Question

A Carnot’s engine has an efficiency of $50\%$ at sink temperature $50°C$. Calculate the temperature of source:
A. $133℃$
B. $143℃$
C. $100℃$
D. $373℃$

Answer 1

Hint: Write the formula for efficiency in terms of work done and heat absorbed. Then, find the relation between temperature and heat. And substitute that relation in the formula for efficiency. Substitute the values and then calculate temperature of source.

Complete answer:
Let ${T}_{1} $ be the temperature of source
      ${T}_{2}$ be the temperature of sink
Given: Efficiency ($\eta$)=$50\%$
Temperature of sink ${T}_{2} = $50°C$= 50+273= 323K$

Equation for efficiency is given by,
$\eta =\dfrac {Net \quad work \quad done \quad per \quad cycle}{Total \quad amount \quad of \quad heat \quad absorbed \quad per \quad cycle}$
$\Rightarrow \eta = \dfrac {W}{{Q}_{1}}$
$\Rightarrow \eta = \dfrac {{Q}_{1}-{Q}_{2}}{{Q}_{1}}$
$\Rightarrow \eta =1 - \dfrac {{Q}_{2}}{{Q}_{1}}$ ...(1)
But, we know heat is directly proportional to temperature.
$\Rightarrow \dfrac {{Q}_{2}}{{Q}_{1}}= \dfrac {{T}_{2}}{{T}_{1}}$ ...(2)
Substituting equation. (2) in equation. (1) we get,
$ \eta =1 - \dfrac {{T}_{2}}{{T}_{1}}$ ...(3)
Now, substituting values in above equation we get,
$\dfrac {50}{100}=1 - \dfrac {323}{{T}_{1}}$ ...(1)
$\dfrac {1}{2}=1 - \dfrac {323}{{T}_{1}}$ ...(1)
$\Rightarrow{T}_{1}=2 \times 323$
$\Rightarrow{T}_{1}= 646K$
$\Rightarrow{T}_{1}= 646-273$
$\Rightarrow{T}_{1}= 373℃$
Therefore, the temperature of the source is $373℃$.

So, the correct answer is “Option D”.

Note:
Make sure you convert the units of temperature from Celsius to Kelvin. And as the efficiency is in percentage, divide the efficiency by 100. And then substitute that value for efficiency. Amount of heat transferred is proportional to change in the temperature. If we have to double the change in temperature, then we have to supply double heat.
Carnot's engine efficiency can be increased if we decrease the temperature of the sink.