Question:
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 4j/s if the ends are maintained at temperatures 200 c and 210 c the rate of heat flow will be
Answer:
T1 = 1000C ; T2 = 1100C ; ∆TA = T2 - T1 = 110 – 100 = 100C ; ∆QA = 4J/s
T3 = 2000C ; T4 = 2100C ; ∆TB = T3 - T4 = 210 – 200 = 100C ; ∆QB = ?
We know that the rate of heat flow is directly proportional to the temperature difference. Hence, ∆QA = 4J/s = T2 - T1 = 100C
∆QA = ∆TB = T3 – T4 = 100C
∆QA / ∆QB = ∆TA / ∆TB
4/x = 10/10 Hence, x = ∆QB = 4 J/s
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