The two ends of a metal rod are maintained at temperatures `100^(@)C` and `110^(@)C` . The rate of heat flow in the rod is found to be `4.0 j//s`. If the ends are maintaind at temperatures `200^(@)C` and `210^(@)C`, the rate of heat flow will be :

Options

A. `16.8 j//s`
B. `8.0 j//s`
C. `4.0 j//s`
D. `44.0 j//s`

Answer

C

Explanation

We know that the rate of flow of heat is proportional to temperature difference between two ends. So
`(dQ)/(dt)prop(T_(2)-T_(1))` or `(dQ)/(dt)=k(T_(2)-T_(1))`
Case (i) `(dQ_(1))/(dt) =k[110 - 100] =kxx10` ….(i)
Case (ii) `(dQ_(2))/(dt) =kxx10` .....(ii)
From (i) and (ii), `(dQ_(1))/(dt) = (dQ_(2))/(dt) =4.0 j//s`