Class 9 Physics Chapter 9 numerical problems are according to the new syllabus of the Punjab Board. Chapter 9 is related to the Transfer of Heat in which you will study the conduction, convection, and radiation applied on different bodies with multiple techniques.

## Class 9 Physics Chapter 9 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

Q/t = kA (T_{1} – T_{2}) / L

### Numerical Problems Chapter 9 – Transfer of Heat

**9.1. The concrete roof of a house of thickness 20 cm has an area 200 m**^{2}. 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 })

^{2}. 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}.

^{ – 1 })

**Solution: **

Thickness of the roof = L = 20 cm = 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 = ΔT = 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 = Q/t =?

Q/t = kA (T_{1} – T_{2}) / L

Q/t = 0.65 x 20 x 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 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 = ΔT = 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 = Q/t =?

Q/t = kA (T1 – T2) / L

*Q* = kA (T1 – T2) / L x t

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

**Class 9 Physics Full Book Numerical Problems with Solution**