AP EAMCET · PHYSICS · Thermodynamics
The efficiency of a Carnot cycle is \(\frac{1}{6}\). By lowering the temperature of sink by 65 K , it increases to \(\frac{1}{3}\). The initial and final temperature of the sink are
- A \(400 \mathrm{~K}, 310 \mathrm{~K}\)
- B \(525 \mathrm{~K}, 65 \mathrm{~K}\)
- C \(309 \mathrm{~K}, 235 \mathrm{~K}\)
- D \(325 \mathrm{~K}, 260 \mathrm{~K}\)
Answer & Solution
Correct Answer
(D) \(325 \mathrm{~K}, 260 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
The efficiency of carnot engine is \(\eta=1-\frac{T_2}{T_1} \Rightarrow \frac{1}{6}=1-\frac{T_2}{T_1} \Rightarrow T_2=\frac{5}{6} T_1\) ....(i) Again, \(\frac{1}{3}=1-\frac{T_2-65}{T_1} \Rightarrow \frac{T_2-65}{T_1}=\frac{2}{3}\) ....(ii) From eq(i) and (ii), we get…
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