Question

A Carnot’s engine works between two temperatures whose difference is $100K$ . If it absorbs $746J$ of heat from the source and gives $546J$ to the sink, calculate the temperatures of the source and the sink.

Answer 1

In the given question, we have been asked to find the temperatures of the source and the sink of a Carnot’s engine and we have been told the temperature difference between them. At first glance, it might seem that the data given in the question is insufficient for the calculation of the required quantities. But we have also been given the heat absorbed from the source and the heat given to the sink. We can use the expression for the efficiency of the engine to find our answer.
Formula Used: $\eta =\dfrac{W}{Q}$ , $\eta =1-\dfrac{{{T}_{\sin k}}}{{{T}_{source}}}$

Step by Step Solution
Let the heat absorbed from the source or the work input to the engine be Q joules.
From the data given in the question, we can say that $(Q)=746J$
From the given question, heat given to the sink $({{Q}_{2}})=546J$
Now, the work obtained from the engine can be given as the heat absorbed by the engine minus the heat given to the sink, that is $W=Q-{{Q}_{2}}$
Substituting the values, the work obtained $(W)=746-546=300J$
Now, the efficiency of the engine can be given as $\eta =\dfrac{W}{Q}$ where the work done has been calculated and the heat input has been given to us; substituting the values, we get
$\eta =\dfrac{300}{746}=0.402$
Now, we have also been told that the temperature difference between the sink and the source is $100K$
So, if we assume the sink temperature to be $TK$ , the source temperature can be given as $(T+100)K$
We can now calculate the efficiency of the engine as $\eta =1-\dfrac{{{T}_{\sin k}}}{{{T}_{source}}}$ where ${{T}_{\sin k}}$ and ${{T}_{source}}$ represent the temperatures of the sink and the source respectively
Substituting the values, we get

& \eta =1-\dfrac{T}{T+100} \\\ & \Rightarrow \eta =\dfrac{100}{T+100} \\\ \end{aligned}$$ We have already found the value of the efficiency of the engine; substituting the value and simplifying further, we get $$\begin{aligned} & 0.402=\dfrac{100}{T+100} \\\ & \Rightarrow T+100=\dfrac{100}{0.402} \\\ & \Rightarrow T+100=248.76K \\\ \end{aligned}$$ **Hence the source temperature is $$248.76K$$** **The sink can now be said to be $$(T)=(248.76-100)K=148.76K$$** **Note** Note that the sink temperature is always lower than the source temperature to allow the flow of heat from the source to the sink; as heat always flows from a body at a higher temperature to a body at a lower temperature. Some students often consider the heat given to the sink as the work done; you should be aware not to make this mistake.