10 gm of ice at – 20°C is dropped into a calorimeter containing 10 gm of water at 10°C; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain
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Heat lost by water to cool from $10^\circ \mathrm{C}$ to $0^\circ C$
$= mS\Delta T = 10 \times 1 \times 10 = 100cal$
Heat gained by ice to rise its temperature from $-20^\circ \mathrm{C}$ to $0^\circ C$
$= m \times S \times \Delta t = 10 \times \frac{1}{2} \times 20 = 100 cal$
Since heat lost = heat gained, therefore the equilibrium temperataure is $0^\circ C$ . Also at $0^\circ C$ there will be ice only. Hence the calorimeter will contain (10 + 10)g i.e 20 g ice.
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