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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.
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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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