A Carnot engine has the same efficiency betwe | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$. For the first case,$T_1 = 800 \ K$ and $T_2 = 500 \ K$. $\

Vedclass · Accepted Answer

$960$

Vedclass · Answer

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$. For the first case,$T_1 = 800 \ K$ and $T_2 = 500 \ K$. $\eta_1 = 1 - \frac{500}{800} = 1 - \frac{5}{8} = \frac{3}{8}$. For the second case,$T_{source} = x \ K$ and $T_{sink} = 600 \ K$. $\eta_2 = 1 - \frac{600}{x}$. Since the efficiencies are equal,$\eta_1 = \eta_2$. $1 - \frac{600}{x} = \frac{3}{8}$. $\frac{600}{x} = 1 - \frac{3}{8} = \frac{5}{8}$. $x = \frac{600 	imes 8}{5} = 120 	imes 8 = 960 \ K$.

$A$ Carnot engine has the same efficiency between $800 \ K$ and $500 \ K$ and between $x \ K$ and $600 \ K$. The value of '$x$' is: (in $K$)

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