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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesHard2 minQB

PhysicsHardnumerical

A gas goes through the cycle: 1 to 2 at constant volume 0.05 $m^{3}$ from 100 kPa to 600 kPa, 2 to 3 isothermal expansion down to 100 kPa, and 3 to 1 at constant pressure. Find the net heat transfer in kJ (one decimal).

Answer:

28.80

Solution:

$Δ U = 0$ , so $Q = W$ . $V_{3} = \frac{600 \times 0.05}{100} = 0.30$ $m^{3}$ . $W_{23} = 600 \times 0.05 \times ln 6 = 53.75$ kJ. $W_{31} = 100 \times (0.05 - 0.30) = - 25$ kJ. $Q = 28.8$ kJ.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

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