The concrete roof of a house of a thickness of 20 cm has an area of 200 m2. The temperature inside the house is 15 °C and the outside is 35°C. Find the rate at which thermal energy will be conducted through the roof. The value of k for concrete is 0.65 W $m^{-1} K ^{-1}$.
$(13000 Js^{-1})$
Difficulty: Medium
Solution: Thickness of the roof = L = 20 cm =$\frac{20}{100}$ = 0.02 m
Area = A = $200 m_{2}$
Temperature outside the house = $T_{1}$ = 35°C = 35 + 273 = 308 K
Temperature inside the house = $T_{2}$ = 15°C = 15 + 273 = 288 K
Change in temperature = $\triangle T = T_{1} - T_{2}$ = 308 – 288 = 20 K
Value of conductivity for concrete = k = $0.65 W m^{-1} K^{-1}$
Rate of conduction of thermal energy =$\frac{Q}{t}$=?
$\frac{Q}{t}$ = $\frac{(〖kA(T〗_{1} – T_{2}))}{L}$
$\frac{Q}{t}$ =$\frac{(0.65 \times 20 \times 20)}{0.02}$ =$\frac{260}{0.02}$= 13000 W
As (1w = $1J s^{-1})$ therefore
$\frac{Q}{t}$ = $1300 J s^{-1}$
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