First slide

Heat Engine; Carnot Engine and refrigerator

Question

The temperature of inside and outside of a refrigerator (working based on carnot cycle) are 273K and 303K respectively. Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly :

Moderate

A

10J
B

20J
C

30J
D

50J

Solution

Given
Temperature of inside $T_{2} = 273 K$
Temperature of outside $T_{1} = 303 K$
$k = \frac{Q_{2}}{W} = \frac{T_{2}}{T_{1} - T_{2}} = \frac{273}{30}$
Heat delivered to surrounding
$∴ Q_{1} = Q_{2} + W = (k + 1) W = (9 + 1) J = 10 J$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App