Introduction to refrigeration question:

Refrigeration is a process of turning the temperature of a substance so low that the substance starts to freeze. Actually, any refrigerator works on the reverse principle of the engine. Now let us discuss how it works and the question based on the refrigeration.

Refrigeration dependency

A refrigerator can be looked as a Carnot heat engine working in reverse direction. Thus, when a Carnot engine works in the opposite direction as a refrigerator, it will absorb an amount of heat Q2 from the sink (contents of the refrigerator) at lower temperature T2. As the heat is to be removed from the sink at lower temperature, an amount of work equal to Q1 Q2 is performed by the compressor of the refrigerator to remove heat from sink and then reject the total heat Q1 = (Q2 + Q1 Q2) to the source (atmosphere) through the radiator which is fixed on the back of the refrigerator. The coefficient of performance of the refrigerator is defined as the ratio of the quantity of the heat removed per cycle from the contents of the refrigerator to the work done by the external agency to remove it. It is denoted by'beta' .

'beta' = Q2 /W = Q2 / (Q1 Q2) = T2 / (T1 T2)

Image of the refrigeration

Example for the refrigerator question

In the refrigerator, heat from inside at 277 K is transferred to a room at 300 K. How many joule of heat will be delivered to the room for each joule of electric energy consumed ideally?

Solution:

Here, T1 = 300 K, T2 = 27 K, W = 1 Joule

The coefficient of the performance of the refrigerator

'beta' = T2 / (T1 T2) = 277 / (300 277) = 277 / 23 = 12.04

Let for each joule of electrical energy consumed, an amount of heat Q2 is removed from inside of the refrigerator. Therefore, amount of heat delivered to the room for each joule of electrical energy consumed is given by

Q1 = Q2 + W = Q2 + 1

Q2 = Q1 1

'beta' = Q2 / (Q1 Q2) = Q2 / W

So, 12.04 = Q1 -1 / 1

Q1 1 = 12.04

Q1 = 13.04 Joule

Solved problems in Electromagentics make use of various predefined laws and theorems. One of the very first is the Coulombs Law for charged particles. Gradually we move into Gauss' Law, Electric Field, Magnetic Force on a Charged Body, Faraday's Electromagnetic Induction etc. It will not be possible to cover the problems based on all the topics here, but we shall show some examples of the problems and their solutions
Solved Problems on Electromagnetics

Problem: A charge of Q1=-1.0C is placed at the origin of the rectangular coordinate system and a second charge, Q2=-10mC is placed on the x-axis at a distance of 50 cm from thee origin. Find the force on Q1 due to Q2, if they are in free space.

Solution: Q1 =-1.0C at (0, 0, 0) and Q2 =-10mC at (0.5, 0, 0)

By Coulombs law,

F12 = Q1Q2a12/4'pi' 'epsi' r2 Where F12 is force between two charges,

Q1,Q2 are two charges,

r is distance between two charges,

a12 is unit vector along the line joining two charges.

r =(0, 0, 0)-(0.5, 0, 0) = -0.5ax

r = 0.5

F12 = ( -1.0 * 10 -6 ) ( -10 * 10-3) 9 * 109 ( -ax)

F12 = - 360 ax Newton ( Answer )