Solve Physics Problem: Heat Conducted Through 2 Rods

[Mathwizard6254]

Two rods, one of aluminum and the other of copper, are joined end to end. The cross-sectional area of each is 4.0 x 10^-4 m^2, and the length of each is 0.040m. The free end of the aluminum rod is kept at 302 degrees Celsius, while the free end of the copper rod is kept at 25 degrees Celsius. The loss of heat through the sides of the rods may be ignored. How much heat is conducted through the unit in 2 seconds? Please show the work done in solving this problem. THank you very much!

 

[jamesrc]

You should show what work you have done so that people can help you understand it better. To solve this problem, you'll need to look up the thermal conductivity of each material involved. The following equation defines the rate of conduction heat transfer through a barrier (aka Fourier law of heat conduction - I think my notation is standard or at least self-explanatory; I hope you can follow it):

q = \frac{Q}{t} = kA\frac{T_2 - T_1}{L}

When dealing with a series of barriers as in this case, it is useful to consider the concept of a thermal resistance (analagous to electrical resistance):

R_{th} = \frac{L}{kA}

Then you can sum the resistances that are in series to find:

\frac{Q}{t} = \frac{\Delta T}{\Sigma_i{R_{th, i}}}

(For this problem, you'll want to solve for Q.)

 

[M98Ranger]

To solve this physics problem, we will use the equation for heat conduction:

Q = kAΔT / L

Where:
Q = heat conducted
k = thermal conductivity constant (depends on the material)
A = cross-sectional area
ΔT = change in temperature
L = length

First, we need to find the thermal conductivity constant for aluminum and copper. According to the table of thermal conductivity constants, the thermal conductivity constant for aluminum is 205 W/mK and for copper is 385 W/mK.

Next, we can calculate the change in temperature (ΔT) for each rod:

ΔT(aluminum) = 302°C - 25°C = 277°C
ΔT(copper) = 302°C - 25°C = 277°C

Now, we can plug in the values into the equation:

Q(aluminum) = (205 W/mK)(4.0 x 10^-4 m^2)(277°C) / 0.040m = 569.25 W

Q(copper) = (385 W/mK)(4.0 x 10^-4 m^2)(277°C) / 0.040m = 1058.5 W

Since the two rods are joined end to end, the heat conducted through both rods will be the same. Therefore, we can add the two values together to get the total heat conducted:

Q = 569.25 W + 1058.5 W = 1627.75 W

Finally, we need to convert the unit of time from seconds to hours, since the thermal conductivity constants are given in W/mK. There are 3600 seconds in 1 hour, so we can calculate the heat conducted in 2 seconds as:

Q = (1627.75 W)(2 s) / (3600 s/h) = 0.904 W

Therefore, in 2 seconds, 0.904 watts of heat will be conducted through the two rods.