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  • A Carnot engine operating between two reservoirs has efficiency . When the temperature of cold r

A Carnot engine operating between two reservoirs has efficiency \frac{1}{3}. When the temperature of cold reservoir raised by x, its efficiency decreases to \frac{1}{6}. The value of x, if the temperature of hot reservoir is 99^{\circ} \mathrm{C}, will be :

Option: 1

66 K


Option: 2

62 K


Option: 3

33 K


Option: 4

16.5 K


Answers (1)

best_answer

Given \eta=\frac{1}{3}
When T_2 \rightarrow\left(T_2+x\right) i.e., temp. of cold reservior
$$ \eta^{\prime}=\frac{1}{6}
Temp. of hot reservior \left(\mathrm{T}_1\right)=99^{\circ} \mathrm{C}
$$ =99+273=372^{\circ} \mathrm{K}
As we know,
$$ \begin{aligned} & \eta=1-\frac{T_2}{T_1}=\frac{1}{3}\, \, \, \, ...(1) \\ & \eta^{\prime}=1-\frac{\left(T_2-x\right)}{T_1}=\frac{1}{6} \, \, \, \, ...(2)\\ & \eta^{\prime}=\frac{T_1-\left(T_2+x\right)}{T_1}=\frac{1}{6} \end{aligned}
From equqation (1)
$$ \frac{1}{3}=1-\frac{\mathrm{T}_2}{372}

$$ \begin{aligned} & \frac{1}{3}=\frac{372-\mathrm{T}_2}{372} \\ & 372-\frac{372}{3}=\mathrm{T}_2 \\ & \mathrm{~T}_2=248 \mathrm{~K} \end{aligned}
By putting the value of T_2 in equation (2)
$$ \begin{aligned} & \frac{T_1-\left(T_2-x\right)}{T_1}=\frac{1}{6} \\ & \frac{372-(248+x)}{372}=\frac{1}{6} \\ & 372-24-x=\frac{372}{6} \\ & 124-x=62 \\ & 124-62=x \\ & x=62 \mathrm{~K} \end{aligned}

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