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Question: A heat engine is operating between \(500K\) to \(400K\). If the engine absorbs \(100J\) heat, then w...

A heat engine is operating between 500K500K to 400K400K. If the engine absorbs 100J100J heat, then which of the following is an impossible amount of heat rejected by the engine?
(A) 80J80J
(B) 85J85J
(C) 90J90J
(D) 10J10J

Explanation

Solution

We know that heat engine is an engine which converts the thermal energy into mechanical energy in thermodynamics. It is operated between a sink temperature and a source temperature. The source is at a higher temperature and the sink is at lower temperature.

Complete step-by-step answer:
As we know that heat engine converts the heat energy to mechanical energy. An ideal reversible heat engine is known as the Carnot engine. It is a theoretical engine. Efficiency is the ratio of useful work to that heat provided. An engine with more efficiency is preferable. Efficiency of a heat engine can be given as η=1T2T1=1Q2Q1=workdoneHeatgiven\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{work\, done}}{{Heatgiven}}. Here T2={T_2} = sink temperature, T1={T_1} = source temperature, Q2={Q_2} = heat rejected ,Q1={Q_1} = heat absorbed by the engine. In the question the value of T1=500K,T2=400K,Q1=100J{T_1} = 500K,{T_2} = 400K,{Q_1} = 100J is given .
Now efficiency can be written as
η=1T2T1=1Q2Q1 1400500=1Q2100 Q2=4005=80J  \Rightarrow \eta = 1 - \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} \\\ \Rightarrow 1 - \dfrac{{400}}{{500}} = 1 - \dfrac{{{Q_2}}}{{100}} \\\ \Rightarrow {Q_2} = \dfrac{{400}}{5} = 80J \\\

So from the above explanation and calculation it is clear to us that the correct answer of the given question is option: (A) 80J80J.

Additional information: It is nearly impossible to design and create a heat engine that will have 100%100\% efficiency. Some examples of heat engines are thermal power stations, internal combustion engines and steam locomotives.

Note: Always remember that the efficiency of a heat engine can be given as η=1T2T1=1Q2Q1=workdoneHeatgiven\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{work\, done}}{{Heatgiven}} , Here T2={T_2} = sink temperature, T1={T_1} = source temperature, Q2={Q_2} = heat rejected ,Q1={Q_1} = heat absorbed by the engine. Always avoid silly mistakes and calculation errors while solving the numerical.