Question

A Carnot’s engine has efficiency 25%. It operates between reservoirs of constant temperatures with temperature difference of ${{80}^{0}}C$. What is the temperature of the low-temperature reservoir?
$\begin{aligned} & (A)-{{25}^{0}}C \\\ & (B){{25}^{0}}C \\\ & (C)-{{33}^{0}}C \\\ & (D){{33}^{0}}C \\\ \end{aligned}$

Answer 1

We will use the formula for efficiency of a Carnot cycle, which is given by the difference in temperatures of the reservoirs at high and low temperatures respectively upon the temperature of the hotter reservoir. Since this is the fractional efficiency, we will convert the percentage efficiency into fractional efficiency and then proceed.

Complete answer:
The percentage efficiency (say $\eta$ ) of the Carnot cycle is given by:
$\Rightarrow \eta =25$
Converting it into fractional efficiency, we get:
$\begin{aligned} & \Rightarrow \eta =\dfrac{25}{100} \\\ & \Rightarrow \eta =0.25 \\\ \end{aligned}$
Now,
Let us assume the temperature of the higher temperature reservoir is given by ${{T}_{H}}$ .
And, the temperature of the lower temperature reservoir is denoted by ${{T}_{C}}$ .
Then, the difference in the temperature of the two reservoirs is given to be:
$\Rightarrow {{T}_{H}}-{{T}_{C}}={{80}^{0}}C$
This could also be written as:
$\Rightarrow {{T}_{H}}-{{T}_{C}}=80K$
Now, the formula for the efficiency of a Carnot is given by:
$\Rightarrow \eta =\dfrac{{{T}_{H}}-{{T}_{C}}}{{{T}_{H}}}$
Putting the values of all the known terms in the above equation, we get:
$\Rightarrow 0.25=\dfrac{80}{{{T}_{H}}}$
$\begin{aligned} & \Rightarrow {{T}_{H}}=\dfrac{80}{0.25} \\\ & \Rightarrow {{T}_{H}}=320K \\\ \end{aligned}$
Thus, the temperature of the high-temperature reservoir comes out to be $320K$ .
Now, the temperature of the low temperature reservoir can be calculated using:
$\begin{aligned} & \Rightarrow {{T}_{H}}-{{T}_{C}}=80 \\\ & \Rightarrow 320-80={{T}_{C}} \\\ & \Rightarrow {{T}_{C}}=240K \\\ \end{aligned}$
Therefore,
$\begin{aligned} & \Rightarrow {{T}_{C}}={{(240-273)}^{0}}C \\\ & \therefore {{T}_{C}}=-{{33}^{0}}C \\\ \end{aligned}$
Hence, the temperature of the low-temperature reservoir comes out to be $-{{33}^{0}}C$ .

Hence, option (C) is the correct option.

Note:
The efficiency of a reservoir can only ideally be equal to one, that is, hundred percent efficient. If we look at the formula for the efficiency of a Carnot engine, we will realize that for efficiency to be equal to one, the temperature of the low temperature reservoir should be equal to 0K, which is practically impossible at normal conditions. In real life cases, there is always some heat loss which tends to reduce the output of a Carnot engine.