The two ends of a metal rod are maintained at temperatures 100 and 110 . The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200 and 210 , the rate of heat flow will be
- 8.0 J/s
- 4.0 J/s
- 44.0 J/s
- 16.8 J/s
Answer
(2) 4.0 J/s
Solution
Rate of heat flow by conduction: H = KA(T1 - T2)L
H (T1 - T2) for the same rod.
Case 1: T1 - T2 = 110 - 100 = 10 ; Case 2: T1 - T2 = 210 - 200 = 10
H2 = H1 = 4.0 J/s
Answer: (2)