Let T₁, T₂, T₃ and T₄ be the temperatures at slab interfaces as shown in the figure.

The rate of heat flow across a slab, with thermal conductivity $$\kappa $$, width w, thickness l, length x, and temperature difference $$\Delta$$T, is given by

$$dQ/dt = \kappa lw(\Delta T/x)$$.

Thus, the heat flow rates through the given slabs are

$${{d{Q_A}} \over {dt}} = {{(2K)(4Lw)({T_1} - {T_2})} \over L} = 8Kw({T_1} - {T_2})$$ ...... (1)

$${{d{Q_B}} \over {dt}} = {{(3K)(Lw)({T_2} - {T_3})} \over {4L}} = {3 \over 4}Kw({T_2} - {T_3})$$ ...... (2)

$${{d{Q_C}} \over {dt}} = {{(4K)(2Lw)({T_2} - {T_3})} \over {4L}} = 2Kw({T_2} - {T_3})$$ ...... (3)

$${{d{Q_D}} \over {dt}} = {{(5K)(Lw)({T_2} - {T_3})} \over {4L}} = {5 \over 4}Kw({T_2} - {T_3})$$ ..... (4)

$${{d{Q_E}} \over {dt}} = {{(6K)(4Lw)({T_3} - {T_4})} \over L} = 24Kw({T_3} - {T_4})$$ ..... (5)

In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., $$d{Q_A}/dt = d{Q_E}/dt$$.

Use equations (1) and (5) to get

$${T_1} - {T_2} = 3({T_3} - {T_4})$$ ..... (6)

Similarly, the steady state condition for the interface at temperature T₂ is

$$d{Q_A}/dt = d{Q_B}/dt + d{Q_C}/dt + d{Q_D}/dt$$,

which gives (by using equations (1)-(4))

$$2({T_1} - {T_2}) = {T_2} - {T_3}$$ ..... (7)

Use equations (6) and (7) to get

$${T_3} - {T_4} = (1/3)({T_1} - {T_2}) = (1/6)({T_2}/{T_3})$$.

Also, equations (2)-(4) give $$d{Q_C}/dt = d{Q_B}/dt + d{Q_D}/dt$$.