Question

A Carnot engine has an efficiency of 50% when its sink is at a temperature of 27℃. The temperature of the source is
(A) 300℃
(B) 327℃
(C) 373℃
(D) 273℃

Answer 1

Hint
The efficiency of the Carnot engine is given by $\left( {1 - \dfrac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}} \right)$ i.e. the Carnot engine operates between two temperatures. ${{\rm{T}}_2}$ is the temperature of the sink and ${{\rm{T}}_1}$ is the temperature of the source.

Complete step by step solution
Given, the efficiency of the Carnot engine is, ${\rm{\eta }} = 50{\rm{\% }} = \dfrac{{50}}{{100}} = 0.5$
The temperature of the sink of the Carnot engine,
The temperature in Kelvin is given by, ${{\rm{T}}_2} = 27 + 273 = 300$ K
The efficiency of the cycle of Carnot engine is given by,
${\rm{\eta }} = 1 - \dfrac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}$
Where ${{\rm{T}}_1}$ is the temperature of the source and ${{\rm{T}}_2}$ is the temperature of the sink.
On putting the values of each term, we get
$0.5 = 1 - \dfrac{{300}}{{{{\rm{T}}_1}}}$
$\dfrac{{300}}{{{{\rm{T}}_1}}} = 0.5$
${{\rm{T}}_1} = \dfrac{{300}}{{0.5}}$
${{\rm{T}}_1} = 600$ K
${{\rm{T}}_1} = {600}^{0}C-{273}^{0}C$
${{\rm{T}}_1} = {327}^{0}C$
The temperature of the source in will be given by,
Therefore, (B) is the required solution.

Additional information
According to Carnot theorem; The Carnot engine work between two temperatures, where the temperature of the hot reservoir is ${{\rm{T}}_1}$ and the temperature of the cold reservoir is ${{\rm{T}}_2}$. And, the efficiency of any engine cannot be more than the Carnot engine. The efficiency of the Carnot engine does not depend on the nature of the working substance. Also, it is not possible to completely convert heat energy into mechanical work. Therefore, the efficiency of an ideal Carnot engine cannot be 100${\rm{\% }}$.

Note
The conversion of temperature should be done in the ‘Kelvin’ scale. And, for the required answer the temperature must be converted in the ‘Celsius’ scale.