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Conduction Law

Two bodies of...

Question

Two bodies of masses $m_{1}$ and $m_{2}$ and specific heat capacities $s_{1}$ and $s_{2}$ are connected by a rod of length l, cross -sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of thr first body is $T_{1}$ and the temperature of the second body is $T_{2} (T_{2} > T_{1})$ . Find the temperature difference between the two bodies at time t.

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Solution

$\frac{Q}{t} = \frac{K A (T_{1} - T_{2})}{L}$

Rise in temperature in

$T_{2} = \frac{K A (T_{1} - T_{2})}{L m s}$

Fall in temperature

$T_{1} = \frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}}$

Final temperature in

$T_{1} - T_{1} - \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}}$

Change in Temperature

= $T_{1} - \frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}}$

$= (T_{2} + \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}})$

$= (T_{1} - T_{2})$

$- [\frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}} + \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}}]$

$l n d T = \frac{K A}{L} (\frac{M_{2} S_{2} + M_{1} S_{1}}{M_{1} S_{1} M_{2} S_{2}})$

So difference in temperature

= $(T_{2} - T_{1}) e^{- λ t}$

where $λ = \frac{K A}{L} \frac{(m_{1} s_{1} + m_{2} s_{2})}{m_{1} s_{1} m_{2} s_{2}}$

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