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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesMedium2 minQB

PhysicsMediumnumerical

40 g of air ( $c_{v} = 0.17$ cal $g^{- 1}$ $C^{\circ- 1}$ ) at $2 7^{\circ}$ C is expanded adiabatically and does 3000 J of work. Find its final temperature in °C (one decimal; 1 cal = $4.18$ J).

Answer:

-78.50

Solution:

$Δ U = - 3000$ J. $m c_{v} = 40 \times 0.17 = 6.8$ cal $K^{- 1}$ = $28.4$ J $K^{- 1}$ . $Δ T = - \frac{3000}{28.4} = - 105.5$ K. $T_{f} = 27 - 105.5 = - 78.5^{\circ}$ C.

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

⏱ 2mℹ️ Source: QB

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