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# A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of ${{227}^{\circ }}C$; the heat is rejected at ${{27}^{\circ }}C$, the cycle is reversible, then what amount of heat is rejected?a) 24 kJ/sb) 223 kJ/sc) 150 kJ/sD) None of the above

Last updated date: 07th Sep 2024
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Hint: We have the temperature of sink as:${{T}_{1}}={{270}^{\circ }}C$ and temperature of source as: ${{T}_{2}}={{27}^{\circ }}C$. Also, heat supplied is: ${{Q}_{1}}=250\text{ }kJ{{s}^{-1}}$. So, by using the formula of efficiency of the heat engine, find the heat rejected: ${{Q}_{2}}$.

Formula used:
$\eta =1-\dfrac{{{T}_{1}}}{{{T}_{2}}}=1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}}$, where $\eta$ is the efficiency of heat engine, ${{T}_{1}}$ is the initial temperature, ${{T}_{2}}$ is the final temperature, ${{Q}_{1}}$ is the heat supplied and ${{Q}_{2}}$ is the heat rejected.

We have:
\begin{align} & {{T}_{1}}={{227}^{\circ }}C \\ & {{T}_{2}}={{27}^{\circ }}C \\ & {{Q}_{1}}=250\text{ }kJ{{s}^{-1}} \\ \end{align}
Now, convert all the temperatures from degree Celsius to degree Kelvin.
We have:
\begin{align} & {{T}_{1}}=227+273 \\ & =500K \end{align}
\begin{align} & {{T}_{2}}=27+273 \\ & =300K \end{align}
Now, by using the formula for efficiency of heat engine, we have:
\begin{align} & \eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}} \\ & =1-\dfrac{500}{300} \\ & =\dfrac{300-500}{300} \\ & =-\dfrac{200}{300} \\ & =-\dfrac{2}{3}......(1) \end{align}
Also, we know that, the efficiency of heat engine is equal to:
$\eta =1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}}$
Now, by substituting the value of efficiency of the heat engine from equation (1), find the value of heat rejected.
We get:
\begin{align} & \Rightarrow -\dfrac{2}{3}=1-\dfrac{250}{{{Q}_{2}}} \\ & \Rightarrow \dfrac{250}{{{Q}_{2}}}=1+\dfrac{2}{3} \\ & \Rightarrow \dfrac{250}{{{Q}_{2}}}=\dfrac{5}{3} \\ & \Rightarrow {{Q}_{2}}=\dfrac{250\times 3}{5} \\ & \Rightarrow {{Q}_{2}}=150kJ{{s}^{-1}} \\ \end{align}
So, the amount of heat rejected by the heat engine is $150kJ{{s}^{-1}}$ .

So, the correct answer is “Option C”.