A Carnot’s engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

Question

A Carnot’s engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is : (A) 420 J (B) 772 J (C) 2100 J (D) 2520 J

Tardigrade Admin · Accepted Answer

Given: T1=250 K ; T2=300 K ; Q2=500 cal Efficiency η=(T2-T1/T1)=(300-250/250)=(50/250)=(1/5) Also, efficiency =(W/Q2) Work done W =Q2 × η=(500 cal /5) W =(500 × 4.184 J /5)=420 J

Q. A Carnot’s engine works as a refrigerator between $250 K$ and $300 K$ . It receives $500 c a l$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

420 J

772 J

2100 J

2520 J

Given: $T_{1} = 250 K; T_{2} = 300 K; Q_{2} = 500 c a l$
Efficiency $η = \frac{T _{2} - T _{1}}{T _{1}} = \frac{300 - 250}{250} = \frac{50}{250} = \frac{1}{5}$
Also, efficiency $= \frac{W}{Q _{2}}$
Work done $W = Q_{2} \times η = \frac{500 c a l}{5}$
$W = \frac{500 \times 4.184 J}{5} = 420 J$

Q. A Carnot’s engine works as a refrigerator between 250K and 300K. It receives 500cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :