Transfer of Heat -… | Class 09 Physics

Transfer of Heat -… | Class 09 Physics | ClassNotes

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Oct 12, 2022

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 $m^{-1} K ^{-1}$. $(13000 Js^{-1})$

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                                  $\frac{Q}{t}$ = $1300 J s^{-1}$

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 ? $(3.6\times10 7 J)$

Solution:        Time = t = 1hour = 3600 s                       Thickness of glass = L = 0.8 cm =$\frac{0.8}{100 }$= 0.008 m                      Area of a glass window = A = $2.0 m\times 2.5$ m = $5 m^{2}$                                  Temperature outside the house = $T_{1}$ = 25°C = 25 + 273 = 298 K                      Temperature inside the house = $T_{2}$ = 5°C = 5 + 273 = 278 K                      Change in temperature = $\triangle T$ = $T_{1} - T_{2}$ = 298 – 278 = 20 K                      Value of conductivity for concrete = k = $0.8 W m^{-1} K^{-1}$                      Rate of conduction of thermal energy =$\frac{Q}{t}$=?                                   $\frac{Q}{t}$ = $\frac{(〖kA(T〗_{1}  – T_{2}))}{L}$                                     Q =$\frac{(〖kA(T〗_{1}  – T_{2}))}{L} \times t$                                $\frac{Q}{t}$= $\frac{(0.8 \times 5 \times 20)}{0.008 \times3600}$= $80/0.008 \times 3600$= 36,000,000 J = $3.6 \times 10$ 7 J

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