Question:
The efficiency of a Carnot engine working between temperatures $127^\circ\text{C}$ and $27^\circ\text{C}$ is
The efficiency of a Carnot engine working between temperatures $127^\circ\text{C}$ and $27^\circ\text{C}$ is
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Remember that for a Carnot engine, the efficiency depends only on the temperatures of the hot and cold reservoirs. Always convert temperatures to Kelvin before applying the formula.
Updated On: Jun 3, 2026
- $25\\%$
- $75\\%$
- $50\\%$
- $20\\%$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_c}{T_h}$ where $T_h$ and $T_c$ are the absolute temperatures of the hot and cold reservoirs respectively. The temperatures must be in Kelvin.
Step 2: Meaning
Efficiency here refers to the ratio of work done by the engine to the heat absorbed from the hot reservoir, expressed as a percentage.
Step 3: Analysis
First, convert the given temperatures into Kelvin: $127^\circ\text{C} = 127 + 273.15 = 400.15 \, \text{K}$ $27^\circ\text{C} = 27 + 273.15 = 300.15 \, \text{K}$ Now apply the Carnot efficiency formula: \[\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300.15 \, \text{K}}{400.15 \, \text{K}}\] Simplify this expression: \[\eta = 1 - \frac{300.15}{400.15} = 1 - 0.7502\] Thus, \[\eta \approx 0.2498 \text{ or } 24.98\%\] Since the options are given in percentages, we round to the nearest whole number: \[\eta \approx 25\%\]
Step 4: Conclusion
The efficiency of the Carnot engine is approximately $25\%$.
Final Answer: (A)
The efficiency of a Carnot engine is given by the formula $\eta = 1 - \frac{T_c}{T_h}$ where $T_h$ and $T_c$ are the absolute temperatures of the hot and cold reservoirs respectively. The temperatures must be in Kelvin.
Step 2: Meaning
Efficiency here refers to the ratio of work done by the engine to the heat absorbed from the hot reservoir, expressed as a percentage.
Step 3: Analysis
First, convert the given temperatures into Kelvin: $127^\circ\text{C} = 127 + 273.15 = 400.15 \, \text{K}$ $27^\circ\text{C} = 27 + 273.15 = 300.15 \, \text{K}$ Now apply the Carnot efficiency formula: \[\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{300.15 \, \text{K}}{400.15 \, \text{K}}\] Simplify this expression: \[\eta = 1 - \frac{300.15}{400.15} = 1 - 0.7502\] Thus, \[\eta \approx 0.2498 \text{ or } 24.98\%\] Since the options are given in percentages, we round to the nearest whole number: \[\eta \approx 25\%\]
Step 4: Conclusion
The efficiency of the Carnot engine is approximately $25\%$.
Final Answer: (A)
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