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 V_{0 }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 0^{0} 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 = 50^{o}C

The temperature in Fahrenheit scale = F =?

F=1.8C +32

F = 1.8 × 50 + 32 = 90 +32

F = 122^{o}F

**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 = 98.6^{o}F

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 = 37^{o}C

**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}

^{-5}

^{-1}

**Answer:**

**Solution:**

Original length of rod = L_{o}= 2 m

Initial temperature = T_{o}= 0 oF = 0+273 = 273 K

Final temperature = T = 20 ^{o}C = 20+273 = 293 K

Change in temperature = △T = T-T_{o}= 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 = ⍺L_{0}△T

△L = 2.5 x 10^{-5}x 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 m**^{3}** air at 15 °C. Find its volume at 40 °C. Thermal coefficient of volume expansion of air is 3.67×10**^{-3}** K**^{-1}

^{3}

^{-3}

^{-1}

**Answer:**

**Solution:**

Original volume = V_{o} = 1.2 m^{3}

Initial temperature = T_{o} = 15 ^{o}C = 15 + 273 = 288 K

Final temperature= T = 40 ^{o}C = 40 + 273 = 313 K

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

Coefficient of volume expansion of air = β =3.67 x 10^{-3} k^{-1}

Volume = V =?

**As**

V = V_{o}(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 m^{3}

**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 = T_{1 }= 10 ^{o}C = 10 + 273 = 283 K

Final temperature = T= 65 ^{o}C = 65 + 273 = 338 K

Change in temperature = △T = T_{2} – T_{1} = 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 = c_{1} = 2100Jkg^{-1}

Latent heat of fusion of ice = L = 336000 JKg^{-1 }

Specific heat of water = c_{2} = 4200 JKg^{-1}k^{-1}

Quantity of heat required = Q =?

**Case I **

Heat gained by ice from -10^{o}C to 0^{o}C

Q_{1}= mc_{1}△T

Q_{1}= 0.1x 2100 x 10 = 2100 J

**Case II**

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

Q_{2}=0.1 x 336000

Q_{2}=33600J

**Case III**

Heat required to raise the temperature of water from 0^{o}C to 10^{o}C

Q_{3} = mc_{3}△T

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

**Total heat required is**

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

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 2.26×10**^{6}** Jkg**^{-1}** **

^{6}

^{-1}

**Solution:**

Mass of water = m = 100 g =0.1 kg

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

The heat required = △Q=?

As we know that

△Q =mH_{v}

△Q = 0.1 x 2.26 x 10^{6}= 0.226 x 10^{6}

△Q = 2.26 x 10^{-1} x 10^{6}= 2.26 x 10^{5}J

**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 = m_{1}= 5g = 0.005 kg

Temperature of stream = T_{1}=100^{o}C

Mass of water = m_{2} = 0.5 kg

Temperature of water =T_{2} = 10 ^{o}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△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△T

Q_{3}=0.5x 4200 x (T_{3}-10)

Q_{3}= 2100 (T_{3}-10)

** According to the law of heat exchange**

Get lost by stream = Heat gained by water

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

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

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

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

34400 = 2121 T_{3}

⇒ T_{3} = 16.2 ^{o}C

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