Question

An ideal heat engine operating at 15 % efficiency exhausts heat into a cold reservoir at 280 K. By how much would the temperature of the hot reservoir need to be raised (in K) in order to increase the efficiency of the heat engine to 35.6 %?

Answer 1

Answer on Question #70455, Physics / Molecular Physics | Thermodynamics

An ideal heat engine operating at $15\%$ efficiency exhausts heat into a cold reservoir at $280\mathrm{K}$. By how much would the temperature of the hot reservoir need to be raised (in K) in order to increase the efficiency of the heat engine to $35.6\%$?

Solution:

Let $T_h$ be the temperature of the "hot" reservoir, and $T_c$ the temperature of the "cold" reservoir

$$\varepsilon = \frac {T _ {h} - T _ {c}}{T _ {h}}$$ $$\varepsilon \times T _ {h} = T _ {h} - T _ {c}$$ $$\varepsilon \times T _ {h} - T _ {h} = - T _ {c}$$ $$T _ {h} (\varepsilon - 1) = - T _ {c}$$ $$T _ {h} = - \frac {T _ {c}}{(\varepsilon - 1)}$$ $$T _ {h} = \frac {T _ {c}}{(1 - \varepsilon)}$$

For $15\%$ efficiency:

$$T _ {h} = \frac {2 8 0 K}{(1 - 0 . 1 5)} = 3 2 9 K$$

For $35.6\%$ efficiency:

$$T _ {h} = \frac {2 8 0 K}{(1 - 0 . 3 5 6)} = 4 3 5 K$$

Finally,

$$\Delta T _ {h} = 4 3 5 K - 3 2 9 K = 1 0 6 K$$

Answer: 106 K in order to increase the efficiency to $35.6\%$

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