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MHT CET · Physics · Thermodynamics

A Carnot engine with efficiency \(50 \%\) takes heat from a source at \(600 \mathrm{~K}\). To increase the efficiency to \(70 \%\), keeping the temperature of the sink same, the new temperature of the source will be

  1. A \(360 \mathrm{~K}\)
  2. B \(1000 \mathrm{~K}\)
  3. C \(900 \mathrm{~K}\)
  4. D \(300 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1000 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \mathrm{T}_{\mathrm{H}}=600 \mathrm{k} \\
& \eta=1-\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{T}_{\mathrm{H}}} \\
& \text { but, } \eta=\frac{1}{2} .... ( Given: \eta=50 \%)
\\
& \Rightarrow \frac{1}{2}=1-\frac{\mathrm{T}_{\mathrm{C}}}{600} \\
\therefore \quad & \mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}
\end{aligned}\)
(Given: \(\eta=50 \%\) )
With \(\mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}\), the efficiency is increased to \(70 \%\)
\(\therefore \quad\) New temperature of the source will be \(\mathrm{T}_{\mathrm{Hac}_{\mathrm{a}}}\)
\(\begin{array}{ll}
\therefore \quad \eta=1-\frac{300}{\mathrm{~T}_{\mathrm{H}_{\mathrm{ec}}}} \\
\quad \frac{300}{\mathrm{~T}_{\mathrm{H}_{\text {sew }}}}=1-\frac{7}{10} \quad \ldots .(\because \eta=70 \%) \\
\therefore \quad \mathrm{T}_{\mathrm{H}_{\text {ecw }}}=\frac{3000}{3}=1000 \mathrm{~K}
\end{array}\)
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