Numerical Problems

 

 

9.1.     The concrete roof of a house of thickness 20 cm has an area 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 W m ^{– 1} K ^{– 1}.(13000 Js ^{– 1} )

Solution:       Thickness of the roof = L = 20 cm = 20/100  = 0.02 m

                      Area = A = 200 m²

                                  Temperature outside the house = T₁ = 35°C = 35 + 273 = 308 K

Temperature inside the house = T₂ = 15°C = 15 + 273 = 288 K

Change in temperature = DT = T₁ – T₂ = 308 – 288 = 20 K

Value of conductivity for concrete = k = 0.65 W m ^{– 1} K ^{– 1}

Rate of conduction of thermal energy = Q/t = ?

Q/t = kA (T1 – T2) / L

Q/t = 0.65 * 20 * 20 / 0.02 = 260 / 0.02 = 13000 W

As (1w = 1J s ^{– 1}) therefore

Q/t = 1300 J s ^{– 1}

 

 

 

Q9.2      How much heat is lost in an hour through a glass window measuring 2.0 m by 2.5 m when inside temperature is 25 °C and that of outside is 5°C, the thickness of glass is 0.8 cm and the value of k for glass is  0.8 W m – 1 K – 1? (3.6×10 7 J)

Solution:        Time = t = 1hour = 3600 s

                       Thickness of glass = L = 0.8 cm = 0.8/100 = 0.008 m

                      Area of a glass window = A = 2.0 m x 2.5 m = 5 m²

                                  Temperature outside the house = T₁ = 25°C = 25 + 273 = 298 K

Temperature inside the house = T₂ = 5°C = 5 + 273 = 278 K

Change in temperature = DT = T₁ – T₂ = 298 – 278 = 20 K

Value of conductivity for concrete = k = 0.8 W m ^{– 1} K ^{– 1}

Rate of conduction of thermal energy = Q/t = ?

Q/t = kA (T1 – T2) / L

Q =  kA (T1 – T2) / L * t

Q / t = 0.8 * 5 * 20 / 0.008 * 3600 = 36,000,000 J = 3.6 x 10 ⁷ J