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Question

10g of ice at 20oC is dropped into a calorimeter containing 10g of water at 10oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain.

A
20g of water
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B
20g of ice
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C
10g ice and 10g water
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D
5g ice and 15g water
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Solution

The correct option is C 10g ice and 10g water
Final temperature of mixture can be greater than 0oC or less than 0oC.

1) Final temperature greater than 0oC
As all ice in caloriemeter absorbs heat and rises to 10oC, Ice at 0oC absorbs latent heat and becomes water at 0oC
θ= required heat =(10×0.5×20)+(10×80)+10×1×θ
Heat lost by water =10(10θ)=10010θ
Required heat > heat lost

2) Final temperature less than 0oC
Let the final temperature be θoC
Heat released by water to attain θoC =10×1×10+10×80+10×0.5×θ
Heat absorbed by ice =10×0.5(10θ)
Since heat absorbed by ice less than heat to be released by water for all temperatures between 0oC and 10oC. The system can't attain a temperature less than 0oC.

3) So, final temperature of mixture can neither greater than 0oC nor less than 0oC. So, the caloriemeter will contain 10g of ice and 10g of water and the final temperature will be 0oC


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