Question

The efficiency of a Carnot engine which operates between the two temperatures $T_1$ = 500 K and $T_2$ = 300 K is
A). 50 %
B). 25 %
C). 75 %
D). 40 %

Answer 1

Hint: The concept of efficiency of the Carnot engine is to be used. Apply the formula of the efficiency of the Carnot engine in terms of the temperature by substituting the given values of the temperature in Kelvin. Converting the temperature scale is necessary and changing the fraction into a percentage by multiplying the fraction with 100.

Complete step by step solution :
The Carnot engine efficiency can be written using different formulae and one of the expressions which we can use here is the efficiency in terms of the temperature as below :
We know the efficiency of the heat engine is given by
$\eta = 1 -\dfrac{T_2}{T_1}$
Here $T_1$ is the temperature of the source and $T_2$ is that temperature of the sink and the efficiency is denoted by $\eta$
Now we are given the values of the temperature as :
The temperature of the source is 500K where K denotes Kelvin
So we get $T_1$ = 500 K
The temperature of the sink is 300K where K denotes Kelvin
So we get $T_2$ = 300 K
Here the temperatures are already given in kelvin and hence we need not convert them into kelvin.
So now the temperatures are known to us so we can now calculate the efficiency as
$\eta = 1 - \dfrac{T_2}{T_1} = 1 -\dfrac{300}{500}$
We get : $\eta = \dfrac{500 - 300}{500} = \dfrac{2}{5}$
Now we have got the fraction of the efficiency we can get the value in the percentage as below :
Efficiency in percentage is given as :
$\eta = \dfrac{2}{5} \times 100 = 40 %$
Hence we have used the concept of the efficiency in terms of the temperatures and finally found the value of the efficiency as 40 %

Note: The possible mistakes that one can make in this kind of problem is the confusion between the temperatures $T_1$ and $T_2$. We need to take care of the definitions of temperatures $T_1$ and $T_2$. Note that $T_2$ refers to the temperature of the sink $T_1$ refers to the temperature of the source.