In a certain process, 400cal of heat is supplied to a system and at the same time 105J of mechanical work was done on the system. The increase in its internal energy is
A
425cal
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B
375cal
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C
505cal
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D
295cal
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Solution
The correct option is A425cal Given, ΔQ=400cal;ΔW=−105J=−1054.2=−25cal From 1st law, ΔQ=ΔU+ΔW ⇒400=ΔU−25 ⇒ΔU=425cal