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$10 \,gm$ of ice at $-20^{\circ} C$ is kept into a calorimeter containing $10 \,gm$ of water at $10^{\circ} C$. The specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain ..........
$20 \,gm$ of water
$20 \,gm$ of ice
$10 \,gm$ ice and $10 \,gm$ of water
$5 \,gm$ ice and $15 \,gm$ water
Solution
(c)
As we know $Q = mc \Delta \theta$
$Q _1=10 \times 1 \times 10=100\,cal$
$Q _2=10 \times 0.5(0-(-20))+10 \times 80=(100+800) \,cal =900 \,cal$
As $Q 1 < Q 2$,
so ice will not completely melt and final temeprature $=0^{\circ}\, C$
As heat given by water in cooling up to $0^{\circ}\, C$ is only just sufficient to increase the temperature of ice from $-20^{\circ}\, C$ to $0^{\circ}\, C$
, hence mixture in equilibrium will consist of $10\, gm$ of ice and $10\, gm$ of water, both at $0 ^{\circ}\, C$.
