Answer:
[tex]90.2^{\circ}C[/tex]
Explanation:
Considering the thermal conductivity of aluminium and brass as [tex]k_{al}=205 W/mK[/tex] and [tex]k_{br}=109 W/mk[/tex] respectively
The temperature at the end of aluminium and brass are given as [tex]T_{al}=150^{\circ}C[/tex] and [tex]T_{br}=20^{\circ}C[/tex] respectively with length of rod L=1.3 m , Length of aluminium [tex]L_{al}=0.8 m[/tex], length of brass [tex]L_{br}=0.5 m[/tex] and letting temperature at steady state be T
At steady state, thermal conductivity of both aluminium and brass are same hence
[tex]H_{br}=H_{al}[/tex]
[tex]k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}[/tex]
Upon re-arranging
[tex]T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}[/tex]
[tex] (205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}[/tex]
[tex]T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}[/tex]
[tex]T=90.2^{\circ}C[/tex]
Therefore, the temperatures at which the metals are joined is [tex]90.2^{\circ}C[/tex]
