Thermal Conductivity Definition

AI Thread Summary
Thermal conductivity measures the rate of heat transfer through a material, defined as the amount of heat passing through a unit area with a temperature difference of 1 degree Celsius. The discussion explores how this property relates to the temperature change of a substance when in contact with a heat source, specifically using Alumina as an example. The user seeks a formula to calculate the rate of temperature change for a given mass in an ideal system without heat loss. Additionally, there is confusion regarding the conversion between units, questioning if 419 W/m K is equivalent to 4.19 W/cm K. Understanding these concepts is crucial for accurately assessing material performance in thermal applications.
Kasper_NYC
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Hi,

I have read that Thermal Conductivity means: “The thermal conductivity of a substance is the quantity of heat in cal/sec passing through a body 1 cm thick with a cross section of 1 sq. cm when the temperature difference between the hot and cold sides of the body is 1 deg. C.” http://www.lib.umich.edu/dentlib/Dental_tables/Thermcond.html

Does it means also how fast a substance can increase or decrease its temperature when is in contact with a heater or colder fount? For example, will it means that in the case of the Alumina (recrystallized), that its index is 38.7 (mcal/s) / (cm^ C/ cm), 1 cubic centimeter will absorbs 0.0387 cal /second, having 1 square cm of surface in contact with the heat fount?

I am interested into find the formula that let's to calculate how fast a given mass of substance warms up or colds down by second, in an ideal system (without losses of heat).

The other question is if “1 (cal/sec) / (cm2 C/cm) = 419 W/m K”; is 419 W/m K equivalent to 4.19 W/m K? (http://hyperphysics.phy-astr.gsu.edu/Hbase/tables/thrcn.html#c1)

Thanks a lot.

PS – I apologize for my English, I am learning it.
 
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Excuse me please,

The other question should be: " If 1 (cal/sec) / (cm2 C/cm) = 419 W/m K”; is it 419 W/m K equivalent to 4.19 W/cm K?"

Thanks again.
 
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