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Questions 4-12 · Q6

Q.Efficiency of a Carnot cycle is 75%. If temperature of the hot reservoir is 727ºC, calculate the temperature of the cold reservoir.

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Step 1. Convert the hot reservoir temperature to kelvin: TH=727°C+273=1000T_H = 727°C + 273 = 1000 K.

Step 2. Carnot efficiency: η=1−TC/TH⇒TC=TH(1−η)=1000×(1−0.75)=1000×0.25=250\eta = 1 - T_C/T_H \Rightarrow T_C = T_H(1-\eta) = 1000 \times (1 - 0.75) = 1000 \times 0.25 = 250 K.

Step 3. Convert back to Celsius: TC=250−273=−23°CT_C = 250 - 273 = -23°C.

Step 4. The textbook's printed answer is "23°C", but working the Carnot formula through carefully gives −23°C (i.e. 250 K), not +23°C (which would be 296 K) — the printed answer appears to have dropped the negative sign. Solving it independently and disclosing this discrepancy, rather than silently reproducing the printed value, gives the physically correct result: −23°C.

✓Final answer

TC = −23°C (250 K) — note: the textbook's printed answer "23°C" appears to be missing its negative sign; the correct value, worked from η = 1 − TC/TH, is −23°C

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