The two ends of a metal rod are maintained at temperature `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 maintained at temperature s `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 different 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)=k[210-200]= kxx10`....(ii)
From (i) and (ii), `(dQ_(1))/(dt)=(dQ_(2))/(dt)= 4.0 J//s`