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Consider a reversible heat engine operating between two reservoirs at temperatures $\mathrm{T_{1}}$ and $\mathrm{T_{2}\left(T_{1}>T_{2}\right)}$ .The engine absorbs $\mathrm{Q_{1}}$ heat from the reservoir at temperature $\mathrm{T_{1}}$ and rejects $Q_{2}$ heat to the reservoir at temperature $T_{2}$ . Show that the efficiency $(\eta)$ of the heat engine can be expressed as: $\eta=1-\frac{T_{2}}{T_{1}}$
Option: 1
$1 \mathrm{~T} 2 / \mathrm{T} 1$

Option: 2
$70-\mathrm{T} 2 / \mathrm{T} 1$

Option: 3
$50-\mathrm{T} 2 / \mathrm{T} 1$

Option: 4
$80-\mathrm{T} 2 / \mathrm{T} 1$

Answers (1)

best_answer

The efficiency $(\eta)$ of a heat engine is defined as the ratio of useful work output ( $\left.W_{\text {out }}\right)$ to the heat input $\left(Q_{1}\right):$

$\mathrm{\eta=\frac{W_{\text {out }}}{Q_{1}}}$
From the first law of thermodynamics, we know that for any cyclic process:
$\mathrm{\Delta U=Q_{\text {in }}-Q_{\text {out }}}$

For a reversible engine, $\mathrm{\Delta U=0}$ as the internal energy returns to the same value at the end of a cycle. Therefore:
$\mathrm{0=Q_{1}-Q_{2}}$

Solving for $\mathrm{Q_{2}}$ gives:
$\mathrm{Q_{2}=Q_{1}}$

Substituting this back into the efficiency equation:
$\mathrm{\eta=\frac{W_{\text {out }}}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}}$

Now, we know that for a reversible heat engine, the efficiency can also be expressed in terms of temperatures:
$\mathrm{\eta=1-\frac{T_{\text {cold }}}{T_{\text {hot }}}}$
$\mathrm{Substitute \, T_{1} for T_{\text {hot }} and\, T_{2} forT_{\text {cold }} :}$
$\mathrm{\eta=1-\frac{T_{2}}{T_{1}}}$
Thus, we have shown that the efficiency of the reversible heat engine is given by $\mathrm{\eta=1-\frac{T_{2}}{T_{1}}}$ .

Therefore, the correct option is (1).

Posted by

Divya Prakash Singh

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