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## Numerical Problems

#### 8.1    Temperature of water in a beaker is 50°C, what is its value in 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 = 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 .                       (1.3)

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 –  = 313 – 288 = 25 K

Coefficient of volume expansion of air β = 3.67

Volume = V = ?

V =  (1+ Δ T)

V = 1.2 (1 + 3.67 1.2(1+ 91.75 )

= 1.2(1 + 0.09175) = 1.2  1.09175

V = 1.3

#### 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 =  = 10°C = 10 + 273 = 283 K

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

Change in temperature = Δ T =  –  = 338 – 283 = 55 K

Heat = Δ Q = ?

Δ Q = mc Δ T

Δ Q = 0.5  2400 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 = 20°C = 20 + 273 = 293 K

Final temperature =  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 * 4200 * 70 / 1000 = 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 = 50000J

Latent heat of fusion of ice  = 336000 J

Amount of ice = m = ?

ΔQf  =  m Hf

Or               m = ΔQf / Hf

m = 50000 / 336000 = 0.1488 kg

=  0.1488 * 1000 =  1488 / 1000 * 1000 = 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 , the specific heat of water is 4200 J  Latent heat of 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

= 0.1 336000

Q2  = 33600 J

Case III:

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

Q3= mc Δ T

Q3 = 0.1  4200 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 * 10^6 jkg^-1

Heat required = ΔQv = ?

ΔQv = mHv

ΔQv = 0.1 * 2.26 * 10^6 = 0.226 * 10^6 = 226 / 1000 * 10^6

= 2.26 * 10^-1 * 10^6 = 2.26 * 10^5  J

#### (Note: Specific heat of water is 4200 Jkg^-1 K^-1, Latent heat of vaporization of water is 2.26*10^6 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 * 2.26 * 10^6 = 11.3 * 10^3  = 11300 J

Case II:

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

Q2 = 0.005 * 4200 * (100 – T3)

Q2 = 21 (100 – T3)

Case III:

Heat gained by water Q3 = m2c Δ T

Q3 = 0.5 * 4200 * (T3 – 10)

Q3 = 2100 (T3 – 10)

According to the law of heat exchange.

Heat lost by stream = heat gained by water

Q1 + Q2 = Q3

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

11300 + 2100 – 21T3  = 2100T3  – 21000

13400 + 21000 – 21T3  = 2100T3  – 21T3

34400 = 2121T3

T3 = 34400 / 2121

T3 = 16.2 °C 