First slide

Heat engine

Question

The efficiency of a Carnot cycle is $\frac{1}{6}$ . By lowering the temperature of sink by 65 K, it increases to $\frac{1}{3}$ . The initial and final temperatures of the sink are

Moderate

A

390 K and 325 K
B

450 K and 410 K
C

350 K and 275 K
D

400 K and 310 K

Solution

$η = \frac{T_{1} - T_{2}}{T_{1}}$ = $\frac{1}{6} - - - - - - - - (i)$
$η^{'} = \frac{T_{1} - (T_{2} - 65)}{T_{1}} = \frac{1}{3}$ ---------(ii)
From equations (i) and (ii)
$\frac{η^{'}}{η} = (\frac{T_{1} - T_{2} + 65}{T_{1}}) (\frac{T_{1}}{T_{1} - T_{2}}) = \frac{(\frac{1}{3})}{(\frac{1}{6})} = 2$
$or \frac{T_{1} - T_{2} + 65}{T_{1} - T_{2}} = 2$
or $T_{1} - T_{2} = 65$ --------(iii)
From equation (i), $\frac{65}{T_{1}} = \frac{1}{6} or T_{1} = 390 K$
and from equation (iii), $T_{2} - T_{1} - 65 = 325 K$

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