Heat of 30 kcal is supplied to a system and 4200 J of external work is done on the system so that its volume decreases at constant pressure.What is change in its internal energy?
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Answer:
ΔQ=30×4200
=126000J
ΔW=−4200J
∴Δu=ΔQ−ΔW
=126000+4200J
=1.302×10
5
J
Explanation:
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