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Physics 9-Ch8 (Thermal Properties of Matter)-Numerical Problems

 Table of Contents

The temperature of water in a beaker is 50°C, what is its value on the Fahrenheit scale? (122°F)

Difficulty: Easy
 Solution:   Temperature in Celsius scale = C = 50°C
                    The temperature on the Fahrenheit scale = F =?
                    F = 1.8C + 32
                    F = 1.8× 50 + 32 = 90 + 32
                    F = 122°F

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

Difficulty: Easy
Solution:   
The temperature on Fahrenheit scale = 98.6°F
The temperature on the Celsius scale =?
The temperature on the Kelvin scale =?
 
F = 1.8C + 32
1.8C = F – 32
1.8C = 98.6 -32
1.8C = 66.6
C =(66.6)(1.8) = 37°C
 
T(K) = C + 273
T(K) = 37 + 273
T(K) = 310K

Calculate the increase in the length of an aluminum bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminum is 2.5×〖10〗−5�−1. (0.1cm)

Difficulty: Easy

Solution:       Original length of rod = �(0=) 2m
                  Initial temperature = �(0=)  0°C = 0 + 273 = 273 K
              Final temperature = T =  20°C = 20 + 273 = 293 K
           Change in temperature = △�   = �−�0 = 293 – 273 = 20 K
         Coefficient of linear expansion of aluminum = α = 2.5×10−5�−1
             Increase in volume △� = ?
                                      △�=��0△ T   
                                        △�=2.5×10−5×20
                                       △�=100×10−5
                △�=0.001�=0.001× 100 = 0.1cm


A balloon contains 1.2�3 air at 15°C. Find its volume at 40°C. The thermal coefficient of volume expansion of air is 3.67×〖10〗−3 �−1. (1.3�(3)

Difficulty: Easy

Solution:       Original volume = �(0=)1.2�3
                      Initial temperature = �0 = 15°C = 15 + 273 = 288 K
                       Final temperature = T = 40°C = 40 + 273 = 313 K
                        Change in temperature = △� = T - �0 = 313 – 288 = 25 K
                        Coefficient of volume expansion of air � = 3.67×10−3�−1 
                        Volume = V = ?
                        V = �0(1+�△�)
                        V = 1.2(1+3.67×10−3×25)= 1.2(1+91.75×10−3)
                            = 1.2(1 + 0.09175) = 1.2×1.09175
                        V = 1.3�3
                                                       

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

Difficulty: Easy

Solution:               Mass of water = m = 0.5 kg
                               Initial temperature = �1 = 10°C = 10 + 273 = 283 K
                               Final temperature = �2 = 65°C = 65 + 273 = 338 K 
                               Change in temperature = △ T = �2−�1 = 338 – 283 = 55 K
                               Heat =△� = ?
                               △� = ��△�
                               △�=0.5×2400×55
                               △� = 115500J

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)

Difficulty: Medium

Solution:     Power = P = 1000 ��−1
                    Mass of water = m = 200 g =2001000  = 0.2 kg
                     Initial temperature = �2= 20°C = 20 + 273 = 293 K
                    Final temperature =  �1= 90°C = 90 + 273 = 363 K
                   Change in temperature = △� =  �2−�1 = 363 – 293 = 70 K
                   Specific heat of water = c = 4200�〖��〗−1�−1
                 Time = t = ?
                 P =�� 
Or            P =��  
Or            P ×� = Q
Or            P ×� = ��△�
Or              t=(��△�)� 
                   t = (0.2×4200×70)1000= 58.8 s

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How much ice will melt by 5000 J of heat? Latent heat of fusion of ice = 336000�〖��〗−1.                                                                                                                       (149 g )
 
                  

 

Difficulty: Easy

Solution:     Amount of heat required to melt ice = 50000J
                   Latent heat of fusion of ice =  = 336000�〖��〗−1
                   Amount of ice = m =?
                       △��=  ���
      Or               m =  (△��)(��)
                         m =50000336000 = 0.1488 kg
                            = 0.1488×1000 =14881000 ×1000 = 148.8�≈149� 

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�〖��〗−1�−1, the specific heat of water is 4200�〖��〗−1�−1, Latent heat of fusion of ice is 336000 J〖��〗−1).

Difficulty: Medium
Solution:   Mass of ice = m = 100 g = 1001000  = 0.1 kg
                  Specific heat of ice = c1 = 2100�〖��〗−1�−1
                   Latent heat of fusion of ice = L = 336000�〖��〗−1
                  Specific heat of water = c = 4200�〖��〗−1�−1
                  Quantity of heat required = Q =?
 
Case I:
            Heat gained by ice from -10°C to 0°C
                                        Q1 = ��△�
                                        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:
           The heat is required to raise the temperature of water from 0°C to 10°C
                                                            Q3= ��△�
Q3 = 0.1×4200×10 = 4200 J 
                                                             Total heat required = Q = Q1 + Q2 + Q3
    Q = 2100 + 33600 + 4200
                     Q = 39900 J

How much heat is required to change 100g of water at 100°C into steam? (Latent heat of vaporization of water is 2.26×〖10〗6�〖��〗−1.                (2.26×〖10〗5�)

Difficulty: Easy

Solution:       Mass of water = m = 100 g = 1001000 = 0.1 kg
              Latent heat of vaporization of water = ��  = 2.26×106 ���−1
                     Heat required = △��= ?
                                               △��= ���
                                              △��= 0.1×2.26×106 = 0.226×106 = 2261000 ×106
                                                        = 2.26×10−1×106  = 2.26×105J

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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�〖��〗−1�−1, Latent heat of vaporization of water is 2.26×〖10〗6 ���−1).
Difficulty: Hard
Solution:    Mass of stream = �1=5� =51000 kg = 0.005 kg
                   Temperature of stream = �1= 100°C
                  Mass of water = �2 = 0.5 kg
                  Temperature of water = �2= 10°C
                  Final temperature = �3= ?
 
Case I:
            Latent heat lost by stream = Q1 = mL
            Q1 = 0.005×2.26×106 = 11.3×103 = 11300 J
 
Case II: 
            Heat lost by stream to attain final temperature Q2 = �1�△�
            Q2 = 0.005×4200×(100−�3) 
            Q2 = 21(100−�3)
 
Case III:
              Heat gained by water Q3 = �2�△�
          Q3 = 0.5×4200×(�3−10) 
          Q3 = 2100(�3−10)
                      According to the law of heat exchange.
                       Heat lost by stream = heat gained by water
                                                   Q1 + Q2 = Q3
                              11300+21(100–�3) = 2100(�3−10)
          11300+2100−21�3 = 2100�3 – 21000
                             13400+21000–21�3 = 2100�3–21�3
                                                       34400 = 2121�3
                                                            �3= 344002121 
                                                            �3= 16.2 °C