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,

E = Energy radiation

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).