JEE Mains · Physics · STD 11 - 11. thermodynamics
A Carnot engine with efficiency \(50\,\%\) takes heat from a source at \(600\,K\). In order to increase the efficiency to \(70\,\%\), keeping the temperature of sink same, the new temperature of the source will be \(.........\,K\)
- A \(360\)
- B \(1000\)
- C \(900\)
- D \(300\)
Answer & Solution
Correct Answer
(B) \(1000\)
Step-by-step Solution
Detailed explanation
\(\text { Initially } \eta=\frac{1}{2}\) \(\eta=1-\frac{ T _2}{ T _1}\) \(\therefore \frac{1}{2}=1-\frac{ T _2}{600} \quad \Rightarrow T _2=300\,K\) \(\Rightarrow \frac{ T _2}{600}=\frac{1}{2} \quad\) Now efficiency is increased to \(70 \%\) and \(T _2=300\) \(K\), Let temp of…
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