Question
A heat engine operating between $200^{\circ} \mathrm{C}$ and $80.0^{\circ} \mathrm{C}$ achieves20.0$\%$ of the maximum possible efficiency. What energyinput will enable the engine to perform 10.0 $\mathrm{kJ}$ of work?
Step 1
We do this by adding 273.15 to each temperature. This gives us $T_C = 80.0^{\circ}C + 273.15 = 353.15K$ and $T_H = 200.0^{\circ}C + 273.15 = 473.15K$. Show more…
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