# Important Questions Class 11 Physics Chapter 11

## Important Questions Class 11 Physics Chapter 11

### Important Questions for CBSE Class 11 Physics Chapter 11 – Thermal Properties of Matter

Heat has an effect on different materials in different ways.Each matter may behave differently when exposed to a specific amount of heat.The thermal propertiesof matter are the properties that matter exhibits when exposed to heat. The properties aid in determining how the matter will behave when subjected to varying degrees of heat. Thermal properties are made up of four major components. They are as follows.

• Heat capacity
• Thermal expansion
• Thermal conductivity
• Thermal stress

This chapter explains several important concepts like:

• Temperature and Heat

The definitions of heat and temperature are discussed in this section. This section will teach you the units used to measure hot and cold temperatures.

• Measurement of Temperature

This section of Thermal Properties of Matter Class 11 NCERT Solutions explains temperature measurement and the instruments used. It also explains the thermometer and its various scales in depth.

• Ideal Gas Equation and Absolute Temperature

This section of Chapter 11 discusses the various properties of a gas and how thermometers that use gas perform better than liquid-filled thermometers. It also explains the variables that describe a gas’s behaviour. The ideal petrol equation, as well as theories from various scientists, are discussed in this chapter.

• Thermal Expansion

Thermal expansion is one of the most important sections of Chapter 11 Physics Class 11. It describes how various elements expand when heated. In the real world, hot water is poured over metal lids in case they become too tight. The reason for this is that the hot water expands the metal, making it easier to open.

• Specific Heat Capacity

Each item has a different boiling point and reacts differently when it absorbs thermal energy. A bowl of hot water, for example, begins to pop bubbles until it becomes turbulent when the water begins to boil. The concept of heat capacity is thoroughly discussed here, and students can learn about its various aspects.

• Calorimetry

Calorimetry is the measurement of heat, and this section of Class 11 Physics Chapter 11 NCERT Solutions goes into great detail about it. Students can quickly grasp this topic with the help of the real-life examples provided here.

• Change of State

Every matter exists in three states: solid, liquid, and gas. The transition of an element from one state to another is known as the “change of state,”and heat plays an important role in this process. Students will learn more about this in the chapter and will also study in detail the topics of triple point, latent heat, and conduction in this segment.

• Heat Transfer

Heat transfer is a common occurrence, and this topic in Chapter 11 explains it in greater detail. The concepts of radiation, convection, and others are discussed here.

• Newton’s Law of Cooling

When left alone, every item gradually cools down, and this chapter on Thermal Properties of Matter goes on to explain the phenomenon in detail.

Thermal Properties of Matter is an important chapter for board exams in Physics.Students should thoroughly understand the concepts since a good number of questions can be asked from this chapter in exams. Extramarks Class 11 Physics Chapter 11 Important Questions will help students understand the types of questions that will be asked in exams. Subject matter experts prepare these concise questions by referring to NCERT Books.

CBSE Class 11 Physics Chapter-11 Important Questions

Study Important Questions Class 11 Physics Chapter-11 – Thermal Properties of Matter

2 Marks Questions

Q1. A 2.5 kg copper block is heated in a furnace to 500 °C before being placed on a large ice block. What is the most ice that can be melted? (Copper specific heat = 0.39 J; water fusion heat = 335 J).

Ans. Copper block mass, m = 2.5 kg = 2500 g

Temperature rise of the copper block, = 500°C

Copper specific heat, C = 0.39 J

Water fusion heat, L = 335 J

Q = mC is the maximum amount of heat that a copper block can lose.

= 2500 0.39 500

= 487500 J

Let g be the amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice, Q = m1L

m1= Q/L = 487500/335 = 1455.22g.

Hence, the maximum amount of ice that can melt is 1.45 kg

Q2. On a hot day, a car is parked in the sun with all windows closed. Explain why it is significantly warmer inside than outside after some time?

Ans. Glass transmits 50% of heat radiation from a hot source such as the Sun but does not allow moderately hot bodies to pass through it.

4 Marks Questions

Q1. A large steel wheel will be mounted on a shaft made of the same material. The outer diameter of the shaft at 27 °C is 8.70 cm, and the diameter of the wheel’s central hole is 8.69 cm. “Dry ice” is used to cool the shaft. What temperature of the shaft causes the wheel to slip on it? Assume the steel’s coefficient of linear expansion is constant over the required temperature range: = α steel=1.20×10-5 K-¹.

Ans.The given temperature, T = 27°C, can be expressed in Kelvin as follows:

27 + 273 = 300 K

Outer diameter of the steel shaft at T, d1 = 8.70 cm

Diameter of the central hole in the wheel at T, d2 = 8.69 cm

Coefficient of linear expansion of steel, = α steel=1.20×10-5 K-¹.

After the shaft is cooled using “dry ice”, its temperature becomes .

The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70

= – 0.01 cm

Temperature T1, can be calculated from the relation:

d= d1 α steel (T1- T)

0.01 = 8.70×1.20×10-5 (T1- 300)

T1- 300= 95.78

Therefore T1= 204.21 K

= 204.21–273.16

= –68.95°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is -69°C.

Q2. In modern thermometry, the triple point of water is a standard fixed point. Why? What’s wrong with using the melting point of ice and the boiling point of water as standard fixed points (as the Celsius scale originally did)?

Ans. Water’s triple point has a one-of-a-kind value of 273.16 K. The triple point of water is always 273.16 K at specific volume and pressure values. The melting and boiling points of ice and water do not have specific values because they are affected by pressure and temperature.

6 Marks Questions

Describe why:

(a) A body with a high reflectivity is an inefficient emitter.

(b) On a cold day, a brass tumbler feels much colder than a wooden tray.

(c) an optical pyrometer (for measuring high temperatures) calibrated for ideal black body radiation gives an incorrect temperature for a red hot iron piece in the open but a correct temperature for the same piece in the furnace.

(d) the earth would be inhospitably cold without its atmosphere.

(e) heating systems based on the circulation of steam are more efficient than those based on the circulation of hot water in warming a building.

Ans.(a) A body with a high reflectivity absorbs light radiations poorly. A poor absorber is also a poor emitter of radiation. As a result, a body with a high reflectivity is a poor emitter.

(b) Brass is a good heat conductor. When one touches a brass tumbler, heat easily transfers from the body to the brass tumbler. As a result, the body’s temperature drops to a lower level, and one feels cooler.

(c) Wood is a poor heat conductor. When one touches a wooden tray, only a small amount of heat is transferred from the body to the wooden tray. As a result, the body’s temperature drops only marginally, and one does not feel cool.

As a result, on a cold day, a brass tumbler feels colder than a wooden tray.

(d) An optical pyrometer calibrated for ideal black body radiation gives an incorrect temperature reading for a red hot iron piece kept in the open.

The equation for black body radiation is:

E= σ(T4-T⁴0)

Where,

T = Temperature of optical pyrometer

T0 = Temperature of open space

σ = Constant

As a result, increasing the temperature of open space reduces radiation energy.

When the same piece of iron is heated, the radiation energy,

E= σT4

(d) Earth would be inhospitably cold without its atmosphere. There will be no extra heat trapped in the absence of atmospheric gases. The entire amount of heat would be radiated back from the earth’s surface.

(e) A heating system based on the circulation of steam is more efficient than one based on the circulation of hot water in warming a building. This is due to the fact that steam contains excess heat in the form of latent heat (540 cal/g).

Q.1 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The inside temperature of a car is much hotter than the outside temperature on a hot sunny day.
Reason: The water vapour and CO2 present inside the air of the car absorb visible and IR radiation.

A-Assertion is true but reason is false.

B-Assertion and reason both are false.

C-Both assertion and reason are true and the reason is the correct explanation of the assertion.

D-Both assertion and reason are true but reason is not the correct explanation of the assertion.

Marks:1

Ans Assertion is true but reason is false.

When the Sun?s rays pass through the car windows, the visible light gets absorbed and re-radiated at longer wavelengths from the inside surfaces in the car. The water vapour and CO2 are transparent for visible radiation but long wavelengths are trapped by car?s windows and the water vapour and CO2 present in the car. So, most of the radiated heat could not leave causes the inside of the car to quickly get hotter than the outside.

Q.2 An iron ball of mass 0.2 kg is heated to 100 oC, when it is put in ice block at 0oC; 25 gram of ice is melted. The specific heat of iron is

1.

||

0.1.

||

0.8.

||

0.08.

Marks:1

Ans

Q.3 Water is boiling in a flask over a burner. To reduce its boiling temperature one must

A-reduce the surrounding temperature

B-connect the mouth of the flask to an evacuating system

C-supply heat from a very intense heat source

D-close the container with an airtight cork

Marks:1

Ans

Water is boiling in a flask over a burner. To reduce its boiling temperature one must connect the mouth of the flask to an evacuating system.

Q.4 The water of mass 75g at 1000C is added to ice of mass 20g at -150C. What is the resulting temperature?Latent heat of ice =80 cal/g and specific heat of ice=0.5

Marks:5

Ans

Q.5 State Newton’s law of cooling.

Marks:3

Ans

Newton’s law of cooling states that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings.

dQ/dt= -k(T2 – T1)

where, T1 is temperature of surrounding medium.

T2 is the temperature of the body.

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### 1. What is thermal expansion?

Thermal expansion is the process by which elements expand when heated.  A common real-world example is using hot water to open a stuck metallic lid.It is one of the most important concepts in this chapter, and students may encounter more than one question from it.

### 2. What is thermal energy?

Chapter 11 of Class 11 Physics deals with thermal energy. The energy possessed by atoms and molecules when a substance is heated is defined as the thermal properties of matter. Temperature has a direct relationship with thermal energy. It means that as the temperature rises, so does the thermal energy of a substance. When a substance is heated, the atoms and molecules begin to vibrate and produce kinetic energy, known as thermal energy. It may differ depending on the substance or material.

### 3. What is heat capacity?

Heat capacity is defined as the amount of heat required to raise the temperature of a body by one degree Celsius. Heat is measured in Joules or Calories, while temperature is measured in Kelvin/Celsius. The following formula is used to calculate a material’s heat capacity.

Q = mCΔT

Where,

Q is the heat capacity in Joules.

m stands for mass in grammes.

C denotes specific heat in J/K.

T = Temperature change in K

### 4. Write the different properties of the thermometric substances.

The thermometric substances have the following properties.

• The substances must have low freezing and boiling points.
• The substance should not have the property of volatility.
• The substance’s thermal conductivity should be adequate.
• The coefficient of expansion should be high for the substance’s sensitivity.
• Pure thermometric substances should be available.
• The rate of expansion should be uniform for easy calibration.
• It should be easy to find.