Conductance

Heat transfer through conduction wall is given by mathematical expression,

%5Cfrac%7Bd%20Q%7D%7Bd%20t%7D%3Dk%20A%20%5Cfrac%7B%5Cleft(T_%7B1%7D-T_%7B0%7D%5Cright)%7D%7Bx%7D

where,

%5Cfrac%7Bd%20Q%7D%7Bd%20t%7D%3D

power transferred through the wall and

x%3D

thickness of the wall Here, side of cube,

a%3D60%20%5C%2Ccm

. Hence, area

A%3D6%20a_%7B%5Ctext%20%7B%2C%20%7D%7D

%5Cbecause

Total area

%3D6

(area of one side of the cube) thickness

x%3D0.1%20%5C%2Ccm

, thermal conductivity,

k%3D4%20%5Ctimes%2010%5E%7B-4%7D%20cal%20s%20%5E%7B-1%7D%20cm%20%5E%7B-1%7D%7B%20%7D%5E%7B%5Ccirc%7D%20C%20%5E%7B-1%7D

, temperature difference,

%5Cleft(T_%7B1%7D-T_%7B0%7D%5Cright)%3D1000%5E%7B%5Ccirc%7D%20C

and potential of DC source

V%3D400%20%5C%2CV

Hence, the power

P%20%3D%5Cfrac%7Bk%20A%20%5Ctimes%204.184%20%5Ctimes%5Cleft(T_%7B1%7D-T_%7B0%7D%5Cright)%7D%7Bx%7D

P%20%3D%5Cfrac%7Bk%206%20a%5E%7B2%7D%20%5Ctimes%204.184%20%5Ctimes%2010%5E%7B3%7D%7D%7Bx%7D

P%20%3D%5Cfrac%7B4%20%5Ctimes%2010%5E%7B-4%7D%20%5Ctimes%206%20%5Ctimes(60)%5E%7B2%7D%20%5Ctimes%2010%5E%7B3%7D%20%5Ctimes%204.184%7D%7B0.1%7D%20J

P%20%3D361.49%20%5C%2CW

Given, voltage supply of

DC%20%3D400%20%5C%2CV

Hence, power generated in a resistance,

P%3D%5Cfrac%7BV%5E%7B2%7D%7D%7BR%7D%3D361.49

%5CRightarrow%20R%3D%5Cfrac%7BV%5E%7B2%7D%7D%7BP%7D

R%3D%5Cfrac%7B400%20%5Ctimes%20400%7D%7B361.49%20%5Ctimes%2010%5E%7B3%7D%7D%3D0.4426%5C%2C%20%5COmega

Chapter Questions
1 MCQs