Answer:
q=2313.04
![W/m^2](https://img.qammunity.org/2020/formulas/physics/middle-school/40i9xiab3sislztnhmu88737oo5kvpcpui.png)
T=690.86°C
Step-by-step explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance
![R=(t)/(KA)](https://img.qammunity.org/2020/formulas/engineering/college/tv77619mfm84u30g0ikkz02j47v9f4dn5t.png)
These are connect in series
![R=\left((t)/(KA)\right)_(fire)+\left((t)/(KA)\right)_(struc)](https://img.qammunity.org/2020/formulas/engineering/college/m0zc7hx429uimqxitcp19xhz22d417mv9l.png)
Heat transfer
![Q=(\Delta T)/(R)](https://img.qammunity.org/2020/formulas/engineering/college/vdb2e71etvzg8gaqpmm7t88581j081vvze.png)
So heat flux
q=2313.04
![W/m^2](https://img.qammunity.org/2020/formulas/physics/middle-school/40i9xiab3sislztnhmu88737oo5kvpcpui.png)
Lets temperature between interface is T
Now by equating heat in both bricks
![(980-T)/((0.2)/(1.6A))=(T-30)/((0.2)/(0.7A))](https://img.qammunity.org/2020/formulas/engineering/college/gg6uo8qmfooglk6lscp0l82knkuvs055ev.png)
So T=690.86°C