The concrete roof of a house of thickness 20 cm has an area of 200 m². The temperature inside the house is 15°C and 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 Wm⁻¹K⁻¹. Ans. (13000 Js⁻¹)


Given:

Thickness of the concrete roof = L =  20 cm = 0.2 m
Area of the roof = A = 200 m² 
Mass of ice  = m = 100 g = 0.1 kg
Temperature outside of home = T₁ = 35°C = (35 + 273) K = 308 K
Temperature inside of home  = T₂ = 15°C = (15 + 273) K = 288 K
Coefficient of Thermal Conductivity of Concrete  = k = 0.65 W m⁻¹K⁻¹

To Find:

Rate of flow of Heat through concrete  = `\frac {Q}{t}`  = ?

Solution: 


We know that 

Rate of flow of heat = `\frac {Q}{t}` = `\frac {kA (T₁-T₂)}{L}`

by putting values

`\frac {Q}{t}` = `\frac {0.65 W m⁻¹K⁻¹ ｘ 200 m² (308 K-288 K)}{ 0.2 m}`

`\frac {Q}{t}` = `\frac {130 W m K⁻¹ (20 K)}{ 0.2 m}`

`\frac {Q}{t}` = `\frac {2,600 W m}{ 0.2 m}`

`\frac {Q}{t}` = 1,300 W

As (1 watts = 1 J s⁻¹ ) so,

`\frac {Q}{t}` = 1,300 J s⁻¹ 

Thus the rate of conduction of energy (heat) through the roof is 1,300 J s⁻¹  

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