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. 8.0 J/s
B. 4.0 J/s
C. 44.0 J/s
D. 16.8 J/s

Answer

B

Explanation

Rate of heat flow through a rod is given by `(dQ)/(dt) = - KA (dT)/(dx)`
Let length of the rod be L.
CaseI: `(dT)/(dx) = (Delta T)/(Delta x) = (110-100)/(L) = (10)/(L)`
`:. (dQ_(1))/(dt) = - KA (10)/(L)`…(i)
Also, `(dQ_(1))/(dt) = 4J s^(-1)` ...(ii)
case II: `(dT)/(dx) = (Delta T)/(Delta x) = (210-200)/(L) = (10)/(L)`
`:. (dQ_(2))/(dt) = - KA (10)/(L)` ...(iii)
so, from equation (i), (ii) and (iii)
`(dQ_(2))/(dt) = (dQ_(1))/(dt) = 4J s^(-1)`