Question

A piece of ice of mass $50\, g$ exists at a temperature of $-20^{\circ} C$. Determine the total heat required to convert it completely to steam at $100^{\circ} C$. (Specific heat capacity of ice $=0.5\, cal / g -{ }^{\circ} C ;$ specific latent heat of fusion for ice $=$ $80\, cal / g$ and specific latent heat of vaporisation for water $=540\, cal / g )$

• A. 26500 cal
• B. 12400 cal
• C. 36500 cal
• D. 46500 cal

Answer 1

Answer: (C)

For changing ice at $-20^{\circ} C$ to steam at $100^{\circ} C$ the complete process is achieved in four different stages:
(i) The rise in temperature of ice from $-20^{\circ} C$ to $0^{\circ} C$ (i.e., melting point of ice).
Heat required for this process
$\Delta Q_{1}=(50\, g )\left(0.5\, cal / g -{ }^{\circ} C \right)(0+20)^{\circ} C =500\, cal$
(ii) The melting of ice at $0^{\circ} C$ Heat required for this process
$\Delta Q_{2}=(50\, g )(80\, cal / g )=4000\, cal$
(iii) The rise in temperature of melted ice (i.e., water) from $0^{\circ} C$ to $1000^{\circ} C$. (i.e., B.P. of water) Heat required for this process
$\Delta Q_{3}=(50 g )\left(1\, cal / g -{ }^{\circ} C \right)\left(100^{\circ} C \right)=5000\, cal$
(iv) The vaporisation of water at $100^{\circ} C$ Heat required for this process
$\Delta Q_{4}=(50\, g )(540\, cal / g )=27000\, cal$
$\therefore $ Total heat required $\Delta Q=\Delta Q_{1}+\Delta Q_{2}+\Delta Q_{3}+\Delta Q_{4}$
$=36500\, cal$