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One end of a brass rod 2m long and havin...

One end of a brass rod `2m` long and having `1cm` radius is maintained at `250^(@)C`. When a steady state is reached , the rate of heat flow acrss any cross-section is `0.5 cal s^(-1)`. What is the temperature of the other end `K=0.26"cal" s^(-1) cm^(-1). ^(@)C^(-1)`.

Text Solution

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`Q/t = 0.5 cal s^(-1), r=1cm`
` therefore Area A =pir^(2)= 3.142xx1cm^(2)= 3.142 cm^(2)`
L=length of rod =`2m = 200cm, T_(1) = 250^(@)C, T_(2)=?` We know `Q/t = (KA(T_(1)-T_(2)))/L ` or `(T_(1)-T_(2))= Q/txx(Deltax)/(kA)= (0.5xx200)/(0.26C^(-1)xx3.142) = 122.4^(@)C`
`therefore T_(2) =250^(@)C-122.4^(@)C=127.6^(@)C`
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