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)`.

Explanation

`Q/t = 0.5 cal S^(-1) , r = 1 cm`
`:.` Area ` A = pir^2 = 3.142 xx 1/cm^2 = 3.142 cm^2`
L = Length of rod = 2m = 200 cm,
`T_1 = 250^@C," "T_2 = ?`
We know `Q/t = (KA(T_1 - T_2))/L`
`or (T_1 - T_2) = Q/t xx (Deltax)/(kA)`
`=(0.5cals^(-1) xx 200cm)/(0.26 cal s^(-1)""^@C^(-1)xx 3.142 cm^2)= 122.4^@C`
`:. T_2 = 250^@C- 122.4^@C = 127.6^@C`