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Question

Efficiency of a cyclic process is 50% and heat interactions are Q1=+1915 J, Q2=40 J, Q3=+125 J and Q4=Q J. Find the value of Q.

A
1080 J
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B
1980 J
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C
1080 J
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D
980 J
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Solution

The correct option is D 980 J
We know that efficiency(η) of a cyclic process is given by

η=work done (W)Heat absorbed per cycle ...(1)

Here, heat absorbed,

Qab=Q1+Q3=1915+125

Qab=2040 J .......(2)

From first law of thermodynamics, work done (W) for a cyclic process is given by

W=total heat absorbedtotal heat rejected

W=(Q1+Q3)(Q2+Q4)

W=(1915+125)(40+Q)

W=2000Q .......(3)

From Eqs. (1), (2) and (3),

η=2000Q2040

0.5=2000Q2040

Q=980 J

So, (d) is the correct answer.

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