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Carnot cycle — efficiency

A Carnot engine operates between T_H = 500 K (hot source) and T_C = 300 K (cold source).\nIt absorbs Q_H = 1000 J per cycle. Calculate: (a) efficiency, (b) work produced, (c) heat rejected.
Given data
T_H = 500 KT_C = 300 KQ_H = 1000 J
Review the theory: Termodinamica
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  1. What is the maximum efficiency of the Carnot engine between T_H = 500 K and T_C = 300 K?
  2. How much work does the engine produce per cycle?
  3. How much heat is rejected to the cold source per cycle?
Full worked solution
  1. What is the maximum efficiency of the Carnot engine between T_H = 500 K and T_C = 300 K?
    η=1−TCTH=1−300500\eta = 1 - \dfrac{T_C}{T_H} = 1 - \dfrac{300}{500}
    η=1−TC/TH=1−300/500=1−0.6=0.4=40%\eta = 1 - T_C/T_H = 1 - 300/500 = 1 - 0.6 = \mathbf{0.4} = \mathbf{40\%}. This is the maximum possible efficiency between these two temperatures — no real heat engine can exceed this limit.
  2. How much work does the engine produce per cycle?
    W=η⋅QH=0.4⋅1000W = \eta \cdot Q_H = 0.4 \cdot 1000
    W=η QH=0.4×1000=400 JW = \eta\,Q_H = 0.4 \times 1000 = \mathbf{400\,J} per cycle. Only 40% of the absorbed heat is converted into useful work.
  3. How much heat is rejected to the cold source per cycle?
    QC=QH−W=1000−400Q_C = Q_H - W = 1000 - 400
    QC=QH−W=1000−400=600 JQ_C = Q_H - W = 1000 - 400 = \mathbf{600\,J}. Verification: QC/QH=600/1000=0.6=TC/TH=300/500Q_C/Q_H = 600/1000 = 0.6 = T_C/T_H = 300/500 ✓ — consistent with Carnot's relation QC/QH=TC/THQ_C/Q_H = T_C/T_H.
Result:η = 40%, W = 400 J, Q_C = 600 J. The Carnot cycle is the most efficient possible for given temperatures.
TheoryTermodinamicaFormula sheetPhysics 1

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