MHT CET · Physics · Thermodynamics
Heat engine operating between temperature \(T_1\) and \(T_2\) has efficiency \(\frac{1}{6}\). When \(T_2\) is lowered by 62 K , its efficiency increases to \(\frac{1}{3}\). Then \(\mathrm{T}_1\) and \(\mathrm{T}_2\) respectively are
- A \(372 \mathrm{~K}, 310 \mathrm{~K}\)
- B \(372 \mathrm{~K}, 330 \mathrm{~K}\)
- C \(330 \mathrm{~K}, 268 \mathrm{~K}\)
- D \(310 \mathrm{~K}, 248 \mathrm{~K}\)
Answer & Solution
Correct Answer
(A) \(372 \mathrm{~K}, 310 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
Given,
\(\eta_1=\frac{1}{6}=1-\frac{T_2}{T_1}=\Rightarrow T_2=\frac{5}{6} T_1\)
Also,
\(\eta_2=\frac{1}{3}=1-\frac{\left(T_2-62\right)}{T_1}\)
Substituting (i),
\(\begin{array}{ll}
& \frac{\left(\frac{5}{6} T_1-62\right)}{T_1}=1-\frac{1}{3} \\
& \frac{5}{6} T_1-62=T_1 \frac{2}{3} \\
\therefore \quad & T_1=372 \mathrm{~K} \\
\therefore \quad & T_2=\frac{5}{6} \times 372=310 \mathrm{~K}
\end{array}\)
\(\eta_1=\frac{1}{6}=1-\frac{T_2}{T_1}=\Rightarrow T_2=\frac{5}{6} T_1\)
Also,
\(\eta_2=\frac{1}{3}=1-\frac{\left(T_2-62\right)}{T_1}\)
Substituting (i),
\(\begin{array}{ll}
& \frac{\left(\frac{5}{6} T_1-62\right)}{T_1}=1-\frac{1}{3} \\
& \frac{5}{6} T_1-62=T_1 \frac{2}{3} \\
\therefore \quad & T_1=372 \mathrm{~K} \\
\therefore \quad & T_2=\frac{5}{6} \times 372=310 \mathrm{~K}
\end{array}\)
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