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Set 54 Problem number 11


Problem

What is the thermal conductivity k of a material if a wall of the material with area 11 m^2, whose thickness is .22 m and whose inside and outside walls are at 171 and 20 Celsius, conducts thermal energy at the rate of 91.5 watts? 

Solution

Conductivity is defined by the following statement: The rate at which thermal energy is conducted is equal to the product of the conductivity, the area of the wall, and the temperature gradient. The conductivity is thus equal to the rate of conduction divided by the area and the gradient.

The conductivity is thus

Generalized Solution

The rate of flow of thermal energy is proportional to c.s. area and to temperature gradient, with proportionality constant k.

This proportionality constant is called the thermal conductivity of the material.  The proportionality for the average rate `dQ/ `dt at which thermal energy Q is conducted through a wall is thus

This makes sense because it ensures that all other things being equal, the rate at which thermal energy is transferred is proportional to the c.s. area of the wall and to the temperature gradient, as we might expect, as as we can confirm by experiment. 

We conclude that

thermal conductivity = k = (`dQ / `dt) / (A ( `dT / `dx) ) = rate of thermal energy flow / (c.s. area * temperature gradient).

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