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Question

If heat absorbed by the working substance from the source is Q1=550 J, heat rejected to the sink is Q2=200 J then work done by the working substance in a cyclic process is

A
250 J
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B
350 J
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C
750 J
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D
550 J
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Solution

The correct option is B 350 J
Heat absorbed by working substance,
Q1=+550 J
Heat rejected to the sink,
Q2=200 J
For a cyclic process, change in internal energy ΔU=0, applying first law of thermodyamics
Q=ΔU+W ...(i)
For the cyclic process, Q=Q1+Q2
(550+(200))=0+W
W=350 J

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