Physics 9-Ch9 (Transfer of Heat)-Numerical Problems

Table of Contents

• The concrete roof of a house of a thickness of 20 cm has an area
• How much heat is lost in an hour through a glass window measuring

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 ${�}^{-1}{�}^{-1}$. 
$(13000�{�}^{-1})$

Solution:       Thickness of the roof = L = 20 cm =$\frac{20}{100}$  = 0.02 m
                      Area = A = $200{�}_2$
                                  Temperature outside the house = ${�}_1$ = 35°C = 35 + 273 = 308 K
                      Temperature inside the house = ${�}_2$ = 15°C = 15 + 273 = 288 K
                      Change in temperature = $\mathrm{△}�={�}_1-{�}_2$ = 308 – 288 = 20 K
                      Value of conductivity for concrete = k = $0.65�{�}^{-1}{�}^{-1}$
                      Rate of conduction of thermal energy =$\frac{�}{�}$=?
                                    $\frac{�}{�}$ = $\frac{(\text{〖}��(�{\text{〗}}_1\text{ –}{�}_2))}{�}$ 
                               $\frac{�}{�}$ =$\frac{(0.65\times 20\times 20)}{0.02}$ =$\frac{260}{0.02}$= 13000 W
                               As (1w = $1�{�}^{-1})$ therefore
                                  $\frac{�}{�}$ = $1300�{�}^{-1}$

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 ? $(3.6\times 107�)$

Difficulty: Hard

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 = $2.0�\times 2.5$ m = $5{�}^2$

                                  Temperature outside the house = ${�}_1$ = 25°C = 25 + 273 = 298 K

                      Temperature inside the house = ${�}_2$ = 5°C = 5 + 273 = 278 K

                      Change in temperature = $\mathrm{△}�$ = ${�}_1-{�}_2$ = 298 – 278 = 20 K

                      Value of conductivity for concrete = k = $0.8�{�}^{-1}{�}^{-1}$

                      Rate of conduction of thermal energy =$\frac{�}{�}$=?

                                   $\frac{�}{�}$ = $\frac{(\text{〖}��(�{\text{〗}}_1\text{ –}{�}_2))}{�}$

                                     Q =$\frac{(\text{〖}��(�{\text{〗}}_1\text{ –}{�}_2))}{�}\times �$

 

                               $\frac{�}{�}$= $\frac{(0.8\times 5\times 20)}{0.008\times 3600}$= $80\mathrm{/}0.008\times 3600$= 36,000,000 J = $3.6\times 10$ 7 J