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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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