Question

A Carnot engine whose low temperature reservoir is at $7^??, C$ has an efficiency of $50\%$ . It is desired to increase the efficiency to $70\%$ . By how many degrees should the temperature of the high temperature reservoir be increased ?

• A. $840\,K$
• B. $280\,K$
• C. $560\,K$
• D. $380\,K$

Answer 1

Answer: (D)

$380\,K$

Answer 2

Carnot engine is a reversible device which draws heat from a hotter reservoir at temperature $T_{1}$, produces a positive amount of work in the surroundings and discharges rest amount of heat into a colder reservoir at temperature $T_{2}$

Efficiency is defined as
$\eta =1-\frac{T_{2}}{T_{1}}$
$\frac{50}{100} =1-\frac{273+7}{T_{1}}$
$\Rightarrow \, \frac{1}{2} =1-\frac{280}{T_{1}}$
$\Rightarrow \, \frac{280}{T_{1}} =\frac{1}{2}$
$T_{1} =560\, K$
Let new temperature of high temperature reservoir is $T_{1}'$.
Then, $\frac{70}{100}=1-\frac{280}{T_{1}'}$
$\Rightarrow\, \frac{280}{T_{1}}'=\frac{3}{10}$
$\Rightarrow\, T_{1}'=\frac{280 \times 10}{3}=933 \,K$
$\therefore$ Increase in temperature
$=933-560=373 \,K \approx 380\, K$