Question

A heat transformer is device that transfers a part of the heat. Supplied to it at an intermediate temperature, to a high temperature reservoir while rejecting the remaining part to a low temperature heat sink. In such a heat transformer, $100kJ$ of heat is supplied at $350K$. the maximum amount of heat in $kJ$ theta can be transferred at $400K$ when the rest is rejected to a heat sink at $300K$ is

• A. $57.14$
• B. $14.29$
• C. $12.50$
• D. $33.33$

Answer 1

The correct option is A $57.14$

Method I:


$Q_1+Q_2=100kJ$
$T=350K$
$T_1=400K$
$T_2=300K$
Entropy change of the heat temperature device:
$\Delta S=\frac{(Q_1+Q_2)}{350}=-\frac{100}{350}$
$=-\frac{2}{7}kJ/K$
Entropy change of reservoir-1,
$\Delta S_1=\frac{Q_2}{T_2}=\frac{100-Q_1}{300}kJ/K$
Entropy change of reservoir-2,
$\Delta S_2=\frac{Q_1}{T_1}=\frac{Q_1}{400}kJ/K$
Total entropy,
$\Delta S_{total}=0$
$\Delta S+\Delta S_1+\Delta S_2=0$
$-\frac{2}{7}-\frac{100-Q_1}{300}+\frac{Q_1}{400}=0$
$-\frac{2}{7}+\frac{1}{3}-\frac{Q_1}{300}+\frac{Q_1}{400}=0$
$-\frac{400Q_1+300Q_1}{120000}=\frac{2}{7}-\frac{1}{3}=\frac{6-7}{21}=-\frac{1}{21}$
$-\frac{100Q_1}{120000}=-\frac{1}{21}$
or $Q_1=57.14kJ$

Method II:



Where, Q is the maximum amount of heat rejected.
$\therefore$ maximum heat transfer, transformer must be reversibel.
$\therefore$ Clausisus inequality.
$\oint \frac{\delta Q}{T}=0$
$\frac{100}{350}-\frac{100-Q}{300}-\frac{Q}{400}=0$
$\frac{10}{35}=\frac{100-Q}{300}+\frac{Q}{400}$
After solving, $Q=57.142kJ$