Two walls of thickness in the ratio 1:3 and thermal conductivities in the ratio 3:2 form a composite wall of a building. If the free surfaces of the wall be at temperatures `30^@C` and `20^@C`, respectively, what is the temperature of the interface?

Explanation

1 st method : If the temperature diffrence across first wall be `Delta T^(@)C`, then that across that second wall will `(10- Delta T)^(@) C` (note)
In steady state, the rate of heat flow across the two walls will be the same.
So, considering an area A normal to the flow of heat,
`(dQ)/(dt)=(K_(1)A(Delta T))/(l_(1))=(K_(2)A(10-Delta T))/(l_(2))`
where `K_(1)=K,K_(2)=3K` and `l_(1)=3l, l_(2)=2l`
`(KA Delta T)/(3l)=(3KA(10-Delta T))/(2l)rArr 2 Delta T = 9(10 - Delta T)`
`Delta T = 8.18^(@)C`
So, the temperature of the interface will be
`30 - Delta T=30 - 8.18 = 21.82^(2)C`
2nd method : Using the equation, temperature of the interface
`T=((K_(1)T_(1)//l_(1)+K_(2)T_(2)//l_(2)))/((K_(1)//l_(1)+K_(2)//l_(2)))`
We have `T=((K(30))/(3l)+(3K(20))/(2l))//((K)/(3l)+(3K)/(2l))`
`= ((10+30))/(((1)/(2)+(3)/(2)))=21.82^(@)C`
3rd method : `R_1=(l_1)/(K_1A)=(3l)/(KA)` and `R_2=(l_2)/(K_2A)=(2l)/(3KA)`
Effective thermal resistance `R=R_1+R_2=(11)/(3)(l)/(KA)`
Now, thermal current `i=(DeltaT)/(R )=(30-20)/((11l)/(3KA))=(30KA)/(11l)`
Also, thermal current through the first slab will be
`i=((30-T))/(R_1)`
i.e., `(30KA)/(11l)=((30-T)KA)/(3l)implies90=330-11T`
`impliesT=21.82^@C`