A piece of ice of mass 50 g exists at a temperature of ${\text{-20}}^0\mathrm{C}$. Determine the total heat required to convert it completely to steam at ${100}^0\mathrm{C}$. (Specific heat capacity of ice = 0.5 cal/$\mathrm{g}{-}^0\mathrm{C}$; 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 $-{20}^0\mathrm{C} \mathrm{to} \mathrm{steam} \mathrm{at} {100}^0\mathrm{C} \mathrm{the}$ complete process is achieved in four different stages:

(i) The rise in temperature of ice from $-{20}^0\mathrm{C} \mathrm{to} 0^0\mathrm{C}$(i.e., melting point of ice).

Heat required for this process

$∆{\mathrm{Q}}_1 = (50 \mathrm{g})(0.5 \mathrm{cal}/{\mathrm{g}}^0\mathrm{C}){(0+20)}^0\mathrm{C} = 500 \mathrm{cal}$

(ii) The melting of ice at $0^0\mathrm{C}$

Heat required for this process

$∆{\mathrm{Q}}_2 = (50 \mathrm{g})(80 \mathrm{cal}/\mathrm{g}) = 4000 \mathrm{cal}$

(iii) The rise in temperature of melted ice (i.e., water from $0^0\mathrm{C} \mathrm{to} {1000}^0\mathrm{C}.(\mathrm{i}.\mathrm{e}., \mathrm{B}.\mathrm{P}. \mathrm{of} \mathrm{water})$

Heat required for this process

$∆{\mathrm{Q}}_3 = (50 \mathrm{g})(1 \mathrm{cal}/{\mathrm{g}}^0\mathrm{C})\left(100{ }^0\mathrm{C}\right) = 5000 \mathrm{cal}$

(iv) The vaporisation of water at ${100}^0\mathrm{C}$

Heat required for this process

$∆{\mathrm{Q}}_4 = (50 \mathrm{g})(540 \mathrm{cal}/\mathrm{g}) = 27000 \mathrm{cal}$

$\therefore \mathrm{Total} \mathrm{heat} \mathrm{required} ∆\mathrm{Q} = ∆{\mathrm{Q}}_1 + ∆{\mathrm{Q}}_2+∆{\mathrm{Q}}_3+∆{\mathrm{Q}}_{4_{ }}$

= 36500 cal