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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 ? |
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Answer» SOLUTION :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` |
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