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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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