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# In Carnot engine efficiency is $40\%$, when the temperature of hot reservoirs is $T$. For efficiency $50\%$ what will be the temperature of the hot reservoir?${\text{A}}.{\text{ }}\dfrac{T}{5}$${\text{B}}{\text{. }}\dfrac{{2T}}{5}$${\text{C}}{\text{. 6T}}$${\text{D}}{\text{. }}\dfrac{{6T}}{5}$

Last updated date: 21st Jun 2024
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Hint: The efficiency of $\eta$is equal to the work done $\left( W \right)$ divided by the heat input $\left( Q \right)$ .First we have to use the carnot engine efficiency value to find the values by using the formula. Also we have to use the given data to find the required solution. We can find out the value is the temperature of hot reservoir $\left( T \right)$

Formula Used:
The efficiency,$\eta = \dfrac{{{\text{work done}}}}{{{\text{heat input}}}} = \dfrac{W}{Q}$
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$

The efficiency of $\eta$ is equal to Work done is divided by heat in put,
Work done is $W$, and heat in put is$Q$ ,
The efficiency,$\eta = \dfrac{{{\text{work done}}}}{{{\text{heat input}}}} = \dfrac{W}{Q}$
$\implies \eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Where,
${T_2}$= Temperature of sink, and
$\implies {T_1}$= temperature of hot reservoir
Substitute the values, so we can write it as,
$\implies \dfrac{{40}}{{100}} = 1 - \dfrac{{{T_1}}}{{{T_2}}}$
Replacing the term and we can write it as,
$\implies \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{40}}{{100}}$
Taking LCM as RHS and we get,
$\implies \dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{100 - 40}}{{100}}$
On subtracting we get,
$\implies \dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{60}}{{100}}$
On dividing the terms we get,
$\implies \dfrac{{{T_2}}}{{{T_1}}} = 0.6$
Taking the cross multiplication we get,
$\Rightarrow {T_2} = 0.6{T_1}$
Now, for efficiency $50\%$
So we can apply the formula and we get,
$\implies \dfrac{{50}}{{100}} = 1 - \dfrac{{{T_2}}}{{{T_1}^\prime }}$
On dividing the term we get,
$\Rightarrow 0.5 = 1 - \dfrac{{{T_2}}}{{{T_1}^\prime }}$
Putting the value for ${T_2}$ and we get,
$\dfrac{{0.6{T_1}}}{{{T_1}^\prime }} = 0.5$
Taking the term as replace we get,
$\dfrac{{0.6}}{{0.5}}{T_1} = {T_1}^\prime$
On simplification we get,
$\Rightarrow {T_1}^\prime = \dfrac{6}{5}{T_1}$
Hence, the correct answer is option $D$.

Note: A thermodynamic cycle is said to have occurred, when a system is taken throughout a series of different states and finally returned to its initial state. A Carnot principle states that the efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. Also it is used only for cyclical devices like heat engines.