Question

Two bars of same length and same cross-sectional area but of different thermal conductivities $K_{1}$ and $K_{2}$ are joined end to end as shown in the figure. One end of the compound bar is at temperature $T_{1}$ and the opposite end at temperature $T_{2}$ (where $T_{1} > T_{2}$) The temperature of the junction is

• A. (a) $\frac{\ K_{1}T_{1} + K_{2}T_{1}}{K_{1} + K_{2}}$
• A. (b)$\frac{\ K_{1}T_{2} + K_{2}T_{1}}{K_{1} + K_{2}}$
• A. (c)$\frac{\ K_{1}(T_{1} + T_{2})}{K_{2}}$
• A. (d)$\frac{\ K_{2}(T_{1} + T_{2})}{K_{1}}$

Answer 1

(a)

Let L and A be length and area of cross sections of each bar respectively.

$\therefore$Heat current through the bar 1 is

$H_{1} = \frac{K_{1}A(T_{1} - T_{0})}{L}$

Hence $T_{0}$is junctions temperature.

Heat current through the bar 2 is

$H_{2} = \frac{K_{2}A(T_{0} - T_{2})}{L}$

At steady state, $H_{1} = H_{2}$

$\therefore\frac{K_{1}A(T_{1} - T_{0})}{L} = \frac{K_{2}A(T_{0} - T_{2})}{L}$

$K_{1}(T_{1} - T_{0}) = K_{2}(T_{0} - T_{2})$

$K_{1}T_{1} - K_{1}T_{0} = K_{2}T_{0} - K_{2}T_{2}$

$\mathrm { K } _ { 1 } \mathrm {~T} _ { 0 } + \mathrm { K } _ { 2 } \mathrm {~T} _ { 0 } = \mathrm { K } _ { 1 } \mathrm {~T} _ { 1 } + \mathrm { K } _ { 2 } \mathrm {~T} _ { 2 }$ $T_{0}(K_{1} + K_{2}) = K_{1}T_{1} + K_{2}T_{2}$

$T_{0} = \frac{K_{1}T_{1} + K_{2}T_{2}}{(K_{1} + K_{2})}$ ….. (i)