Class 9 Physics Chapter 8 numerical problems are according to the new syllabus of the Punjab Board. Chapter 8 is related to the Thermal Properties of Matter in which you will study the working principles of how matter is converted into different substances using multiple phenomenons.

## Class 9 Physics Chapter 8 Numerical Problems

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

F = 1.8C + 32

T(K) = C + 273

Δ L = α Δ T

V = V_{o}(1 + βΔ T)

Δ Q = mc Δ T

P = W / t

P = Q / t

ΔQ_{f} = m H_{f}

Q = mL

### Numerical Problems Chapter 8 – Thermal Properties of Matter

**8.1 Temperature of water in a beaker is 50°C, what is its value in the Fahrenheit scale? (122°F) **

**Solution: **

Temperature in Celsius scale = C = 50°C

Temperature in Fahrenheit scale = F = ?

F = 1.8C + 32

F = 1.8 + 32

F = 90 + 32

F = 122°F

**8.2 Normal human body temperature is 98.6°F, convert it into Celsius scale and kelvin scale. (37°C,310K)**

**Solution: **

Temperature in Fahrenheit scale = 98.6°F

Temperature in Celsius scale = ?

Temperature in Kelvin scale = ?

F = 1.8C + 32

1.8C = F – 32

1.8C = 98.6 -32

1.8C = 66.6

C = = 37°C

T(K) = C + 273

T(K) = 37 + 273

T(K) = 310K

**8.3 Calculate the increase in the length of an aluminium bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminium is 2.5. (0.1cm)**

**Solution: **

Original length of rod = 2m

Initial temperature = 0°C = 0 + 273 = 273 K

Final temperature = T = 20°C = 20 + 273 = 293 K

Change in temperature = Δ T = T – = 293 – 273 = 20 K

Coefficient of linear expansion of aluminum = α = 2.5

Increase in volume Δ L = ?

Δ L = α Δ T

Δ L = 2.5

Δ L = 100

Δ L = 0.001 m = 0.001 100 = 0.1cm

**8.4 A balloon contains 1.2 air at 15°C. Find its volume at 40°C. Thermal coefficient of volume expansion of air is 3.67 . (1.3m**^{3})

^{3})

**Solution: **Original volume = 1.2

Initial temperature = = 15°C = 15 + 273 = 288 K

Final temperature = T = 40°C = 40 + 273 = 313 K

Change in temperature = Δ T = T – T_{o} = 313 – 288 = 25 K

Coefficient of volume expansion of air β = 3.67

Volume = V =?

V = V_{o}(1 + βΔ T)

V = 1.2 (1 + 3.67 x 10^{-2} x 25)

V = 1.2(1 + 0.09175)

V = 1.2(1.09175)

V = 1.3m^{3}

**8.5 How much heat is required to increase the temperature of 0.5 kg of water from 10°C to 65°C? (115500 J)**

**Solution:**

Mass of water = m = 0.5 kg

Initial temperature = T_{1} = 10°C = 10 + 273 = 283 K

Final temperature = T_{2} = 65°C = 65 + 273 = 338 K

Change in temperature = Δ T = T_{2} – T_{1} = 338 – 283 = 55 K

Heat = Δ Q =?

Δ Q = mc Δ T

Δ Q = 0.5 x 2400 x 55

Δ Q = 115500J

**8.6 An electric heater supplies heat at the rate of 1000J per second. How much time is required to raise the temperature of 200 g of water from 20°C to 90°C? (58.8 s)**

**Solution:**

Power = P = 1000 Js^{-1}

Mass of water = m = 200 g = 200/1000 = 0.2 kg

Initial temperature = T_{2}= 20°C = 20 + 273 = 293 K

Final temperature = T_{1} = 90°C = 90 + 273 = 363 K

Change in temperature = Δ T = T_{2} – T_{1} = 363 – 293 = 70 K

Specific heat of water = c = 4200 Jkg^{-1} K^{-1}

** **Time = t =?

P = W / t

Or P = Q / t

Or P t = Q

Or P t = mc Δ T

Or t= mc Δ T / P

t = 0.2 x 4200 x 70 / 1000

t = 58.8 s

**8.7 How much ice will melt by 5000 J of heat? Latent heat of fusion of ice = 336000 J. (149 g )**

**Solution: **

Amount of heat required to melt ice = ΔQ_{f} = 50000J

Latent heat of fusion of ice = H_{f} = 336000 J

Amount of ice = m =?

ΔQ_{f} = m H_{f}

Or m = ΔQ_{f} / H_{f}

m = 50000 / 336000 = 0.1488 kg

m = 0.1488 x 1000 g

m = 1488 x 1000 / 1000

m = 148.8 g ≈ 149 g

**8.8 Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C. (39900 J) **

**(Note: Specific heat of ice is 2100 J, and the specific heat of water is 4200 J Latent heat of the fusion of ice is 336000 J .**

**Solution: **

Mass of ice = m = 100 g = ** ** = 0.1 kg

Specific heat of ice = c_{1} = 2100 J

Latent heat of fusion of ice = L = 336000 J

Specific heat of water = c = 4200 J

Quantity of heat required = Q =?

**Case I:**

Heat gained by ice from -10°C to 0°C

Q_{1} = mc Δ T

Q_{1} = 0.1 2100 10 = 2100 J

**Case II:**

Heat required for ice to melt = Q_{2} = mL

Q_{2} = 0.1 336000

Q_{2} = 33600 J

**Case III:**

Heat is required to raise the temperature of water from 0°C to 10°C

Q_{3}= mc Δ T

Q_{3} = 0.1 x 4200 x 10 = 4200 J

Total heat required = Q = Q_{1} + Q_{2} + Q_{3}

Q = 2100 + 33600 + 4200

Q = 39900 J

**8.9 How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is 2.26 . (2.26 J)**

**Solution: **

Mass of water = m = 100 g = 100 / 1000 = 0.1 kg

Latent heat of vaporization of water = H_{v} = 2.26 x 10^{6} jkg^{-1}

Heat required = ΔQ_{v} =?

ΔQ_{v} = mH_{v}

ΔQ_{v} = 0.1 x 2.26 x 10^{6} = 0.226 x 10^{6} = 226 x 10^{6} / 1000

ΔQ_{v} = 2.26 x 10^{-1} x 10^{6} = 2.26 x 10^{5} J

**8.10 Find the temperature of the water after passing 5 g of steam at 100°C through 500 g of water at 10°C. (16.2°C)**

**(Note: Specific heat of water is 4200 Jkg**^{-1} K^{-1}, Latent heat of vaporization of water is 2.26 x 10^{6} jkg^{-1} K^{-1}).

^{-1}K

^{-1}, Latent heat of vaporization of water is 2.26 x 10

^{6}jkg

^{-1}K

^{-1}).

**Solution: **

Mass of stream = m_{1} = 5 g = 5 / 1000 kg = 0.005 kg

Temperature of stream = T_{1} = 100°C

Mass of water = m_{2} = 0.5 kg

Temperature of water = T_{2} = 10°C

Final temperature = T_{3} = ?

**Case I:**

Latent heat lost by stream = Q_{1} = mL

Q_{1} = 0.005 x 2.26 x 10^{6} = 11.3 x 10^{3} = 11300 J

**Case II:**

Heat lost by stream to attain final temperature = Q_{2} = m_{1}c Δ T

Q_{2} = 0.005 x 4200 x (100 – T_{3})

Q_{2} = 21 (100 – T_{3})

**Case III:**

Heat gained by water Q_{3} = m_{2}c Δ T

Q_{3} = 0.5 x 4200 x (T_{3} – 10)

Q_{3} = 2100 (T_{3} – 10)

According to the law of heat exchange.

Heat lost by stream = heat gained by the water

Q_{1} + Q_{2} = Q_{3}

11300 + 21 (100 – T_{3}) = 2100 (T_{3} – 10)

11300 + 2100 – 21T_{3} = 2100T_{3} – 21000

13400 + 21000 = 2100T_{3} + 21T_{3}

34400 = 2121T_{3}

T_{3} = 34400 / 2121

T_{3} = 16.2 °C

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