Determining Engine Efficiency
The efficiency of a heat engine is determined by the temperatures at which it takes heat (source temperature, $T_H$) and rejects heat (sink temperature, $T_L$). For a given source temperature, the engine that rejects heat at the lowest temperature will have the highest efficiency. The maximum possible efficiency for a heat engine operating between two temperatures is given by the Carnot efficiency formula:
$\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$
where temperatures must be in absolute units (Kelvin).
Temperature Conversion
First, convert the given Celsius temperatures to Kelvin (K) by adding 273.15. $T_H = 130^{\circ}\text{C} = 130 + 273.15 = 403.15 \text{ K}$
| Engine | Rejection Temperature ($T_L$) in $^{\circ}\text{C}$ | Rejection Temperature ($T_L$) in K |
|---|---|---|
| A | $20^{\circ}\text{C}$ | $20 + 273.15 = 293.15 \text{ K}$ |
| B | $40^{\circ}\text{C}$ | $40 + 273.15 = 313.15 \text{ K}$ |
| C | $50^{\circ}\text{C}$ | $50 + 273.15 = 323.15 \text{ K}$ |
Efficiency Comparison
All three engines operate with the same source temperature ($T_H = 403.15 \text{ K}$). To find the most efficient engine, we need to identify the one with the lowest rejection temperature ($T_L$).
- Engine A rejects heat at $T_{LA} = 293.15 \text{ K}$.
- Engine B rejects heat at $T_{LB} = 313.15 \text{ K}$.
- Engine C rejects heat at $T_{LC} = 323.15 \text{ K}$.
Comparing the rejection temperatures: $293.15 \text{ K} < 313.15 \text{ K} < 323.15 \text{ K}$ So, $T_{LA} < T_{LB} < T_{LC}$.
Conclusion
Since Engine A has the lowest rejection temperature ($T_L$) while operating at the same source temperature ($T_H$) as Engines B and C, it will have the highest Carnot (maximum possible) efficiency.
Therefore, the most efficient engine is A.
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