#### Question

The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm^{2}) is maintained at 20°C and the right end is maintained at 80°C. Neglecting any loss of heat through radiation, find (a) the temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = 385 W m^{−1}°C^{−1}.

#### Solution

Area of cross section*, A* = 0.2 cm^{2} = 0.2 × 10^{−4} m^{2}

Thermal conductivity, *k* = 385 W m^{–1} °C^{–1 }

`"Rate of flow of heat" = "temperature diffrences"/ "Thermal resistance"`

`(DeltaQ)/(Deltat) = (KA (T_1 - T_2))/ l`

`(DeltaQ)/(Deltat) = ((80 - 20 )/ 0.2) xx 385xx0.2xx10^-4 `

= 2310 × 10^{-3}

= 2.31 J/sec

Let te temperature of point C be *T*, which is at a distance of 11 cm from the left* *end*. *

Rate of flow of heat is given by

`(DeltaQ)/(Deltat) = (KA DeltaT)/l`

⇒ `(DeltaT)/l= ((DeltaQ)/(Deltat))xx1/(KA)`

`(T-20)/(11xx10^-2) = 2.31/(383xx0.2xx10^-4)`

⇒ T = 33 + 20

⇒ T = 53° c