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Question: A carnot engine operates with sources at \({{127}^{o}}C\) and sinks at \({{27}^{o}}C\). If the sourc...

A carnot engine operates with sources at 127oC{{127}^{o}}C and sinks at 27oC{{27}^{o}}C. If the source supplies 40kJ40kJ of heat energy, the work done by the engine is
(A). 30kJ30kJ
(B). 10kJ10kJ
(C). 4kJ4kJ
(D). 1kJ1kJ

Explanation

Solution

A carnot engine converts heat into works and operates between two reservoirs of which one is at higher temperature and the other is at lower temperature. The efficiency of a carnot engine is the difference of one and the ratio of temperatures of the sink and source. Efficiency is also the ratio of work done by the engine to the heat supplied by the engine. Equating both relations, we can calculate the work done.

Formulas used:
η=1−T2T1\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}
η=WH\eta =\dfrac{W}{H}

Complete step-by-step solution:
A carnot engine operates between a source and a sink and a working substance. The source supplies the heat which is converted to work and the residual heat is given to the sink. The source is a body at higher temperature while the sink is a body at lower temperature. The performance of a carnot engine is determined by its efficiency. The efficiency is given by-
η=1−T2T1\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}} - (1)
Here, η\eta is the efficiency of the engine
T2{{T}_{2}} is the temperature of the sink
T1{{T}_{1}} is the temperature of the source
The efficiency is also given as-
η=WH\eta =\dfrac{W}{H} - (2)
Here, WW is the total work done by the engine
HH is the heat supplied by the engine
Therefore, from eq (1) and eq (2), we solve the problem as,
1−T2T1=WH1-\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{W}{H} - (3)
Given, T1=127oC{{T}_{1}}={{127}^{o}}C, T2=27oC{{T}_{2}}={{27}^{o}}C, H=40kJH=40kJ
Substituting given values in eq (3), we get,
1−27+273127+273=W40 ⇒1−300400=W40 ⇒100400=W40 ∴W=10kJ \begin{aligned} & 1-\dfrac{27+273}{127+273}=\dfrac{W}{40} \\\ & \Rightarrow 1-\dfrac{300}{400}=\dfrac{W}{40} \\\ & \Rightarrow \dfrac{100}{400}=\dfrac{W}{40} \\\ & \therefore W=10kJ \\\ \end{aligned}
Therefore, the work done by the engine is 10kJ10kJ. Hence, the correct option is (B).

Note: The efficiency of a carnot engine has no unit as it is a ratio of similar units. The temperature is to be converted into kelvin before calculation. Carnot engines work on the principle that heat engines working between the same reservoirs sme less efficiency than a carnot engine.