A heat engine is involved with exchange of heat of 1915 J, – 40J, + 125 J and –Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is
Solution
$\eta$ = <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;"> Work done</div>
<div style="text-align: center;"> Heat supplied </div>
</span>
<br><br>$\Rightarrow$ $\frac{1}{2}$ = $\frac{1915-40+125-Q}{1915+125}$
<br><br>$\Rightarrow$ $\frac{1}{2}$ = $\frac{2000-Q}{2040}$
<br><br>$\Rightarrow$ 2040 = 4000 – 2Q
<br><br>$\Rightarrow$ 2Q = 1960
<br><br>$\Rightarrow$ Q = 980 J
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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