How Does Copper's Thermal Conductivity Affect Heat Transfer?

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SUMMARY

The discussion focuses on the thermal conductivity of copper, specifically its value of 401 W/(m·K), and its impact on heat transfer through a copper bar with a length of 0.10 m and a cross-sectional area of 1.0 x 10-6 m2. The participants worked through calculations for heat conduction, temperature gradient, and scenarios involving bars in series and parallel configurations. The correct application of the heat conduction equation, P = KA(ΔT/d), is crucial for accurate results, particularly in determining the rate of heat transfer and temperature at junctions.

PREREQUISITES
  • Understanding of thermal conductivity and its units (W/(m·K))
  • Familiarity with the heat conduction equation P = KA(ΔT/d)
  • Knowledge of temperature gradients and their calculation
  • Concepts of series and parallel thermal resistance in heat transfer
NEXT STEPS
  • Research the derivation and applications of Fourier's law of heat conduction
  • Learn about thermal resistance in series and parallel configurations
  • Explore practical applications of thermal conductivity in engineering materials
  • Study temperature gradient calculations in different materials
USEFUL FOR

Students in thermodynamics, engineers working with heat transfer systems, and anyone interested in the practical applications of thermal conductivity in materials science.

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Homework Statement


A copper bar of thermal conductivity 401 W/(m·K) has one end at 102°C and the other end at 23°C. The length of the bar is 0.10 m and the cross sectional area is 1.0 multiplied by 10-6 m2.
(a) What is the rate of heat conduction along the bar?

(b) What is the temperature gradient in the bar?

(c) If two such bars were placed in series (end to end) between the same temperature baths, what would script p be?

(d) If two such bars were placed in parallel (side by side) with the ends in the same temperature baths, what would script p be?

(e) In the series case, what is the temperature at the junction where the bars meet?


Homework Equations


P=KA(deltaT/d)


The Attempt at a Solution


I'm stuck on part (a). I used the above equation and got: (401)(1x10^-6)(102+23/.1)=.50125 which is apparently wrong.
 
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Did you plug in the correct value for delta T?Delta T should be the temperature difference.
 
opps...yeah that was suppose to be a negative sign. Thanks
 

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