In the 1^{text{st}} case,a Carnot engine oper | vedclass

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(A) First case: The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.For the first case,$\eta = 1 - \frac{100}{300} = 1 - \frac{

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same as the $1^{	ext{st}}$ case

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(A) First case: The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.For the first case,$\eta = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3}$.Second case: Two engines are connected in series. The net efficiency $\eta_{	ext{net}}$ of two engines in series is given by $\eta_{	ext{net}} = \eta_1 + \eta_2 - \eta_1 \eta_2$.For the first engine $(E_1)$ operating between $300 \, K$ and $200 \, K$,$\eta_1 = 1 - \frac{200}{300} = 1 - \frac{2}{3} = \frac{1}{3}$.For the second engine $(E_2)$ operating between $200 \, K$ and $100 \, K$,$\eta_2 = 1 - \frac{100}{200} = 1 - \frac{1}{2} = \frac{1}{2}$.Now,calculating the net efficiency: $\eta_{	ext{net}} = \frac{1}{3} + \frac{1}{2} - (\frac{1}{3} 	imes \frac{1}{2}) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2+3-1}{6} = \frac{4}{6} = \frac{2}{3}$.Thus,the efficiency in the $2^{	ext{nd}}$ case is the same as in the $1^{	ext{st}}$ case.

In the $1^{\text{st}}$ case,a Carnot engine operates between temperatures $300 \, K$ and $100 \, K$. In the $2^{\text{nd}}$ case,as shown in the figure,a combination of two engines is used. The efficiency of this combination (in the $2^{\text{nd}}$ case) will be:

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