Q.
A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy
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a
Increases by 600 J
b
Decreases by 800 J
c
Increases by 800 J
d
Decreases by 50 J
answer is C.
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Detailed Solution
ΔQ=ΔU+ΔW∵ΔQ=200cal=200×4.2=840J and ΔW=40J⇒ΔU=ΔQ−ΔW=840−40=800J
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