The two ends of a rod of length `L` and a uniform cross-sectional area `A` are kept at two temperature `T_(1)` and `T_(2)` `(T_(1) gt T_(2))`. The rate of heat transfer. `(dQ)/(dt)`, through the rod in a steady state is given by

Options

A. `(dQ)/(dt) = (KL(T_(1)-T_(2)))/(A)`
B. `(dQ)/(dt) = (KL(T_(1)-T_(2)))/(LA)`
C. `(dQ)/(dt) = KLA(T_(1)-T_(2))`
D. `(dQ)/(dt)=(KA(T_(1)-T_(2)))/(L)`

Answer

D

Explanation

For a rod of length `L` and area of cross-section `A` whose faces are maintained at temperature `T_(1)` and `T_(2)` respectively.
Then in steady state the rate of heat flowing form one face to the other face in time `t` is given by
`(dQ)/(dt) = (KA(T_(1)-T_(2)))/(L)`.