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Q. 10 gm of ice cubes at $0^{\circ}C $ are released in a tumbler (water equivalent 55 g) at $40^{\circ }C . $ Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearely (L = 80 cal/g)

AIPMTAIPMT 1988Kinetic Theory

Solution:

Let $\theta$ be the temperature when thermal equilibrium has reached. Heat gained by ice to be converted to water at $\theta^{\circ} C$ $=m L+m \times c \times(T-0)=10 \times 80+10 \times 1 \times T$
$65 t=2200-800=1400$
$T=\frac{1400}{65} \approx 21.5^{\circ} C \approx 22^{\circ} C$