WQ_1=η_engine=0.3 .......(1) nbsp; Q_2W=COP_ref=5.........(2) nbsp; Multiplying equation ( 1) and (2) Q_2Q_1=0.3×5=1.5 Q_1=Q_21.5=10001.5kJ=666.67 kJ ...

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Standard XII

Chemistry

Internal Energy

An engine of ...

Question

An engine of 30% thermal efficiency is used to drive a refrigerator of COP 5. What is the heat input into the engine for each MJ removed from the cold body by the refrigerator?

5 MJ

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200 kJ

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333.33 kJ

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666.67 kJ

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Solution

The correct option is D 666.67 kJ

$\frac{W}{Q_{1}} = η_{engine} = 0.3 . . . . . . . (1)$

$\frac{Q_{2}}{W} = C O P_{ref} = 5......... (2)$

Multiplying equation ( 1) and (2)

$\frac{Q_{2}}{Q_{1}} = 0.3 \times 5 = 1.5$

$Q_{1} = \frac{Q_{2}}{1.5} = \frac{1000}{1.5} k J = 666.67 k J$

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An engine of 30% thermal efficiency is used to drive a refrigerator of COP 5. What is the heat input into the engine for each MJ removed from the cold body by the refrigerator?

The correct option is D 666.67 kJ WQ1=ηengine=0.3.......(1) Q2W=COPref=5.........(2) Multiplying equation ( 1) and (2) Q2Q1=0.3×5=1.5 Q1=Q21.5=10001.5kJ=666.67kJ

The correct option is D 666.67 kJ

$\frac{W}{Q_{1}} = η_{engine} = 0.3 . . . . . . . (1)$

$\frac{Q_{2}}{W} = C O P_{ref} = 5......... (2)$

Multiplying equation ( 1) and (2)

$\frac{Q_{2}}{Q_{1}} = 0.3 \times 5 = 1.5$

$Q_{1} = \frac{Q_{2}}{1.5} = \frac{1000}{1.5} k J = 666.67 k J$