JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2023
- Medium
- Numerical
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 \(.........\,K\)
Pick an option to check your answer.
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 source \(T _1= T\)
\(\Rightarrow \frac{7}{10}=1-\frac{300}{T}\)
\(\Rightarrow \frac{300}{ T }=1-\frac{7}{10}\)
\(\Rightarrow \frac{300}{ T }=\frac{3}{10} \quad \therefore T =1000\,K\)
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