A) 1
B) 1/2
C) 2/3
D) 1/3
Correct Answer: D
Solution :
Let the temperature of common interface be \[{{T}^{o}}C\]Rate of heat flow \[H=\frac{Q}{t}=\frac{KA\Delta T}{l}\] \[\therefore \] \[{{H}_{1}}={{\left( \frac{Q}{t} \right)}_{1}}=\frac{2KA(T-{{T}_{1}})}{4x}\] And \[{{H}_{2}}={{\left( \frac{Q}{t} \right)}_{2}}=\frac{KA({{T}_{2}}-T)}{x}\] In steady state, the rate of heat flow should be same in whole system i.e., \[{{H}_{1}}={{H}_{2}}\] \[\Rightarrow \]\[\frac{2KA(T-{{T}_{1}})}{4x}=\frac{KA({{T}_{2}}-T)}{x}\] \[\Rightarrow \]\[\frac{T-{{T}_{1}}}{2}={{T}_{2}}-T\] \[\Rightarrow \]\[T-{{T}_{1}}=2{{T}_{2}}-2T\] \[\Rightarrow \]\[T=\frac{2{{T}_{2}}+{{T}_{1}}}{3}\] Hence, heat flow from composite slab is \[H=\frac{KA({{T}_{2}}-T)}{x}\] \[=\frac{KA}{x}\left( {{T}_{2}}-\frac{2{{T}_{2}}+{{T}_{1}}}{3} \right)=\frac{KA}{3x}({{T}_{2}}-{{T}_{1}})\] ?.(i) [from Eq. (i)] Accordingly, \[H=\left[ \frac{A({{T}_{2}}-{{T}_{1}})K}{x} \right]f\] ?(iii) By comparing Eqs. (ii) and (iii), we get \[f=\frac{1}{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
