required for this process∆Q1 = (50 g)(0.5 cal/g0C)(0+20)0C = 500 cal(ii) The melting of ice at 00CHeat required for this process∆Q2 = (50 g)(80 cal/g) = 4000 cal(iii) The rise in temperature of melted ice (i.e., water from 00C to 10000C.(i.e., B.P. of water) Heat required for this process∆Q3 = (50 g)(1 cal/g0C)(100 0C) = 5000 cal(iv) The vaporisation of water at 1000CHeat required for this process∆Q4 = (50 g)(540 cal/g) = 27000 cal∴ Total heat required ∆Q = ∆Q1 + ∆Q2+∆Q3+∆Q4 = 36500 cal