Question

Two parallel plates $A$ and $B$ are joined together to form a compound plate. The thickness of the plates are $4.0\text{cm}$ and $2.5\text{cm}$ respectively and the area of cross-section is $100{\text{cm}}^2$ for each plate.
The thermal conductivities are $K_A=200\text{W}{/}^{\circ }\text{C m}$ for the plate $A$ and $K_B=400\text{W}{/}^{\circ }\text{C m}$ for the plate $B$. The outer surface of the plate $A$ is maintained at ${100}^{\circ }C$ and the outer surface of the plate B is maintained at $0^{\circ }C$. Find the rate of heat flow through any cross-section.

• A. $2810\text{W}$
• B. $3540\text{W}$
• C. $3810\text{W}$
• D. $4240\text{W}$

Answer 1

The correct option is C $3810\text{W}$

Given:
Area of each plate, $A={10}^{-2}{\text{m}}^2$
Thickness of the plate $A,l_A=0.04\text{m}$
Thickness of the plate $B,l_B=0.025\text{m}$
Thermal conductivity of plate $A,K_A$=$200{\text{Wm}}^{-1\circ }{C}^{-1}$
Thermal conductivity of plate $B,K_B=400{\text{Wm}}^{-1\circ }{C}^{-1}$
Temperature of one surface, $T_1={100}^{\circ }C$
Temperatue of other surface , $T_2=0^{\circ }C$

Now,
Thermal resistance of the plate $A$ is,
$R_A=\frac{l_A}{k_{A}A}=\frac{0.04}{200\times {10}^{-2}}$

$=0.02\text{W}{}^{\circ }{C}^{-1}$

Thermal resistance of the plate $B$ is,
$R_B=\frac{l_B}{k_{B}A}=\frac{0.025}{400\times {10}^{-2}}$

$=0.00625{\text{W}}^{\circ }{C}^{-1}$

Since the both plates are in series combination.

Equivalent resistance can be given as, $R=R_A+R_B$
$\Rightarrow R=0.02+0.00625$
$\Rightarrow R=0.02625{\text{W}}^{\circ }{C}^{-1}$

$\therefore$Rate of heat flow, $\frac{\Delta Q}{\Delta t}=\frac{T_1-T_2}{R}=\frac{100-0}{0.02625}=3810\text{W}$

Hence, $\left(C\right)$ is the correct answer.