কৃতকাজ

In a heat engine, the temperature of the source and sink is 500K500 K and 375K375 K. If the engine consumes 25×105 25\times{ 10 }^{ 5 } per cycle, the work is done per cycle is:

হানি নাটস

Given,

T1=500KT_1=500K

T2=375KT_2=375K

Q1=25×105JQ_1=25\times 10^5J

Efficiency of the engine,

η=1T2T1\eta=1-\dfrac{T_2}{T_1}

η=1375500\eta=1-\dfrac{375}{500}

η=10.75\eta=1-0.75

η=0.25=25\eta=0.25=25%

We know that, η=WQ1\eta=\dfrac{W}{Q_1}

W=ηQ1W=\eta Q_1

W=0.25×25×105W=0.25\times 25\times 10^5

W=6.25×105JW=6.25\times 10^5J

The correct option is A.

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