Board of Intermediate and Secondary Education Rawalpindi Class 9 physics notes pdf download short questions, MCQs, and Problems.
MCQs Thermal Properties of Matter G9
i) Water freezes at
- A. 0 °F
- B. 32 °F
- C. -273 K
- D. 0 K
ii) Normal human body temperature is
- A. 15 °C
- B. 37 °C
- C. 37 °F
- D. 98.6 °C
iii) Mercury is used as thermometric material because it has
- A. uniform thermal expansion
- B. low freezing point
- C. small heat capacity
- D. all the above properties
iv) Which of the following materials has large specific heat?
- A. copper
- B. ice
- C. water
- D. mercury
v) Which of the following materials has a large value of temperature coefficient of linear expansion?
- A. aluminum
- B. gold
- C. brass
- D. steel
vi) What will be the value of β for a solid for which α has a value of 2 x10-5 K-1?
- A. 2 x 10-5 K-1
- B. 6 x 10-5 K-1
- C. 8 x 10-15 K-1
- D. 8 x 10-5 K-1
vii) A large water reservoir keeps the temperature of nearby land moderate due to
- A. low temperature of the water
- B. low specific heat of water
- C. less absorption of heat
- D. large specific heat of water
viii) Which of the following affects evaporation?
- A. temperature
- B. surface area of the liquid
- C. wind
- D. all of the above
Short Questions Physic bise Rawalpindi class 9
Why does heat flow from hot body to cold body?
Energy flows from higher energy states to lower energy states. The kinetic energy of the molecules of a hot body is greater than the kinetic energy of the molecules of a cold body. Thus the fast-moving molecules impart their kinetic energy to the slow-moving molecules of a colder body.
Q.3) Define the terms heat and temperature.
Answer:
HEAT
Heat is a form of energy. It is the kinetic energy of the molecules of a hot body.
TEMPERATURE
It is the degree of coldness or hotness of a body.
Q.4) What is meant by the internal energy of a body?
Answer:
The sum of all types of energy associated with the atoms and molecules of a body is called internal energy.
Q.5) How does heating affect the motion of molecules of a gas?
Answer:
Temperature is directly proportional to kinetic energy. It means that the larger the temperature of a gas, the faster the molecules will move. Due to this reason, when the heat of a system increases, the motion of the molecules also increases.
Q.6) What is a thermometer? Why is mercury preferred as a thermometric substance?
Answer:
The device used to measure the temperature is called a thermometer.
Mercury is used as the thermometric substance in the thermometer because it is
Opaque
Low heat capacity
Good conductor
Does not wet the walls of the tube
Very low freezing point
Q.7) Explain the volumetric thermal expansion.
Answer:
The volume of a solid also changes with the temperature change and this is known as volumetric thermal expansion or cubical thermal expansion.
Mathematically
Where V0 is the initial volume, V is the final volume, △T is the change in temperature and β is the temperature coefficient of volume expansion.
Q.8) Define specific heat. How would you find the specific heat of a solid?
Answer:
Specific heat of a substance is the amount of heat required to raise the temperature of 1 kg mass of that substance through 1 K.
Specific heat of a solid can be calculated by the formula
△Q = c m △T
Where △Q is the amount of heat absorbed, c is the specific heat capacity of the solid, m is the mass of the solid, and △T is the temperature change.
Q.9) Define and explain the latent heat of fusion.
Answer:
LATENT HEAT OF FUSION
The heat energy required to change the unit mass of a substance from solid to liquid state at its melting point without change in its temperature is called the latent heat of fusion.
Hf denotes it
The latent heat of fusion encompasses the process of adding heat to melt a solid and the process of subtracting heat to freeze a liquid.
For example, ice changes into the water at 00 C.
Q.10) Define latent heat of vaporization.
Answer:
LATENT HEAT OF VAPORIZATION
The quantity of heat that changes the unit mass of a liquid completely into a gas at its boiling point without any change in its temperature is called the latent heat of vaporization.
Hv denotes it
Q.11) What is meant by evaporation? On what factors the evaporation of a liquid depends? Explain how cooling is produced by evaporation
Answer:
Evaporation is the changing of a liquid into vapors (gaseous state) from the surface of the liquid without heating it.
FACTORS AFFECTING THE RATE OF EVAPORATION
i) TEMPERATURE
Evaporation is faster at a higher temperature than at a low temperature.
ii) SURFACE AREA
The larger the surface area of a liquid, the greater the number of molecules has the chance to escape from its surface.
iii) Nature of liquid
Different liquids have different rates of evaporation.
iv) WIND
Wind blowing over the surface of liquid increases the rate of evaporation.
Due to evaporation, fast-moving molecules of the liquid escape, and therefore the average kinetic energy of the liquid decreases. This is the reason, cooling is produced due to evaporation.
Problems
Q.1) Temperature of water in a beaker is 50°C. What is its value on the Fahrenheit scale?
Answer:
Solution:
Temperature in Celsius scale = 50oC
The temperature in Fahrenheit scale = F =?
F=1.8C +32
F = 1.8 × 50 + 32 = 90 +32
F = 122oF
Q.2) Normal human body temperature is 98.6°F. Convert it into the Celsius scale and Kelvin scale.
Answer:
Solution:
Temperature in Fahrenheit scale = 98.6oF
i) Temperature in Celsius scale =?
ii) Temperature in Kelvin scale =?
i)
F = 1.8 C + 32
1.8 C= F – 32
1.8 C = 98.6 – 32
1.8 C = 66.6
⇒ C = 37oC
ii)
T = C+273
T = 37+273
T = 310K
Q.3) Calculate the increase in the length of an aluminum bar 2m long when heated from 0°C to 20°C. If the thermal coefficient of linear expansion of aluminum is 2.5 x 10-5 x K-1
Answer:
Solution:
Original length of rod = Lo= 2 m
Initial temperature = To= 0 oF = 0+273 = 273 K
Final temperature = T = 20 oC = 20+273 = 293 K
Change in temperature = △T = T-To= 293 – 273 = 20 K
Coefficient of linear expansion of aluminum = ⍺ = 2.5×10-5 K-1
Increase in volume △L=?
By using the formula
△L = ⍺L0△T
△L = 2.5 x 10-5x 2 x 20
△L = 100 x 10-5
△L = 0.001 m = 0.001 x 100 = 0.1 cm
Q.4) A balloon contains 1.2 m3 air at 15 °C. Find its volume at 40 °C. Thermal coefficient of volume expansion of air is 3.67×10-3 K-1
Answer:
Solution:
Original volume = Vo = 1.2 m3
Initial temperature = To = 15 oC = 15 + 273 = 288 K
Final temperature= T = 40 oC = 40 + 273 = 313 K
Change in temperature= △T= T – To = 313 – 288 = 25 K
Coefficient of volume expansion of air = β =3.67 x 10-3 k-1
Volume = V =?
As
V = Vo(1+β △T)
V = 1.2(1+3.67 x 10-3 x 25)
V = 1.2(1+91.75 x 10-3)
V = 1.2(1+0.09175)
V = 1.2 x 1.09175
V= 1.3 m3
Q.5) How much heat is required to increase the temperature of 0.5 kg of water from 10 °C to 65 °C?
Answer:
Solution:
Mass of water = m = 0.5 kg
Initial temperature = T1 = 10 oC = 10 + 273 = 283 K
Final temperature = T= 65 oC = 65 + 273 = 338 K
Change in temperature = △T = T2 – T1 = 338 – 283 = 55 K
Heat = △Q =?
As we know that
△Q = mc△T
△Q = 0.5 x 2400 x 55
△Q =115500
Q.8) Find the quantity of heat needed to melt 100g Of ice at -10 °C into the water at 10 °C.
Solution:
Mass of ice = m = 100 g = 0.1 kg
Specific heat of ice = c1 = 2100Jkg-1
Latent heat of fusion of ice = L = 336000 JKg-1
Specific heat of water = c2 = 4200 JKg-1k-1
Quantity of heat required = Q =?
Case I
Heat gained by ice from -10oC to 0oC
Q1= mc1△T
Q1= 0.1x 2100 x 10 = 2100 J
Case II
Heat required for ice to melt = Q2 = mL
Q2=0.1 x 336000
Q2=33600J
Case III
Heat required to raise the temperature of water from 0oC to 10oC
Q3 = mc3△T
Q3 = 0.1 x 4200 x 10 = 4200 J
Total heat required is
Q = Q1 + Q2 + Q3
Q = 2100 + 33600 + 4200
Q = 39900 J
Q.9) How much heat is required to change 100 g of water at 100°C into steam? (Latent heat of vaporization of water is 2.26×106 Jkg-1
Solution:
Mass of water = m = 100 g =0.1 kg
Latent heat of vaporization of water = Hv= 2.26 x 106 Jkg-1
The heat required = △Q=?
As we know that
△Q =mHv
△Q = 0.1 x 2.26 x 106= 0.226 x 106
△Q = 2.26 x 10-1 x 106= 2.26 x 105J
Q.10) Find the temperature of the water after passing 5 g of steam at 100 °C through 500 g of water at 10 °C.
Solution:
Mass of stream = m1= 5g = 0.005 kg
Temperature of stream = T1=100oC
Mass of water = m2 = 0.5 kg
Temperature of water =T2 = 10 oC
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△T
Q2=0.005 x 4200 x (100 – T3)
Q2= 21 (100– T3)
Case III Heat gained by water Q3= m2c△T
Q3=0.5x 4200 x (T3-10)
Q3= 2100 (T3-10)
According to the law of heat exchange
Get lost by stream = Heat gained by water
Q1 + Q2 = Q3
11300 + 21 (100 – T3) = 2100(T3-10)
11300+2100-21 T3 = 2100 T3 – 21000
13400 + 21000 = 2100 T3 + 21 T3
34400 = 2121 T3
⇒ T3 = 16.2 oC
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