An ideal gas heat engine operates in carnot cycle between 22 7 ∘ C and 12 7 ∘ C . It absorbs 6 × 1 0 4 cal of heat at higher temperature. Amount of heat converted into work, is:NEET 2005

For a carnot engine efficiency e = 1 − T 1 T 2 = 1 − 227 + 273 127 + 273 = 1 − 500 400 = 5 1 Now, e = Heat Output Work Output = 6 × 1 0 4 W W = e × 6 × 1 0 4 = 5 1 × 6 × 1 0 4 = 1.2 × 1 0 4 cal .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoSecond Law of ThermodynamicsHard2 minPYQ_2005

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An ideal gas heat engine operates in carnot cycle between $22 7^{\circ} C$ and $12 7^{\circ} C$ . It absorbs $6 \times 1 0^{4}$ cal of heat at higher temperature. Amount of heat converted into work, is:NEET 2005

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For a carnot engine $efficiency e = 1 - \frac{T _{2}}{T _{1}} = 1 - \frac{127 + 273}{227 + 273} = 1 - \frac{400}{500} = \frac{1}{5}$ Now, $e = \frac{Work Output}{Heat Output}$ $= \frac{W}{6 \times 1 0 ^{4}}$ $W = e \times 6 \times 1 0^{4} = \frac{1}{5} \times 6 \times 1 0^{4} = 1.2 \times 1 0^{4} cal .$

Stream:NEETSubject:PhysicsTopic:Heat & ThermoSubtopic:Second Law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2005

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