Question

The efficiency of the Carnot engine is 1/6. On decreasing the temperature of the sink by 65K, the efficiency increases to 1/3. Find the temperature of the sink and the source respectively.

• A. 325 K and 390 K
• B. 320 K and 390K
• C. 325 K and 395K
• D. 320 K and 380 K

Answer 1

The correct option is A 325 K and 390 K

We have,
$\eta =\frac{T_2-T_1}{T_2},$ where $T_1$ and $T_2$ are the temperatures of sink and source respectively.
$\therefore \eta =\frac{T_2-T_1}{T_2}=\frac{1}{6}$ .....(i)
Now the temperature of the sink is reduced by 65 K.
$\therefore$ temp. of the sink $=(T_1-65)$
$\therefore$ $\eta =\frac{T_2-(T_1-65}{T_2}=\frac{1}{3}$ ....(ii)
On solving eqns. (i) and (ii), we get,
$T_1=325K$
$T_2=390K$