Physics · The second law of thermodynamics: reversible and irreversible processes
A reversible engine takes heat from a reservoir at and gives out heat to at C. H
A reversible engine takes heat from a reservoir at \( 527^{\circ} \mathrm{C} \) and gives out heat to \( a \sin k \) at \( 127^{\circ} \) C. How many Joules/s must it take from the reservoir to perform useful mechanical work at the rate of \( 750 W ? \)
- A. 257.14
- B. 1500
- C. 457.14
- D. 557.14
Step-by-step solution
Convert temperatures to Kelvin: T1 = 527°C + 273 = 800 K, T2 = 127°C + 273 = 400 K. Efficiency of reversible engine η = 1 - T2/T1 = 0.5. Work output W = 750 W. Heat input Q1 = W/η = 750/0.5 = 1500 J/s.
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