Question

The $\left(\frac{W}{Q}\right)$ of a carnot-engine is $\frac{1}{6}$, now the temperature of sink is reduced by ${62}^{o}C$, then this ratio becomes twice, therefore the initial temperature of the sink and source are respectively.

• A. ${33}^{o}C,{67}^{o}C$
• B. ${37}^{o}C,{99}^{o}C$
• C. ${67}^{o}C,{33}^{o}C$
• D. $97K,37K$

Answer 1

The correct option is B ${37}^{o}C,{99}^{o}C$

$\frac{W}{Q}=\frac{T_2-T_1}{T_1}$

$T_1=$ Temperature of source =$T_1°C+273$

$T_2=$ Temperature of sink =$T_2°C+273$

$\therefore$$\frac{W}{Q}=\frac{t_1-t_2}{t_1+273}$

Now,When $\frac{W}{Q}=\frac{1}{6}$

then,$\frac{1}{6}=\frac{t_1-t_2}{t_1+273}-\left(i\right)$

if $\frac{W}{Q}=\frac{2}{6}=\frac{1}{3}$

then,Sink Temperature=$t_2-62$

$\therefore$$\frac{1}{3}=\frac{t_1-(t_2-62)}{t_1+273}$

$=\frac{1}{3}=\frac{t_1-t_2+62}{t_1+273}$

$=\frac{t_1-t_2}{t_1+273}+\frac{62}{t_1+273}$

$\frac{1}{3}=\frac{1}{6}+\frac{62}{t_1+273}$

$\therefore$$\frac{62}{t_1+273}+\frac{1}{6}$

$\therefore$$t_1=99°C$

$\therefore$$t_2=37°C$.