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

A heat engine operates between a cold reservoir at temperature $\;T_{2}=600 K\;$ & a hot reservoir at temperature $\;T_{1}\;.$ It takes 1000 J of energy from hot reservoir & gives 200 J . What can be the minimum temperature of the hot reservoir

$(a)\;3000\;K\qquad(b)\;1500\;K\qquad(c)\;900\;K\qquad(d)\;1200\;K$

1 Answer

Answer : (a) $\;3000\;K$
Explanation :
$W=1000 J-200 J$
$=800 J$
$Efficiency= \large\frac{800J}{1000J}=0.8$
Since no engine has efficiency more that cannot engine $\;9.8 \leq 1- \large\frac{T_{2}}{T_{1}}$
$T_{1} \geq - \large\frac{T_{2}}{T_{1}}$
$T_{1} \geq 3000 K$
answered Mar 12, 2014 by yamini.v
 

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