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Question
The left end of a copper rod (length = 20 cm, area of cross section = 0.20 cm2) 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.
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Solution
Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
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
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