Question

A wall of dimension $2.00m$ by $3.50m$ has a single - pane window of dimensions $0.75m$ by $1.20m$.If the inside temperature is ${20}^{o}C$ and the outside temperature is $-{10}^{o}C$, effective thermal resistance of the opaque wall and window are $2.10m^{2}K{W}^{-1}$ and $0.21m^{2}K{W}^{-1}$ respectively. The heat flow through the entire wall will be

• A. $215W$
• B. $205W$
• C. $175W$
• D. $110W$

Answer 1

Answer: (A)

$215W$

The wall and the window are in parallel arrangement ; so net heat flow is the sum of the heat flow through the wall and the window.
The temperature difference,
$T_H-T_C=30K$
Area of window,
$A_1=\left(0.75\right)\left(1.20\right)=0.90m^2$
heat flow through window pane,
$\left(\frac{dQ}{dt}\right)=\frac{T_H-T_C}{R_1}=\left(0.90\right)\frac{30}{0.21}=128.6W$
The area of the wall,
$A_2=\left(2.00\right)\left(3.50\right)-\left(0.75\right)\left(1.20\right)=6.10m^2$
Heat flow through window pane,
${\left(\frac{dQ}{dt}\right)}_1=\frac{T_H-T_{T}C}{R_1}=\left(0.90\right)\frac{30}{0.21}=128.6W$
The area of the wall,
$A_2=\left(2.00\right)\left(3.50\right)-\left(0.75\right)\left(1.20\right)=6.10m^2$
Heat flow through the wall,
${\left(\frac{dQ}{dt}\right)}_2=\frac{T_H-T_C}{R_2}=\frac{\left(6.10\right)\left(30\right)}{\left(2.10\right)}=87W$
Net heat flow,$\frac{dQ}{dt}=128.6+87=215.6$