A heat engine is involved with exchange of heat of $1915 J, - 40 J, + 125 J and - Q J$ , during one cycle achieving an efficiency of $50.0$ %. The value of $Q$ is

980 J

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400 J

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640 J

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40 J

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Solution

The correct option is A $980 J$
$Efficiency, η = \frac{Work done}{Heat absorbed}$

$\frac{Q_{1} + Q_{2} + Q_{3} + Q_{4}}{Q_{1} + Q_{3}} = 0.5$

$Here, Q_{1} = 1915 J, Q_{2} = - 40 J and Q_{3} = 125 J$

$∴ \frac{1915 - 40 + 125 + Q_{4}}{1915 + 125} = 0.5$

$\Rightarrow 1915 - 40 + 125 + Q_{4} = 1020$
$\Rightarrow Q_{4} = 1020 - 2000$
$\Rightarrow Q_{4} = - Q = - 980 J \Rightarrow Q = 980 J$

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A heat engine is involved with exchange of heat of 1915 J,−40 J,+125 J and −Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is

The correct option is A 980 JEfficiency,η=Work doneHeat absorbed Q1+Q2+Q3+Q4Q1+Q3=0.5 Here,Q1=1915 J,Q2=−40 J and Q3=125 J ∴1915−40+125+Q41915+125=0.5 ⇒1915−40+125+Q4=1020 ⇒Q4=1020−2000 ⇒Q4=−Q=−980 J⇒Q=980 J

A heat engine is involved with exchange of heat of $1915 J, - 40 J, + 125 J and - Q J$ , during one cycle achieving an efficiency of $50.0$ %. The value of $Q$ is