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# An ideal heat engine exhausting heat at $77{}^\circ C$ is to have a 30% efficiency. It must take heat atA) $127{}^\circ C$ B) $227{}^\circ C$ C) $327{}^\circ C$ D) $673{}^\circ C$

Last updated date: 08th Sep 2024
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Hint: Efficiency is the ratio of work done and heat taken to do that work. We have a direct equation for efficiency in terms of temperature of source and sink. Here temperature of sink and efficiency is given. We just need to substitute the values in the equation to found out the temperature of source.

Formula used:
Efficiency of heat engines,
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Where temperature used is in kelvin scale.

Heat engines convert heat to mechanical energy which is used to do mechanical work. There is a heat reservoir from which heat is taken and some work is done and remaining heat is transferred to a cold reservoir. Carnot engine is an ideal heat engine. Refrigerators and heat pumps are heat engines that work in reverse order.

Efficiency of a heat engine is the ratio of work done to that of heat taken to do that work. Usually, heat engines have 30% to 50% efficiency. It is impossible for a heat engine to achieve 100% efficiency.

Given, efficiency of heat engine,
$\eta =\frac{30}{100}$
and
${{T}_{2}}=77{}^\circ C=77+273=350K$

We have to find out what is $T_1$. On substituting the values, We get,
$\frac{30}{100}=1-\frac{350}{{{T}_{1}}}$
On further solving we get temperature as ${{T}_{1}}=500K=227{}^\circ C$

Therefore, the answer is option (B)

Note: Like all other questions here, sign conversion is important and also be careful that efficiency is given in percentage. This is a direct question but still while using the temperature connecting equation for efficiency remember that the temperature of sink is always less than that of source.