Question

A Carnot's engine works as a refrigerator between 250K and 300 K. If it receives 700 calories of heat from the reservoir at the lower temperature, then amount of heat rejected at the higher temperature is:

• A. 900 cals
• B. 625 cals
• C. 725 cals
• D. 1000 cals

Answer 1

The correct option is A 900 cals

$T_1=300K$ ,$T_2=250K$ ,$Q_1=?$ ,$Q_2=700cals$

Efficiency, $\eta$=$\frac{usefulworkdone}{heatabsorbed}$ ..(1)

Change in internal energy $\Delta U=0$

$\Delta U=Q_1-Q_2-W=0$

$W=Q_1-Q_2$ ..(2)

From (1) and (2),

$\eta$ =$\frac{W}{Q_1}$=$\frac{Q_1-Q_2}{Q_1}$=$1-\frac{Q_2}{Q_1}$ ..(3)

Also,$\eta =\frac{T_1-T_2}{T_1}=1-\frac{T_2}{T_1}$ ...(4)

On Comparing (3) and (4),

$\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$

$Q_1=\frac{T_1}{T_2}\times Q_2=\frac{300}{250}\times 700\approx 900cals$