For an ideal heat engine, the temperature of the source is 127$^\circ$C. In order to have 60% efficiency the temperature of the sink should be ___________$^\circ$C. (Round off to the Nearest Integer)
Answer (integer)
-113
Solution
<p>Given, temperature of source, T<sub>H</sub> = 127$^\circ$C</p>
<p>= 273 + 127 = 400 K</p>
<p>Efficiencies, $\eta$ = 60% = 0.6</p>
<p>The efficiency of Carnot (ideal) heat engine is given by</p>
<p>$\eta = \left( {1 - {{{T_L}} \over {{T_H}}}} \right)$</p>
<p>where, T<sub>L</sub> = temperature of sink,</p>
<p>$$0.6 = \left( {1 - {{{T_L}} \over {{T_H}}}} \right) \Rightarrow {{{T_L}} \over {{T_H}}} = 1 - 0.6$$</p>
<p>${{{T_L}} \over {{T_H}}} = 0.4 \Rightarrow {T_L} = 0.4 \times {T_H}$</p>
<p>$= 0.4 \times 400 = 160K$</p>
<p>$= 160 - 273 = - 113^\circ C$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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