Heat flow through aluminium rod

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SUMMARY

The discussion centers on calculating heat flow through a solid aluminium rod with a diameter of 40mm and a length of 750mm, given a temperature difference of 100°C. The thermal conductivity of aluminium is specified as 235 Wm-1K-1. The correct formula for heat transfer is applied, but confusion arises regarding the calculation of the cross-sectional area. The final answer for heat flow is confirmed to be 39.4 Js-1.

PREREQUISITES
  • Understanding of thermal conductivity and its units (Wm-1K-1)
  • Knowledge of the formula for heat transfer: k * A * ΔT / L
  • Ability to convert units from millimeters to meters
  • Familiarity with the geometry of a solid cylinder for calculating cross-sectional area
NEXT STEPS
  • Study the derivation and application of the heat transfer formula in different materials
  • Learn how to calculate the cross-sectional area of various geometric shapes
  • Explore the concept of thermal resistance in solid materials
  • Investigate the effects of temperature gradients on heat flow in conductive materials
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This discussion is beneficial for physics students, engineers, and anyone involved in thermal analysis or materials science, particularly those working with heat transfer in solid objects.

em3
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Homework Statement


An Aluminium rod has a diameter of 40mm. and is 750mm long. If the temperature difference between the ends of the rod is 100C .

Take the thermal conductivity of Aluminium to be 235 Wm-1K-1

Homework Equations


k * A * ΔT / L

The Attempt at a Solution


235 * 0.097 m * 100 / 0.75

The answer is supposed to be 39.4 Js-1 but I can't get it right? So I don't know if I'm doing something wrong, and I've tried so many times. I don't know what I'm doing wrong.
Thanks
 
Last edited:
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Hi em3,

Welcome to Physics Forums!

em3 said:
235 * 0.095 m * 100 / 0.75
How did you arrive at 0.095 m for the cross sectional area?
 
40mm. and is 750mm into m, 0.04m and 0.75m. Then did A=2πrL+2πr2

2 * ∏ * 0.02 * 0.75 + 2∏0.022≈ 0.097
Sorry, made a typo in first post
 
em3 said:
40mm. and is 750mm into m, 0.04m and 0.75m. Then did A=2πrL+2πr2
You want the cross sectional area, not the total surface area.
 
How do I find that? The question only gives me the diameter and length, no inner or outer diameter.
 
em3 said:
How do I find that? The question only gives me the diameter and length, no inner or outer diameter.
Assume that a rod is a solid cylinder.
 
Oh ok thanks, I got it. Thanks a lot!
 

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