KCET · Physics · Thermodynamics
The efficiency of Carnot's heat engine is \(0.5\) when the temperature of the source is \(\mathrm{T}_{1}\) and that of \(\operatorname{sink}\) is \(\mathrm{T}_{2}\). The efficiency of another Carnot's heat engine is also 0.5. The temperatures of source and sink of the second engine are respectively
- A \(2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}\)
- B \(2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}\)
- C \(\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5\)
- D \(\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}\)
Step-by-step Solution
Detailed explanation
Efficiency of Carnot's heat engine, \(\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\)
Efficiency remains same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor.
Efficiency remains same when both \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are increased by same factor.
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