Test Index

Thermodynamics Part 1

Section: Physics

© examsnet.com

Question : 68 of 100

Marks: +1, -0

A Carnot engine, having an efficiency of

\eta = \frac{1}{10}

as heat engine, is used as a refrigerator . If the work done on the system is

10\ \mathrm{J}

, the amount of energy absorbed from the reservoir at lower temperature is

100 J
99 J
90 J
1 J

Solution:

The efficiency

(\eta)

of a Carnot engine and the coefficient of performance

(\beta)

of a refrigerator are related as

\beta = \frac{1-\eta}{\eta}

\;\;

Here,

\eta = \frac{1}{10}

\;\; \therefore \beta = \frac{1-\frac{1}{10}}{\frac{1}{10}} = 9

Also, Coefficient of performance

(\beta)

is given by

\beta = \frac{Q_2}{W}

, where

Q_2

is the energy absorbed from the reservoir.

or,

9 = \frac{Q_2}{10}

\therefore Q_2 = 90\ \mathrm{J}\; \text{. }\;

© examsnet.com

Go to Question:

Prev Question

Next Question