NEET-XII-Physics
28: Heat Transfer
- #24An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm2 are welded together as shown in figure (28-E3). One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m-1°C-1 and of copper = 390 W m-1°C-1.
Figure
Ans :
q1 and q2 are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.
Applying KVL at the hot junction, we get
q = q1 + q2
Rate of heat flow, q = `` \frac{KA∆T}{l}``
As q = q1 + q2,
`` \frac{{K}_{\,\mathrm{\,P\,}}A\left({T}_{1}-{T}_{2}\right)}{l}=\frac{{K}_{1}A\left({T}_{1}-{T}_{2}\right)}{l}+\frac{{K}_{2}A\left({T}_{1}-{T}_{2}\right)}{l}``
`` Kp={K}_{1}+{K}_{2}``
`` =390+200=590\,\mathrm{\,W\,}/\,\mathrm{\,m\,}°\,\mathrm{\,C\,}``
`` \Rightarrow q=\frac{{K}_{P}A\left({T}_{1}-{T}_{2}\right)}{l}``
`` q=\frac{590\times {10}^{-4}\left(60-20\right)}{1}``
`` q=2.36\,\mathrm{\,W\,}``
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