Question

# 10 g of ice at −20∘C is dropped into a calorimeter containing 10 g of water at 10∘C; the specific heat of water is twice that of ice. Ehen equilibrium is reached, the calorimeter will contain

A

20 g of water

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B

20 g of rice

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C

10 g ice and 10 g of water

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D

5 g ice and 15 g of water

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Solution

## The correct option is C 10 g ice and 10 g of water Q1 = 10×1×10 = 100 cal Q2 = 10×0.50[0−(−20)]+10×80 = (100 + 800) cal = 900 cal As Q1 < Q2, so ice will not completely melt and final temperature = 0∘C. As heat given by water in cooling up to 0∘C is only just sufficient to increase the temperature of the ice from −20∘C to 0∘C, hence mixture in equilibrium will consist of 10 g of ice and 10 g of water, both at 0∘C.

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