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Question

Find the change in the internal energy when 15 gm of air is heated from 0oC to 5oC. The specific heat of air at constant volume is 0.2 cal/gmoC.

A
75 Cal
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B
30 Cal
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C
15 Cal
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D
105 Cal
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Solution

The correct option is C 15 Cal
heat = ΔQ=Mass × specific heat×temperature change=15×0.2×5 cal=15 cal
Also, isochoric process implies work done =0
Applying first law, Change in internal energy= ΔQ =15 cal

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