#### Question

Consider the situation shown in the figure . The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross section of the bent part if the total heat taken out per second from the end at 100°C is 130 J.

#### Solution

`R_{BC} = 1/{KA }= 1/{KA}`

`R_{CD} = 1/{KA }= 60/{KA}`

`R_{CD} = 5/{KA} , R_{AB} = 20/M , R_{EF} = 20/{KA}`

Let ;

`R_1 = R_{BC} + R_{CD} + R_{DE} = 70/{KA}`

Let :

`R_{BE} = 60/{KA} = R_2*q* = *q*_{1} +* **q*_{2} ...............(1)

R_{1} and R_{2} are in parallel, so total heat across R_{1} and R_{2} will be same.

⇒ `q_1R_2 = q_2R_2`

`q_1xx 70/{KA} = q_2xx 60/{KA}`

`7q_1 = 6q_2`

`{7q_1}/{6}= q_2`

From equation (1) and (2),

q = q_1 + {7q_1}/6`

`q = {13q_1}/6`

`q=130 J`

`130 = 13Qq_1`

q_{1} = 60 J/sec