Practice question
Question
A system in a cyclic process absorbs 940 J of heat and rejects 360 J . What is the net work done?
Explanation
**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). For cyclic: Δ U = 0 , Q_net = W . Q_net = Q_absorb - Q_reject = 940 - 360 = 580 J . W = 580 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R
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