## Class 9 Physics Chapter 8 Numerical Problems 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 = Vo(1 + βΔ T)

Δ Q = mc Δ T

P = W / t

P =  Q / t

ΔQf  =  m Hf

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 .                       (1.3m3)

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 – To = 313 – 288 = 25 K

Coefficient of volume expansion of air β = 3.67

Volume = V =?

V = Vo(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 = 1.3m3

#### 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 = T1 = 10°C = 10 + 273 = 283 K

Final temperature = T2 = 65°C = 65 + 273 = 338 K

Change in temperature = Δ T = T2 – T1  = 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 = T2= 20°C = 20 + 273 = 293 K

Final temperature = T1 = 90°C = 90 + 273 = 363 K

Change in temperature = Δ T = T2 – T1  = 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 = ΔQf = 50000J

Latent heat of fusion of ice = Hf = 336000 J

Amount of ice = m =?

ΔQf  =  m Hf

Or               m = ΔQf / Hf

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 = c1 = 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

Q1 = mc Δ T

Q1 = 0.1  2100 10 = 2100 J

Case II:

Heat required for ice to melt = Q2 = mL

Q2 = 0.1 336000

Q2 = 33600 J

Case III:

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

Q3= mc Δ T

Q3 = 0.1 x 4200 x 10 = 4200 J

Total heat required = Q = Q1 + Q2 + Q3

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 = Hv = 2.26 x 106 jkg-1

Heat required = ΔQv =?

ΔQv = mHv

ΔQv = 0.1 x 2.26 x 106 = 0.226 x 106 = 226 x 106 / 1000

ΔQv = 2.26 x 10-1 x 106 = 2.26 x 105  J

#### (Note: Specific heat of water is 4200 Jkg-1 K-1, Latent heat of vaporization of water is 2.26 x 106 jkg-1 K-1).

Solution:

Mass of stream = m1 =  5 g = 5 / 1000 kg = 0.005 kg

Temperature of stream = T1 = 100°C

Mass of water = m2 = 0.5 kg

Temperature of water = T2 = 10°C

Final temperature = T3 = ?

Case I:

Latent heat lost by stream = Q1 = mL

Q1 = 0.005 x 2.26 x 106 = 11.3 x 103  = 11300 J

Case II:

Heat lost by stream to attain final temperature = Q2 = m1c Δ T

Q2 = 0.005 x 4200 x (100 – T3)

Q2 = 21 (100 – T3)

Case III:

Heat gained by water Q3 = m2c Δ T

Q3 = 0.5 x 4200 x (T3 – 10)

Q3 = 2100 (T3 – 10)

According to the law of heat exchange.

Heat lost by stream = heat gained by the water

Q1 + Q2 = Q3

11300 + 21 (100 – T3) = 2100 (T3 – 10)

11300 + 2100 – 21T3 = 2100T3 – 21000

13400 + 21000 = 2100T3 + 21T3

34400 = 2121T3

T3 = 34400 / 2121

T3 = 16.2 °C

Class 9 Physics Full Book Numerical Problems with Solution

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