Question:
If an ideal heat engine with an efficiency of $40\%$ rejects heat at $27^\circ$C, then it should have absorbed heat at
If an ideal heat engine with an efficiency of $40\%$ rejects heat at $27^\circ$C, then it should have absorbed heat at
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For Carnot engines:
- Always use Kelvin scale
- $\eta = 1 - \frac{T_c}{T_h}$
Updated On: Apr 30, 2026
- $377^\circ$C
- $500^\circ$C
- $227^\circ$C
- $427^\circ$C
- $460^\circ$C
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
For an ideal (Carnot) engine:
\[
\eta = 1 - \frac{T_c}{T_h}
\]
where temperatures are in Kelvin.
Step 1: Convert temperature to Kelvin.
\[ T_c = 27^\circ\text{C} = 300\ \text{K} \]
Step 2: Substitute efficiency.
\[ 0.4 = 1 - \frac{300}{T_h} \]
Step 3: Solve for $T_h$.
\[ \frac{300}{T_h} = 0.6 \Rightarrow T_h = \frac{300}{0.6} = 500\ \text{K} \]
Step 4: Convert back to Celsius.
\[ T_h = 500 - 273 = 227^\circ\text{C} \]
Step 1: Convert temperature to Kelvin.
\[ T_c = 27^\circ\text{C} = 300\ \text{K} \]
Step 2: Substitute efficiency.
\[ 0.4 = 1 - \frac{300}{T_h} \]
Step 3: Solve for $T_h$.
\[ \frac{300}{T_h} = 0.6 \Rightarrow T_h = \frac{300}{0.6} = 500\ \text{K} \]
Step 4: Convert back to Celsius.
\[ T_h = 500 - 273 = 227^\circ\text{C} \]
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