Two metal cubes A and B of the same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperature. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are \[300W/m{/^0}C\] and \[200W/m{/^0}C\] respectively. After a steady state is reached, find the temperature t of the interface.

Hint:In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time. Here, using the formula for heat flow we are going to find the solution. When the two rods are kept in contact with each other then, the temperature at the junction of 2 rods is known as junction temperature.

Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where,
A is cross-sectional area of metal cube
\[\Delta T\] is temperature difference between two ends of the metal cubes
L is length of the metal cube
K is thermal conductivity

Complete step by step solution:

Image: Two metal cubes of the same size.

Consider two metal cubes A and B of the same size. The extreme ends of the combination are maintained at the indicated temperature and are thermally insulated. If the coefficients of thermal conductivity of A and B are \[300W/m{/^0}C\] and \[200W/m{/^0}C\]. After it reaches steady, we need to find the temperature t of the interface. The rate of flow of heat for both metal cubes is,
\[{\left( {\dfrac{Q}{t}} \right)_1} = {K_1}A\dfrac{{\Delta T}}{L}\]
And, \[{\left( {\dfrac{Q}{t}} \right)_2} = {K_2}A\dfrac{{\Delta T}}{L}\]

As we know that, at steady-state,
\[{\left( {\dfrac{Q}{t}} \right)_1} = {\left( {\dfrac{Q}{t}} \right)_2} \\ \]
\[\Rightarrow {K_1}A\dfrac{{\Delta T}}{L} = {K_2}A\dfrac{{\Delta T}}{L} \\ \]
\[\Rightarrow {K_1}A\dfrac{{\left( {{t_1} - t} \right)}}{L} = {K_2}A\dfrac{{\left( {t - {t_2}} \right)}}{L} \\ \]
\[\Rightarrow {K_1}A\left( {{t_1} - t} \right) = {K_2}A\left( {t - {t_2}} \right) \\ \]
\[\Rightarrow {K_1}\left( {{t_1} - t} \right) = {K_2}\left( {t - {t_2}} \right)\]

Given, \[{t_1} = {100^0}C\], \[{t_1} = {0^0}C\], \[{K_1} = 300W/m{/^0}C\] and \[{K_2} = 200W/m{/^0}C\]
Then, above equation will become,
\[300\left( {100 - t} \right) = 200\left( {t - 0} \right) \\ \]
\[\Rightarrow 30000 - 300t = 200t \\ \]
\[\Rightarrow 500t = 30000 \\ \]
\[\Rightarrow t = \dfrac{{30000}}{{500}} \\ \]
\[\therefore t = {60^0}C\]

Therefore, the temperature t of the interface is \[{60^0}\].

Note:Here, we have considered two rods, in such cases, two rods have different thermal conductivities and are maintained at different temperatures. We found the value of temperature of the common junction by using the heat flow formula.