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Question

10 g of ice at 20C is dropped into a calorimeter containing 10 g of water at 10C; 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 = 0C. As heat given by water in cooling up to 0C is only just sufficient to increase the temperature of the ice from 20C to 0C, hence mixture in equilibrium will consist of 10 g of ice and 10 g of water, both at 0C.


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