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,–


w4.PNG


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-


w5.PNG


Now,


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


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


w5.PNG


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


=


=


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