A metal bar is used to conduct heat.When the temperature at one end is 88.0°C and at the other is 15.0°C,heat is transferred at a rate of 19.0 J/s.The bar is then stretched uniformly to twice its original length.If again it has ends at 88.0°C and 15.0°C,at what rate will heat be transferred between it ends?
A)4.75 J/s
B)9.50 J/s
C)19.0 J/s
D)38.0 J/s
E)76.0 J/s

Question

Quizplus · Accepted Answer

The rate of heat transfer through a material is given by the equation $Q/t = kA(T_1 - T_2)/L$, where $Q/t$ is the rate of heat transfer, $k$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $T_1$ and $T_2$ are the temperatures at the two ends, and $L$ is the length of the material. When the bar is stretched to twice its original length, its length $L$ doubles, and assuming its volume remains constant, its cross-sectional area $A$ halves. Since the rate of heat transfer is inversely proportional to $L$ and directly proportional to $A$, doubling $L$ would halve the rate, and halving $A$ would also halve the rate again, leading to a quarter of the original rate. Therefore, if the original rate was 19.0 J/s, the new rate will be $19.0 J/s / 4 = 4.75 J/s$.

A Metal Bar Is Used to Conduct Heat