Question

A Carnots engine, with its cold body at ${17}^{o}C$ has $50$ efficiency. If the temperature of its hot body is now increased by ${145}^{o}C$, then efficiency becomes

• A. $55$
• B. $60$
• C. $40$
• D. $45$

Answer 1

The correct option is B $60$

$\text{Given:-}$ Temperature = ${17}^{\circ }C$

= 273.15 + 17 = 290 K

Efficiency = 50%

$\text{Solution:-}$

Efficiency of Carnot engine,

$\Rightarrow \eta =1-\frac{290}{T_H}$

$\Rightarrow 0.5=1-\frac{290}{T_H}$

$T_H=580K$

Now

${\eta }^{\prime }=1-\frac{290}{T_H+145}$

${\eta }^{\prime }=1-\frac{290}{580+145}$

${\eta }^{\prime }=1-\frac{290}{725}$

$\Rightarrow {\eta }^{\prime }=\frac{435}{725}\times 100$

${\eta }^{\prime }=60$ %

$\text{Hence the correct option is B}$