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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

A heat engine operates between a cold reservoir at temperature $\;T_{1}=300 K\;$ & a hot reservoir at temperature $\;T_{2}=600 K\;$ . It takes 500 J heat from hot reservoir and delivers x J of heat to cold reservoir in a cycle . Then x = ?

$(a)\;100\qquad(b)\;150\qquad(c)\;200\qquad(d)\;250$

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1 Answer

Answer : (b) $\;150$
Explanation :
Work done by engine = (500-x) J
Efficiency $\;\eta=\large\frac{(500-x)}{500}=1-\large\frac{\pi}{T_{2}}$
$\large\frac{500-x}{500}=1-\large\frac{300}{600}$
$x=150\;J$
answered Mar 12, 2014 by yamini.v
 

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