Example thermal conductivity and coeifficient of the thermal conductivity. A copper bar of the thermal conductivity 401 W (mk) has one end at `104^(@)C`. The length of the bar is `0.10` m and the cross-sectional area is `1.0xx10^(-6) m ^(-2)`. What is the rate of heat conduction along the bar ?

Explanation

The ability to conduct heat in solids is called " thermal conductivity ".
Consider a bar with rectangular crossection as show in the figure . The faces ABCD and EFGH are maintained at `theta_(1) and theta _(2)` respectively `(theta_(0) gt theta_(2))`. Heat passes from one end to the other.

The amount of heat conducted (Q) depends on,
(1) Amount of heat conducted Q is proportional to area of cross -section A prependicular to flow.
`therefore Qprop A " " ........................(1)`
(2) Is proportional to temperature diffrence between the two ends
`therefore Qprop (theta_(2)-theta_(1))" " ........................(2)`
(3) Is proportional to the time of flow.
`therefore Qprop t" " ........................(3)` and
(4) Is inversely proportional to the length of the conductor.
`therefore Qprop (1)/(l)" " ........................(4)`
From the above laws `Q prop (A(theta_(2)- theta _(1))t)/(l) or Q=(KA(theta _(2) -theta _(1))t)/(l) ` where K= constant called coefficient of thermal conductivity .
Coefficient of thermal conductivity (K): It is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient. S.I. Unit w m `K^(-1)` Dimensional formula`=[M^(1)L^(1)T^(-3) theta ^(-1)]`
Problem:
Thermal conductivity of copper , `K_(c)=401 W//m-K `
Temperatureat one end , `theta _(2)=104^(@) C`
Temperature of 2nd end, `theta _(1)=24^(@)C`
Length of copper bar, `l=0.1m, ` Area, `A=1.0xx10^(-6)m^(-2)`
Rate of conduction. `(Q)/(t)=(KA(theta _(2)- theta _(1)))/(l) =(401xx1xx10^( -6)(104-24))/(0.1)=401xx80xx10^(-5)=0.3208 J//S`