Question 121: Thermodynamics & KTG

Question

A piece of ice (heat capacity = 2100 J kg-1 oC-1 and latent heat = 3.36 $$ 	imes $$ 105 J kg-1 ) of mass m grams is at
- 5 oC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the
ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat
exchange in the process, the value of m is

Accepted Answer

In the final state, ice-water mixture is in equilibrium. Thus, the temperature of m grams of ice is raised from $$-$$5 $$^\circ$$C to 0 $$^\circ$$C. The heat absorbed in this process is
Q
1
= mS$$\Delta$$T. ...... (1)
The state of m
1
= 1 g of ice is changed from solid to liquid. The heat absorbed in the melting process is
Q
2
= m
1
L. ........ (2)
The heat supplied is Q = 420 J. By energy conservation, Q = Q
1
+ Q
2
. Substitute Q
1
and Q
2
from equations (1) and (2) to get
$$m = {{Q - {m_1}L} \over {S\Delta T}} = {{420 - ({{10}^{ - 3}})(3.36 	imes {{10}^5})} \over {(2100)(5)}}$$
$$ = 8 	imes {10^{ - 3}}$$ kg = 8 g.

Thermodynamics & KTG

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