Question
An engine is supposed to operate between two reservoirs at temperature $727^{\circ} \mathrm{C}$ and $227^{\circ} \mathrm{C}$. The maximum possible efficiency of such an engine is(A) $(3 / 4)$(B) $(1 / 4)$(C) $(1 / 2)$(D) 1
Step 1
The formula to convert Celsius to Kelvin is K = C + 273.15. Therefore, the temperatures in Kelvin are: \[ T_1 = 727^{\circ}C + 273.15 = 1000K \] \[ T_2 = 227^{\circ}C + 273.15 = 500K \] Show more…
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