First slide

Change of state

Question

A piece of ice of mass 50 g exists at a temperature of ${-20}^{0} C$ . Determine the total heat required to convert it completely to steam at $100^{0} C$ . (Specific heat capacity of ice = 0.5 cal/ $g -^{0} C$ ; specific latent heat of fusion for ice = 80 cal/g and specific latent heat of varporisation for water =540 cal/g).

Moderate

A

26500 cal
B

12400 cal
C

36500 cal
D

46500 cal

Solution

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

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