Q. 27

# Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J s–1m–1 °C–1, whereas it is 390 J s–1m–1°C–1for the straight part. Calculate the ratio of the rate of hat flow through the bent part to the rate of heat flow through the straight part.

Answer :

From previous question,–

Now, its given bent part has k = 780 J s–1m–1 °C–1 and straight part has k =390 J s–1m–1°C–1for

Also Resistance equivalent circuit was as follows-

Now,

RAB = , RBC = , RCD = RDE =, REF= and RBE =

After reducing the equivalent resistance across B and E, our circuit becomes

Where

R1 = RBC+RCD + RDE

=

=

Since length is in cm and conductivity in meters, so multiply with 10-2

R1 =

Again, since R1 and RBE are in parallel, so total heat across R1 and RBE will be same.

ie, q1×R1 = q2 ×RBE

=

=

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Work done in Thermodynamic Process43 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses