How Does Heat Transfer Through Different Metals?

In summary: But if you're sure, then I'll just go with that.In summary, two 0.525 m rods, one lead and the other copper, with a square cross section of 1.50 cm on a side, are connected between metal plates held at 2.00°C and 106°C. Using the relevant equations, the amount of heat flowing through the two rods in 1.0 s can be calculated by setting up an equation for each rod and adding the results together. The correct interpretation of "square cross section" is that the cut ends look like a square, with each side measuring 1.50 cm.
  • #1
prettykitty
8
0
Two 0.525 m rods, one lead and the other copper, are connected between metal plates held at 2.00°C and 106°C. The rods have a square cross section of 1.50 cm on a side. How much heat flows through the two rods in 1.0 s? Assume no heat is exchanged between the rods and the surroundings.

Relevant equations
Q=kA(ΔT/L)t
k for Pb is 34.3
k for Cu is 395

My attempt
I set up the equation for the rods in the exact same manner.
Q=(34.3)(1.77e-4)(104/.0150)(1)
Q=(395)(1.77e-4)(104/.0150)(1)

I know what is incorrect about these equations is the area I am using. I took the square cross section to mean diameter, which evidently it is not. The answers given in the text are:
Q=(34.3)(.0151)^2(104/.0150)(1)
Q=(395)(.0151)^2(104/.0150)(1)

How should I interpret square cross section in the future?
To finish this problem I am aware that Qtotal=QPb+QCu
Thanks!
 
Physics news on Phys.org
  • #2
A cross section is the surface you get when an object is cut (or intersected by a plane). If you cut a round rod at 90 degrees to the axis the cross section is circular. If you cut it at say 30 degrees to the axis the cross section would be an elipse. If in doubt assume it's cut at 90 degrees.

So when it says the rods have a "square cross section", that means the cut ends look like a square. When it says "1.50 cm on a side" that means the cross section has sides (so can't be round).

If it had said "the rods have a triangular cross section of 1.50 cm on a side" then the cross section would be an equilateral triangle of area 1.255 cm2.
 
  • #3
The answers given in the text are:
Q=(34.3)(.0151)^2(104/.0150)(1)
Q=(395)(.0151)^2(104/.0150)(1)

Are you sure?

They seem to have L = 0.0150m = 1.5cm but the problem says L = 0.525m.
 

1. How does heat flow by conduction?

Heat flow by conduction is the transfer of thermal energy between two objects that are in direct contact with each other. The heat energy flows from the object with higher temperature to the object with lower temperature, until both objects reach the same temperature.

2. What factors affect heat flow by conduction?

The rate of heat flow by conduction is affected by several factors, including the thermal conductivity of the materials, the temperature difference between the two objects, the distance between the two objects, and the surface area of contact between the objects.

3. How is heat flow by conduction different from heat flow by convection and radiation?

Heat flow by conduction is different from convection and radiation in that it requires direct contact between the objects, whereas convection and radiation can occur without direct contact. Convection involves the transfer of heat through the movement of fluids, while radiation involves the transfer of heat through electromagnetic waves.

4. Why is heat flow by conduction important?

Heat flow by conduction is important in many applications, such as cooking, heating and cooling systems, and insulation. It also plays a crucial role in the Earth's processes, such as the transfer of heat from the core to the surface, which drives plate tectonics and other geologic activities.

5. How is heat flow by conduction measured?

The rate of heat flow by conduction can be measured using a device called a thermal conductivity meter. This device measures the thermal conductivity of a material, which is a measure of how well a material conducts heat. The higher the thermal conductivity, the faster heat will flow through the material.

Similar threads

Replies
1
Views
900
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
9K
  • Introductory Physics Homework Help
Replies
3
Views
7K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
14K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
3K
Replies
1
Views
1K
Back
Top