Q: A copper rod with a length of 1.5 meters and a cross-sectional area of 0.02 square meters is initially at a temperature of 100°C. One end of the rod is placed in contact with a substance at 20°C, while the other end is maintained at a temperature of 100°C. The thermal conductivity of copper is 400 W/m-K. Determine the rate of heat transfer through the rod.

A: Given data: Length of the rod (L) = 1.5 m Cross-sectional area of the rod (A) = 0.02 m² Temperature at one end of the rod (T1) = 100°C = 373 K Temperature at the other end of the rod (T2) = 20°C = 293 K Thermal conductivity of copper (k) = 400 W/m-K

The rate of heat transfer through the rod can be determined using the formula:

Q = (k * A * ΔT) / L

Where: Q = rate of heat transfer k = thermal conductivity A = cross-sectional area ΔT = change in temperature L = length of the rod

First, calculate the change in temperature (ΔT) using the given temperatures: ΔT = T1 - T2 ΔT = 373 K - 293 K ΔT = 80 K

Now, we can substitute the given values into the formula: Q = (400 W/m-K * 0.02 m² * 80 K) / 1.5 m Q = (640 W) / 1.5 Q = 426.67 W

Therefore, the rate of heat transfer through the rod is 426.67 watts.