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Question

A system in a cyclic process absorbs 780 J of heat and rejects 300 J . What is the net work done?

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Explanation

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h <1, impossible 100% efficient, Carnot efficiency η_Carnot =1 - T_c/T_h maximum possible between T_h and T_c. For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 780 - 300 = 480 J . W = 480 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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