#### Question

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

#### Solution

Resistance of any branch, `R = l/{KA}`

Here, *K* is the thermal conductivity, *A* is the area of cross section and *l* is the length of the conductor.

`R_{BC} = 1/{780.A} = {5xx10^-2}/{780.A}`

`R_{CD} = {60xx10^-2}/"780.A"`

`R_{DE} ={5xx10^-2}/{780.A}`

`R_{AB} = {20xx10^-2}/{390.A}`

`R_{EF} = {20xx10^-2}/{390.A}`

`R_{BE} = {60xx10^-2}/{390.A}`

`R_(BE) = R_2 = (60xx10^-2)/(390xxA)`

since R_{1}_{ }and R_{2}_{ }are in parallel, the amount of heat flowing through them will be same.

`{q_1}/{q_2} = {R_2}/{R_1}`

`= {60xx10^-2xx780xxA}/{390xxAxx70xx10}`

`= 12/7`

`⇒ {q_1}/{q_2 }= 12/7`