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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 R1 and R2 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`
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