Conduction of Heat through a Material

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SUMMARY

The discussion focuses on calculating the heat conduction through a combination of ice and aluminum. The thermal conductivities are specified as 2.34 W/m-K for ice and 237 W/m-K for aluminum. The interface temperature between the two materials is determined to be approximately -24.956 degrees C, leading to a calculated heat flux of 7000 W/m². The formula used for this calculation is Q = (kA ΔT t) / L, where the variables represent thermal conductivity, area, temperature change, time, and thickness.

PREREQUISITES
  • Understanding of heat conduction principles
  • Familiarity with the formula Q = (kA ΔT t) / L
  • Knowledge of thermal conductivity values for different materials
  • Basic algebra for solving equations
NEXT STEPS
  • Research the impact of temperature on thermal conductivity in various materials
  • Learn about the concept of heat flux and its applications in engineering
  • Explore advanced heat transfer topics, such as convection and radiation
  • Study the effects of material thickness on heat conduction rates
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Students in physics or engineering, thermal engineers, and anyone interested in understanding heat transfer mechanisms in materials.

Jared944
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Homework Statement



.005 m of Ice is in contact with .0015 m of Aluminum. The temp of the Ice is -10.0 degrees C, and the temp of the Aluminum is -25.0 degrees C.

I need to calculate the heat per second per square meter that is conducted through the Ice - Aluminum combination

Homework Equations



The equation for the conduction of heat through a Material is (see attachment) Q= ((kA Change in T)t)/L

I know that the answer needs to be in J/(s*m^2), but I am unsure how to get to that point.

Thanks for any help!

Jared
 

Attachments

  • formula.gif
    formula.gif
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I am assuming that the temperature is -10 C at the side of the ice away from the aluminum and the temperature is -25 C at the side of the aluminum away from the ice. The thermal conductivity of the ice is 2.34 W/m-K and the thermal conductivity of aluminum is 237 W/m-K in this temperature range. If T is the interface temperature between the ice and the aluminum, then $$q=2.34\frac{(-10-T)}{0.005}=237\frac{[T-(-25)]}{0.0015}$$where q is the heat flux.

If we solve for the interface temperature T, we obtain -24.956 C. This is very close to the aluminum temperature on the far boundary. Substituting into the equation gives a value of 7000 W/m^2 for the heat flux q.
 

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