A cylinderical brass rod with a radius of 2.10 cm and length of 1.50 m is placed in a fire with temperature of 490 deg. C. If the other end of the rod is at a temperature of 33 deg. C, what is the rate of heat transfer through the rod (ignore any heat loss from the sides of the rod).
Brass constant = 110
I assume you mean for Brass constant, the thermal conductivity for brass is 100Watts/mK
Heat flow=100watt/m k * Area*(deltaT)/length
= 100watt/mK * .0105^2m^2(490-33)K/1.5m
= you do the math.
To calculate the rate of heat transfer through the brass rod, we can use the formula for heat conduction:
Q = (k * A * ΔT) / L
Where:
Q = Rate of heat transfer
k = Thermal conductivity of the material (in this case, brass constant = 110)
A = Cross-sectional area of the rod
ΔT = Temperature difference between the two ends of the rod
L = Length of the rod
First, let's calculate the cross-sectional area of the rod:
A = π * r^2
Where:
r = Radius of the rod
Plugging in the values:
r = 2.10 cm = 0.021 m (converting cm to m)
A = π * (0.021)^2 = π * 0.000441 = 0.00138034 m^2
Next, let's calculate the temperature difference:
ΔT = T2 - T1
Where:
T1 = Temperature at one end of the rod (33°C)
T2 = Temperature at the other end of the rod (490°C)
Plugging in the values:
ΔT = 490°C - 33°C = 457°C
Finally, let's put all the values into the formula to calculate the rate of heat transfer:
Q = (110 * 0.00138034 * 457) / 1.50
Calculating:
Q ≈ 0.37034 J/s or 0.37034 Watts
Therefore, the rate of heat transfer through the rod is approximately 0.37034 J/s or 0.37034 Watts.