NEET-XII-Physics
28: Heat Transfer
- #18A metal rod of cross sectional area 1.0 cm2 is being heated at one end. At one time, the temperatures gradient is 5.0°C cm-1 at cross section A and is 2.5°C cm-1 at cross section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J°C-1, thermal conductivity of the material of the rod = 200 W m-1°C-1. Neglect any loss of heat to the atmosphereAns :
Let the temperatures at the ends A and B be TA and TB, respectively.
Rate of flow of heat at end A of the rod is given by
`` \frac{{\,\mathrm{\,dQ\,}}_{\,\mathrm{\,A\,}}}{\,\mathrm{\,d\,}t}=K\,\mathrm{\,A\,}.\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,d\,}l}\left({T}_{\,\mathrm{\,A\,}}\right)``
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Rate of flow of heat at end B of the rod is given by
`` \frac{d{Q}_{\,\mathrm{\,B\,}}}{\,\mathrm{\,d\,}t}=K\,\mathrm{\,A\,}.\frac{\,\mathrm{\,d\,}}{\,\mathrm{\,d\,}l}\left({T}_{\,\mathrm{\,B\,}}\right)``
Heat absorbed by the rod = ms∆T
Here, s is the specific heat of the rod and ∆T is the temperature difference between ends A and B.
Rate of heat absorption by the rod is given by
`` \frac{\,\mathrm{\,d\,}Q}{\,\mathrm{\,d\,}t}=ms\frac{\,\mathrm{\,d\,}T}{\,\mathrm{\,d\,}t}``
`` \therefore ms\frac{\,\mathrm{\,dT\,}}{\,\mathrm{\,d\,}t}=K\,\mathrm{\,A\,}.\frac{{\,\mathrm{\,dT\,}}_{\,\mathrm{\,A\,}}}{\,\mathrm{\,d\,}l}-K\,\mathrm{\,A\,}.\frac{{\,\mathrm{\,dT\,}}_{\,\mathrm{\,B\,}}}{\,\mathrm{\,d\,}l}``
`` \Rightarrow \left(0.4\right).\frac{\,\mathrm{\,dT\,}}{\,\mathrm{\,d\,}t}=200\times 1\times {10}^{-4}\left(5-2.5\right)``
`` \frac{\,\mathrm{\,dT\,}}{\,\mathrm{\,d\,}t}=12.5°\,\mathrm{\,C\,}/\,\mathrm{\,sec\,}``
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