KCET · Physics · Thermodynamics
For which combination of working temperatures, the efficiency of 'Carnot's engine' is the least \( ? \)
- A \( 60 \mathrm{~K}, 40 \mathrm{~K} \)
- B \( 40 \mathrm{~K}, 20 \mathrm{~K} \)
- C \( 80 \mathrm{~K}, 60 \mathrm{~K} \)
- D \( 100 \mathrm{~K}, 80 \mathrm{~K} \)
Answer & Solution
Correct Answer
(D) \( 100 \mathrm{~K}, 80 \mathrm{~K} \)
Step-by-step Solution
Detailed explanation
Efficiency of Carnot's engine is given as
\( \eta=1-\frac{T_{2}}{T_{1}} \)
For \( T_{1}=60 \mathrm{~K} \) and \( T_{2}=40 \mathrm{~K}, \eta=1-\frac{40}{60}=0.333 \)
For \( T_{1}=40 \mathrm{~K} \) and \( T_{2}=20 \mathrm{~K}, \eta=1-\frac{20}{40}=0.5 \)
For \( T_{1}=80 K \) and \( T_{2}=60 K, \eta=1-\frac{60}{80}=0.25 \)
For \( T_{1}=100 \mathrm{~K} \) and \( T_{2}=80 \mathrm{~K}, \eta=1-\frac{80}{100}=0.2 \)
Therefore, the efficiency of Carnot's engine is the least for \( T_{1}=100 \mathrm{~K} \) and \( T_{2}=80 \mathrm{~K} \)
\( \eta=1-\frac{T_{2}}{T_{1}} \)
For \( T_{1}=60 \mathrm{~K} \) and \( T_{2}=40 \mathrm{~K}, \eta=1-\frac{40}{60}=0.333 \)
For \( T_{1}=40 \mathrm{~K} \) and \( T_{2}=20 \mathrm{~K}, \eta=1-\frac{20}{40}=0.5 \)
For \( T_{1}=80 K \) and \( T_{2}=60 K, \eta=1-\frac{60}{80}=0.25 \)
For \( T_{1}=100 \mathrm{~K} \) and \( T_{2}=80 \mathrm{~K}, \eta=1-\frac{80}{100}=0.2 \)
Therefore, the efficiency of Carnot's engine is the least for \( T_{1}=100 \mathrm{~K} \) and \( T_{2}=80 \mathrm{~K} \)
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