Question

A composite rod consists of two different materials, A and B, each of length 0.5L. The thermal conductivity of Material A is half that of Material B, that is

Answer 1

Answer:

The thermal conductivity of A is given by the formula $Ha=kAb(T2b-T1b)$

Explanation:

Thermal conductivity is the ability of a material to transfer a quantity of heat.

Since thermal conductivity is given by the letter H, let's say the thermal conductivity of A is Ha, and for B, Hb. Ha is half that of Hb, that is $Ha=\frac{1}{2} Hb$ .

Thermal conductivity is given by the formula $H=\frac{kA(T2-T1)}{L}$

Where k is a constant called the thermal conductivity, which depends on the material; A is the cross sectional area of the rod; T2 is final temperature and T1 is initial temperature; L is length of the rod.

Then we can calculate Hb using $Hb=\frac{kAb(T2b-T1b)}{Lb}$

Ab is cross sectional for rod of material B, T2b and T1b are temperatures in material B and Lb is length of material B, which is 0.5.

Finally, we can write Ha in terms of Hb. Since 2 multiplied by 0.5 is 1 we get:

$Ha=\frac{1}{2}\frac{kAb(T2b-T1b)}{0.5} = kAb(T2b-T1b)$