Question

What would be the efficiency of a Carnot engine operating with boiling water as one reservoir and a freezing mixture of ice and water as the other reservoir?
A. $27\%$
B. $77\%$
C. $20\%$
D. $67\%$

Answer 1

In the given question, it is stated that the hot reservoir of the Carnot engine consists of boiling water and the cold reservoir of the Carnot engine is of a mixture of ice and water. Hence, we have to use the following details and use the equation of efficiency of a Carnot engine to find the respective solution.

Complete step by step answer:
The Carnot engine is a model proposed by Leonard Carnot which proposes the relation between the work done by an engine in the form of heat processes. He concluded the formula of Carnot engine efficiency as $\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ where ${T_2}$ is the temperature from the cold reservoir and ${T_1}$ is the temperature from the hot reservoir.

In the given question it is said that the hot reservoir consists of boiling water.
So, the temperature of boiling water is ${100^ \circ }{\text{ }}C$.
Converting it into Kelvin scale we get, ${T_1} = 100 + 273 = 373{\text{ }}K$
And the cold reservoir contains a freezing mixture of water and ice.
Thus, the freezing point of water is ${0^ \circ }{\text{ }}C$.
In the Kelvin scale we get, ${T_2} = 0 + 273 = 273{\text{ }}K$.
Thus, the efficiency of the Carnot engine is,
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}} \\\ \Rightarrow \eta = 1 - \dfrac{{273}}{{373}} \\\ \Rightarrow \eta \approx 0.27$
The efficiency of the Carnot Engine in percentage is $0.27 \times 100\% = 27\%$

Hence, the correct option is A.

Note: It must be noted that the efficiency of the Carnot cycle is considered to be $100\%$. But practically no heat engine can have an efficiency of $100\%$. Consider every calculation of temperature in Kelvin scale, if we consider the Celsius scale in this given question we will not have an answer. It is mandatory for the calculation of temperature to be in Kelvin Scale.