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. `44 Js^(-1)`
B. `16.8 Js^(-1)`
C. `8 Js^(-1)`
D. `4 Js^(-1)`

Answer

D

Explanation

Here, `DeltaT_(1)` = 110-100 = `10^(@)`C
`(dQ_(1))/(dt)= 4Js^(-1) rArr DeltaT_(2)` = 210 -200 = `10^(@)`C
`(dQ_(1))/(dt)`=?
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same, i.e., `10^(@)`C, So, the same rate of heat will flow in the second case.
Hence, `(dQ_(2))/(dt)` = 4J`s^(-1)`