Thermodynamics Cubes of Metal Problem

[mneox]

Homework Statement

Three cubes of equal lengths, lead, iron, and copper are arranged between heat boxes at 100C and 0C. The heat current between boxes is 155W.

1) What is the side length of the cube?
2) What is the temperature between the lead and iron cubes?

Some k values are also given for the materials.

I know that we have to use H = [kA (T2-T1)] / L and the teacher partially explained it but moved along too fast for me to catch.

Homework Equations

H = [kA (T2-T1)] / L

The Attempt at a Solution

I realize that since it's a cube, A = L^2.

So therefore:

H = k L (100 - T1) = k L (T1 - T2) = k L (T2 - 0)

And then I get stuck because I want to find L, but my temperatures are still unknown. Isn't there some way to find the length without the temperature? Thank you for any help!

 

[rl.bhat]

$$H = \frac{kA(T_2 - T_1)}{L}$$

In the given problem, it becomes

$$H = \frac{k_{eff}.L^2(T_2 - T_1)}{3L}$$

$$H = \frac{k_{eff}.L(T_2 - T_1)}{3}$$...(1)

If A1 = A2 = A3 = A and L1 = L2 = L3 = L

$$\frac{3}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}$$

Find keff, and substitute in equation 1 and solve for L.

 

[mneox]

Thanks for your reply.. but what is keff??

(Not the value, but what does it stand for?)

 

[rl.bhat]

Thanks for your reply.. but what is keff??

(Not the value, but what does it stand for?)

It is the effective thermal conductivity. If you have a single block of same dimension instead of three different blocks, what would be its thermal conductivity to have the same rate of energy flow.

 

[rwisz]

I would approach this problem by first identifying the key variables and equations needed to solve it. In this case, the key variables are the side length of the cube (L), the temperature difference between the heat boxes (T2-T1), and the thermal conductivity (k) of the different materials. The equation that relates these variables is the heat transfer equation, which you have correctly identified as H = [kA (T2-T1)] / L, where A is the surface area of the cube.

To solve for the side length of the cube, we need to know the values of the other variables. We are given the heat current (H) of 155W, but we need to determine the temperature difference (T2-T1) and the thermal conductivity (k) of each material.

To find the temperature difference, we can use the fact that the heat current is the same for all three cubes. This means that the heat lost by the hot box (100C) must be equal to the heat gained by the cold box (0C). We can set up an equation using this principle:

H = [kA (T2-T1)] / L = 155W

Since we know that A = L^2, we can rearrange this equation to solve for T2-T1:

T2-T1 = 155W L / (k L^2)

T2-T1 = 155W / k L

Now we can plug in the values for the thermal conductivity (k) of each material and solve for T2-T1. Once we have this value, we can use it in the original equation to solve for the side length of the cube (L).

To find the thermal conductivity (k), we can use the given k values for each material and the fact that k is directly proportional to the heat current (H) and inversely proportional to the temperature difference (T2-T1). This means that we can set up an equation like this:

k lead / k iron = (H lead / H iron) * (T2-T1 iron / T2-T1 lead)

We can solve for k lead by plugging in the given values for H and T2-T1 for lead and iron, and then use this value to find k iron. We can repeat this process for copper as well.

After finding the values for T2-T1 and k for each material, we can plug them