Question

Two reversible heat engines A and B are arranged in series. A rejecting heat directly to B. Engine A receives 200 kJ at a temperature of 421 ${}^o$C from a hot source, while B is in communication with a cold sink at temperature 4.4${}^o$C. If the work output of A is twice that of B. The intermediate temperature between A and B is

• A. ${143.3}^{o}C$
• B. ${212.7}^{o}C$
• C. ${108.5}^{o}C$
• D. ${170.5}^{o}C$

Answer 1

The correct option is A ${143.3}^{o}C$



$W_A=2W_B$

Now, $\frac{W_A}{Q_1}=\frac{T_1-T_2}{T_1}......\left(i\right)$

$\Rightarrow W_A=\left[\frac{T_1-T_2}{T_1}\right]\times Q_1.....\left(ii\right)$

$\frac{W_B}{Q_2}=\frac{T_2-T_3}{T_2}$

Now, $W_B=\frac{T_2-T_3}{T2}[Q_1-W_A][\because Q_2=Q_1-W_A].....\left(iii\right)$

Putting value of equation (iii) in equation (i)

$\Rightarrow W_A=2\frac{[T_2-T_3]}{T_2}[Q_1-W_A]$

$\Rightarrow [\frac{T_2}{2(T_2-T_3)}+1]W_A=Q_1$

Putting value of equation (ii) in above equation

$\Rightarrow \left[\frac{T_2}{(T_2-T_3)2+1]}\right[\frac{T_1-T_2}{T_1}]Q_1=Q_1$

$\Rightarrow \frac{T_2}{2T_2-2T_3}=\frac{T_1}{T_1-T_2}-1=\frac{T_2}{T_1-T_2}$

$\Rightarrow T_1-T_2=2T_2-2T_3$

$\Rightarrow T_2=\frac{T_1+2T_3}{3}=\frac{694+2\times 277.4}{3}=416.267K$

$\Rightarrow T_2={143.3}^{o}C$