Conduction between a steel/copper cylinder.

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SUMMARY

The discussion centers on the heat transfer between a steel cylinder and a copper cylinder in thermal contact, with specific dimensions of 7.0 cm radius and 4.0 cm length. The heat transfer equation used is Q = (kAtΔT)/L, where k values are 66.9 for steel and 395 for copper. The calculated temperature at the interface is 29.4°C, leading to an initial heat transfer calculation of Q = 1.6E6 J. However, the correct heat transfer value is confirmed to be Q = 1.17E6 J, indicating an error in the initial calculations or assumptions.

PREREQUISITES
  • Understanding of thermal conductivity and heat transfer principles
  • Familiarity with the heat transfer equation Q = (kAtΔT)/L
  • Basic knowledge of temperature gradients and their effects on heat flow
  • Ability to perform algebraic manipulations to solve for unknowns
NEXT STEPS
  • Review the derivation and application of the heat transfer equation Q = (kAtΔT)/L
  • Investigate the thermal conductivity values for various materials
  • Explore the concept of thermal contact resistance and its impact on heat transfer
  • Learn about steady-state heat conduction in composite materials
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Students studying thermodynamics, engineers working on thermal management systems, and anyone involved in heat transfer analysis in materials science.

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Homework Statement

A steel cylinder of radius 7.0 cm and length 4.0 cm is placed in end-to-end thermal contact with a copper cylinder of the same dimensions. If the free ends of the two cylinders are maintained at constant temperatures of 85°C (steel) and 20°C (copper), how much heat will flow through the cylinders in 19 min?

Homework Equations



Q = (kAtdeltaT)/L

Q=heat
k for steel=66.9
k for copper=395
A=area
t=time
T=temperature
L=length

The Attempt at a Solution



x=temperature in the middle of the cylinders

I set heat transferred through the steel part of the cylinder equal to the heat transferred through the copper part of the cylinder.

(66.9)(85-x) = (395)(x-20)
x=29.4

then using that value, i solved for Q.

Q=((66.9)(.0154)(85-29.4)(1140))/.04
so Q=1.6E6 J

BUT the answer is Q=1.17E6 J
where'd i go wrong?
 
Physics news on Phys.org
Q=1.17E6 J should be right. Perhaps the answer key is wrong?
 

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