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Standard XIII

Physics

Heat Engines

A heat engine...

Question

A heat engine operates between a cold reservoir at temperature $T_{2} = 400 K$ and a hot reservoir at temperature $T_{1} .$ It takes $300 J$ of heat from the hot reservoir and delivers $240 J$ of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be $K$

500.0

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Solution

Efficiency of heat engine is given by
$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$
$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

Here
$Q_{1} = 300 J$
$Q_{2} = 240 J$
$T_{2} = 400 K$

on substitution we get
$\frac{300}{T_{1}} = \frac{240}{400}$
$T_{1} = 500 K$

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A heat engine operates between a cold reservoir at temperature T2=400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be K

Efficiency of heat engine is given by η=1−T2T1=1−Q2Q1 Q1T1=Q2T2 Here Q1=300 J Q2=240 J T2=400 K on substitution we get 300T1=240400 T1=500 K

A heat engine operates between a cold reservoir at temperature $T_{2} = 400 K$ and a hot reservoir at temperature $T_{1} .$ It takes $300 J$ of heat from the hot reservoir and delivers $240 J$ of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be $K$

Efficiency of heat engine is given by
$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$
$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

Here
$Q_{1} = 300 J$
$Q_{2} = 240 J$
$T_{2} = 400 K$

on substitution we get
$\frac{300}{T_{1}} = \frac{240}{400}$
$T_{1} = 500 K$