In a refrigerator, one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from -3^{\circ}C to 27^{\circ}C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Answers (1)

Here refrigerator acts as a Carnot's engine in reverse order with an efficiency \eta


\eta =1-\frac{T_2}{T_1}=1-\frac{270}{300}=110
 

\eta '=0.5\times0.1=0.05

Coefficient of performance \beta =\frac{Q_2}{W}=\frac{1-\eta '}{\eta '}=\frac{1-0.05}{0.05}=19

Q2=19% WD by motor on Refrigerator

=19\times 1kW=19kJs

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series NEET May 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions