A certain engine which operates in a Carnot cycle absorbs 3-347 kJ at 400- C- how much work is done by the engine per cycle- The temperature of sinkis 100- C-A- 1-49 kJB- 2-98 kJC- 1-85 kJD- 2-51 kJ

The correct option is A 1.49 kJη nbsp;= T_2 T_1T_2 = 1 T_1T_2 = 1 373673 = 0.446 η = wq_1 or W = η× q_1 = 0.446 × 3.347 = 1.49 kJ ...

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Standard XII

Chemistry

Carnot Cycle

A certain eng...

Question

A certain engine which operates in a Carnot cycle absorbs 3.347 kJ at $400^{\circ} C,$ how much work is done by the engine per cycle? The temperature of sink is $100^{\circ} C .$

1.49 kJ

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2.51 kJ

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1.85 kJ

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2.98 kJ

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Solution

The correct option is A 1.49 kJ
$η = \frac{T_{2} - T_{1}}{T_{2}} = 1 - \frac{T_{1}}{T_{2}} = 1 - \frac{373}{673} = 0.446 η = \frac{△ w}{q_{1}} o r △ W = η \times q_{1} = 0.446 \times 3.347 = 1.49 k J$

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Carnot Cycle

Standard XII Chemistry

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A certain engine which operates in a Carnot cycle absorbs 3.347 kJ at 400∘C, how much work is done by the engine per cycle? The temperature of sink is 100∘C.

The correct option is A 1.49 kJη=T2−T1T2=1−T1T2=1−373673=0.446η=△wq1 or △W=η×q1=0.446×3.347=1.49kJ

A certain engine which operates in a Carnot cycle absorbs 3.347 kJ at $400^{\circ} C,$ how much work is done by the engine per cycle? The temperature of sink is $100^{\circ} C .$

The correct option is A 1.49 kJ
$η = \frac{T_{2} - T_{1}}{T_{2}} = 1 - \frac{T_{1}}{T_{2}} = 1 - \frac{373}{673} = 0.446 η = \frac{△ w}{q_{1}} o r △ W = η \times q_{1} = 0.446 \times 3.347 = 1.49 k J$