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Thermal Properties of Matter Class 11 Important Questions NEET Physics [Free PDF Download]

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Last updated date: 17th Apr 2024
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Efficient NEET Preparation with Important Questions for Thermal Properties of Matter Class 11 Physics

Thermal Properties of matter is a fundamental chapter of Physics which introduces students to different heat and temperature-related properties of different states of matter such as solid, liquid, and gas. It is a very crucial chapter for students to develop a thorough understanding and foundation. Solids, liquids, and gases, have different thermal properties and hence they react differently to temperature and heat. In order to understand the properties and processes in accordance with temperature, refer to the chapter and solve the important questions. 


These questions have been created by the top subject matter experts at Vedantu. They have provided all the solutions and explanations that are required for the students to grasp the concept of the chapter easily. By solving the Thermal Properties of Matter important questions, students can revise the topic and assess their knowledge about the topic efficiently.

Competitive Exams after 12th Science

Access JEE NEET Important Questions Physics Thermal properties of matter

Single Correct Questions:

1. Why is it permissible to use any low density gas in a constant volume gas thermometer?

(A) All low density gases expand in the same way.

(B) They are simple to manage.

(C) Their pressure variation is related to the square of the temperature change.

(D) The stated assertion is false.


2. Choose the correct statement for a constant volume gas thermometer.

(A) It does not utilise mercury to measure pressure.

(B) The pressure is measured at constant volume, and the scale is calibrated to give the temperature reading

(C) Any gas with a high density can be used.

(D) Pressure ∝ Temp2


3. A hole is located in the centre of a homogeneous metal sheet. When the sheet is equally heated, what happens to the hole?

(A) Its size increases

(B) Its size decreases

(C) It changes shape

(D) Remains the same size


4. What is the calorimeter's operating principle?

(A) The rate of heat loss is related to the temperature differential.

(B) If no heat is lost to the surroundings, the total heat lost by a hot body equals the total heat acquired by a cold body.

(C) Heat loss is solely determined by a substance's specific heat.

(D) The difference in heat flow from a sample to two separate samples of the same mass is a product of their specific heats and temperature differences.


5. What happens to the boiling point of water in a container when the lid completely seals the container?

(A) Increases

(B) Decreases

(C) Remains same

(D) A substance's boiling point is a property.


6. Choose the correct statement

(A) A material transforms from solid to gas solely through the liquid phase when heated continuously.

(B) Variable temperature causes phase change.

(C) Water can exist in all three states.

(D) The boiling point rises as pressure rises.


7. When water is kept in an extremely hot pan, it heats up completely. What mode of heat transport is to blame for this?

(A) Conduction

(B) Convection

(C) Radiation

(D) Convection along with radiation


8. Heat conduction can be expressed quantitatively as?

(A) For a given temperature difference, heat flow

(B) Heat flow rate over time for a given temperature difference

(C) Temperature variation produced by heat flow per unit time

(D) Temperature variation per unit of heat transferred


9. Choose the correct statement.

(A) There is no need for a medium for radiation.

(B) Convection is the transport of heat via vibration of particles at their place.

(C) Conduction is the movement of particles to transfer heat.

(D) The heat of the gas flame that heats the cooking pan is caused by radiation.


10. What happens to the temperature of a metal rod at steady state if we hold it from one end and dip the other in a source of fire?

(A) Be uniform

(B) Decrease with time

(C) Maintain a steady value along the length of the rod

(D) Stay nonuniform


11. $100 \mathrm{~g}$ of water is heated from $30^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4184 \mathrm{~J} / \mathrm{kg} / \mathrm{K}$ (NEET 2011)

(A) $8.4 \mathrm{~kJ}$

(B) $84 k J$

(C) $2.1 \mathrm{~kJ}$

(D) $4.2 \mathrm{~kJ}$


12. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32 V$, the efficiency of the engine is(NEET 2010)

(A) $0.5$

(B) $0.75$

(C) $0.99$

(D) $0.25$


13. One $\mathrm{kg}$ of a diatomic gas is at a pressure of $8 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{3}$. What is the energy of the gas due to its thermal motion? (NEET 2009)

(A) $5 \times 10^{4} \mathrm{~J}$

(B) $6 \times 10^{4} J$

(C) $7 \times 10^{4} J$

(D) $3 \times 10^{4} \mathrm{~J}$


14. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature $T_{0}$, while Box contains one mole of helium at temperature $\left(\dfrac{7}{3}\right) T_{0}$. The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, $T_{f}$ in terms of $T_{0}$ is (NEET 2010)

(A) $T_{f}=\dfrac{3}{7} T_{0}$

(B) $T_{f}=\dfrac{7}{3} T_{0}$

(C) $T_{f}=\dfrac{3}{2} T_{0}$

(D) $T_{f}=\dfrac{5}{2} T_{0}$


15. Two thermally insulated vessels 1 and 2 are filled with air at temperatures $\left(T_{1}, T_{2}\right)$, volume $\left(V_{1}, V_{2}\right)$ and pressure $\left(P_{1}, P_{2}\right)$ respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be(NEET 2004)

(A) $T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right) /\left(P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}\right)$

(B) $\left(T_{1}+T_{2}\right) / 2$

(C) $T_{1}+T_{2}$

(D) $T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right) /\left(P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}\right)$


16. A carnot engine takes $3 \times 10^{6} \mathrm{cal}$. of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is(NEET 2003)

(A) $4.2 \times 10^{6} \mathrm{~J}$

(B) $8.4 \times 10^{6} \mathrm{~J}$

(C) $16.8 \times 10^{6} \mathrm{~J}$

(D) Zero


17. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_{p} / C_{V}$ for the gas is (NEET 2002)

(A) $\dfrac{4}{3}$

(B) 2

(C) $\dfrac{5}{3}$

(D) $\dfrac{3}{2}$


18. 1 mole of a gas with $\gamma=7 / 5$ is mixed with 1 mole of a gas with $\gamma=5 / 3$, then the value of $\gamma$ for the resulting mixture is(NEET 2002)

(A) $7 / 5$

(B) $2 / 5$

(C) $3 / 2$

(D) $12 / 7$


19. At what temperature is the $r . m . s$ velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} C$? (NEET 2002)

(A) $80 K$

(B) $-73 \mathrm{~K}$

(C) $3 K$

(D) $20 K$


20. A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by: (NEET 2002)

(A) $\dfrac{(\gamma-1)}{2 \gamma R} M v^{2} K$

(B) $\dfrac{\gamma M^{2} v}{2 R} K$

(C)$\dfrac{(\gamma-1)}{2 R} M v^{2} K$

(D) $\dfrac{(\gamma-1)}{2(\gamma+1) R} M v^{2} K$


21. A homogeneous metallic rod rotates with constant angular speed about its perpendicular bisector. If it is heated uniformly to slightly boost its temperature 

(A) its speed of rotation increases.

(B) its speed of rotation decreases.

(C) its speed of rotation remains same.

(D) its speed increases because its moment of inertia increases.


22. The time period of a pendulum increases as the temperature rises. 

(A) grows as its effective length increases, despite the fact that its centre of mass remains at the centre of the bob.

(B) drops as its effective length rises, despite the fact that its centre of mass stays at the centre of the bob.

(C) increases as its effective length increases due to the bob's centre of mass shifting below the centre.

(D) diminishes as its effective length remains constant but its centre of mass shifts above the bob's centre.


23. Heat is connected to: 

(A) kinetic energy of random motion of molecules.

(B) kinetic energy of orderly motion of molecules.

(C) total kinetic energy of random and orderly motion of molecules.

(D) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other.


24. A sphere, a cube, and a thin circular plate of the same material and mass are all heated to the same high temperature at the start. 

(A) Plate will cool fastest and cube the slowest

(B) Sphere will cool fastest and cube the slowest

(C) Plate will cool fastest and sphere the slowest

(D) Cube will cool fastest and plate the slowest.


25. 'Gulab Jamuns' (assumed to be round) are to be baked. They come in two sizes, one twice as large (in radius) as the other. Pizzas (presumed to be discs) must also be heated in the oven. They also come in two sizes, one twice as large (in radius) as the other. All four are combined and heated to oven temperature. Select the appropriate option from the following: 

(A) Both size gulab jamuns will get heated in the same time.

(B) Smaller gulab jamuns are heated before bigger ones.

(C) Smaller pizzas are heated before bigger ones.

(D) Bigger pizzas are heated before smaller ones.


Solution:

  1. (A)

Experiments have demonstrated that low density gases have comparable expansion behaviour. If we keep the volume of the gas constant, the pressure rises monotonically as the temperature rises. As a result, any low density gas can be used for a constant volume gas thermometer.


  1. (B)

A mercury manometer is coupled to a gas bulb in a constant volume gas thermometer. The volume of the bulb is kept constant by external techniques as the temperature of the bulb rises, and the pressure of the gas rises. This pressure is used to define the temperature scale. (T = kP, where k is an integer.)


  1. (A)

The circumference of the hole is composed of metal. When the sheet is heated, the circular circumference expands, increasing the size of the hole. The hole will not distort since the sheet is uniform in all directions from the centre of the hole and is likewise uniformly heated.


  1. (B)

The calorimeter principle asserts that heat lost by a hot body is the same as heat absorbed by a cool body if no heat is lost to the surroundings. The opposite assertion, that the quantity of heat flow from a sample to two separate samples of the same mass is a product of their specific heats and temperature differences, is correct, but it is not a calorimetric concept. It is simply a mathematical method of comprehending heat transport.


  1. (A)

When the vapour pressure of a liquid equals the air pressure, boiling occurs. The vapour pressure rises as the temperature rises. When the temperature rises sufficiently, boiling begins. However, as pressure increases, higher vapour pressure must be reached, which is obtained by raising temperature, and therefore the boiling point rises.


  1. (C)

Water can exist in all three phases that are thermally balanced with one another. It is known as the triple point of water. When a substance is heated, it can go from solid to gas. This is known as sublimation. The boiling point rises as pressure falls. At constant pressure, phase change occurs.


  1. (B)

Conduction is the transmission of heat energy caused by the vibration of molecules at their locations, while convection is the transfer of heat energy caused by the motion of molecules themselves, and radiation is the transfer of heat energy caused by electromagnetic waves. When water is placed in an extremely hot pan, the water molecules in direct contact with the pan become heated and migrate away, allowing the colder molecules to settle down and touch the pan, thereby becoming heated. The movement of heated molecules represents convection.


  1. (B)

Conduction heat transfer happens as a result of an initial temperature difference. Heat flow per unit time for a given temperature differential can quantify conduction heat flow. It is important to note that temperature differences induce heat flow, not the other way around.


  1. (A)

Radiation does not require any medium for heat transmission. When a cooking pan is heated on a gas stove, conduction is the process that transfers heat from the flame to the pan. Radiation is the transfer of energy by electromagnetic waves known as radiant energy.


  1. (D)

When one end of a rod is held in fire. The temperature at that end will be equivalent to that of fire. The temperature will steadily drop towards the other end. At steady state, the rate of heat transfer is constant, and the temperature at each location is constant but varies non-uniformly over the length of the rod.


  1.  (A)

$\Delta U=\Delta Q=m c \Delta T$

$=100 \times 10^{-3} \times 4184(50-30) \approx 8.4 k J$


  1.  (B)

$T_{1} V^{\gamma-1}=T_{2}(32 V)^{\gamma-1}$

$\Rightarrow T_{1}=(32)^{\gamma-1} \cdot T_{2}$

For diatomic gas, $\gamma=\dfrac{7}{5}$

$\therefore \gamma-1=\dfrac{2}{5}$

$\therefore T_{1}=(32)^{\dfrac{2}{5}} \cdot T_{2} \Rightarrow T_{1}=4 T_{2}$

Now, efficiency $=1-\dfrac{T_{2}}{T_{1}}$

$=1-\dfrac{T_{2}}{4 T_{2}}$

$=1-\dfrac{1}{4}$

$=\dfrac{3}{4}$

$=0.75 .$

 

  1. (A)

Volume $=\dfrac{\text { mass }}{\text { density }}=\dfrac{1}{4} m^{3}$

$K . E=\dfrac{5}{2} P V$

$=\dfrac{5}{2} \times 8 \times 10^{4} \times \dfrac{1}{4}$

$=5 \times 10^{4} \mathrm{~J}$


  1. (C)

Heat lost by $\mathrm{He}=$ Heat gained by $N_{2}$

$n_{1} C v_{1} \Delta T_{1}=n_{2} C v_{2} \Delta T_{2}$

$\dfrac{3}{2} R\left[\dfrac{7}{3} T_{0}-T_{f}\right]$

$=\dfrac{5}{2} R\left[T_{f}-T_{0}\right] \Rightarrow T_{f} $

$=\dfrac{3}{2} T_{0}$


  1. (A)

Here $Q=0$ and $W=0$. Therefore from first law of thermodynamics $\Delta U=Q+W=0$

$\therefore$ Internal energy of the system with partition $=$ Internal energy of the system without partition.

$n_{1} C_{v} T_{1}+n_{2} C_{v} T_{2}=\left(n_{1}+n_{2}\right) C_{v} T $

$\therefore T=\dfrac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}}$

But $n_{1}=\dfrac{P_{1} V_{1}}{R T_{1}}$ and $n_{2}=\dfrac{P_{2} V_{2}}{R T_{2}}$

$\therefore T=\dfrac{\dfrac{P_{1} V_{1}}{R T_{1}} \times T_{1}+\dfrac{P_{2} V_{2}}{R T_{2}} \times T_{2}}{\dfrac{P_{1} V_{1}}{R T_{1}}+\dfrac{P_{2} V_{2}}{R T_{2}}} $

$=\dfrac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}$


  1. (B)

$\eta=1-\dfrac{T_{2}}{T_{1}}$

$=1-\dfrac{(273+27)}{(273+627)}$

$=1-\dfrac{300}{900}$

$=1-\dfrac{1}{3}=\dfrac{2}{3}$

But $\eta=\dfrac{W}{Q}$

$\therefore \dfrac{W}{Q}=\dfrac{2}{3} \Rightarrow W$

$=\dfrac{2}{3} \times Q$

$=\dfrac{2}{3} \times 3 \times 10^{6}$

$=2 \times 10^{6} \mathrm{cal}$

$=2 \times 10^{6} \times 4.2 \mathrm{~J}$

$=8.4 \times 10^{6} \mathrm{~J}$


  1. (D)

$P \propto T^{3} \Rightarrow P T^{-3}=$ constant .... $(i)$

But for an adiabatic process, the pressure temperature relationship is given by $P^{1-\gamma} T^{\gamma}=$ constant $\Rightarrow P T^{\dfrac{\gamma}{1-\gamma}}=$ constant. .... $(i i)$

From $(i)$ and $(i i) \dfrac{\gamma}{1-\gamma}=-3 \Rightarrow \gamma=-3+3 \gamma \Rightarrow \gamma=\dfrac{3}{2}$


  1. (C)

If $n_{1}$ moles of adiabatic exponent $\gamma_{1}$ is mixed with $n_{2}$ moles of adiabatic exponent $\gamma_{2}$ then the adiabatic component of the resulting mixture is given by

$\dfrac{n_{1}+n_{2}}{\gamma-1}=\dfrac{n_{1}}{\gamma_{1}-1}+\dfrac{n_{2}}{\gamma_{2}-1} $

$\dfrac{1+1}{\gamma-1}=\dfrac{1}{\dfrac{7}{5}-1}+\dfrac{1}{\dfrac{5}{3}-1}$

$\therefore \dfrac{2}{\gamma-1}=\dfrac{5}{2}+\dfrac{3}{2}=4$

$\therefore 2=4 \gamma-4 \Rightarrow \gamma=\dfrac{6}{4}=\dfrac{3}{2}$


  1. (D)

$v_{r m s}=\sqrt{\dfrac{R T}{M}}$

For $v_{r m s}$ to be equal $\dfrac{T_{H_{2}}}{M_{H_{2}}}=\dfrac{T_{O_{2}}}{M_{O_{2}}}$

Here $M_{H_{2}}=2 ; M_{O_{2}}=32$

$T_{O_{2}}=47+273=320 K$

$\therefore \dfrac{T_{H_{2}}}{2}=\dfrac{320}{32}$

$\Rightarrow T_{H_{2}}=20 K$

$v_{r m s}=\sqrt{\dfrac{R T}{M}}$

For $v_{r m s}$ to be equal $\dfrac{T_{H_{2}}}{M_{H_{2}}}=\dfrac{T_{O_{2}}}{M_{O_{2}}}$

Here $M_{H_{2}}=2 ; M_{O_{2}}=32$

$T_{O_{2}}=47+273=320 K$

$\therefore \dfrac{T_{H_{2}}}{2}=\dfrac{320}{32}$

$\Rightarrow T_{H_{2}}=20 K$


  1. (C)

Here, work done is zero.

So, loss in kinetic energy $=$ change in internal energy of gas

$\dfrac{1}{2} m v^{2}=n C_{v} \Delta T=n \dfrac{R}{\gamma-1} \Delta T $

$\dfrac{1}{2} m v^{2}=\dfrac{m}{M} \dfrac{R}{\gamma-1} \Delta T $

$\therefore \Delta T=\dfrac{M v^{2}(\gamma-1)}{2 R} K$


  1. (B)

The moment of inertia of a regular solid rod about its perpendicular bisector is given by $(I)=m L^{2}$, where $m$ is the rod's mass and $L$ is the rod's length.

The angular momentum of a rotating object is given by $L=I \omega$ where $\omega$ is the object's angular velocity.

In the absence of any external torque on the body, like in the case of the provided question, the angular momentum remains constant, implying that as the moment of inertia grows, so must the angular velocity.

We've been told that heat is being delivered to the rod at a consistent pace, indicating that the rod's temperature will rise and expand linearly. As we have established a relationship between the two values above, the rod's moment of inertia will grow as the length of the rod increases. When heat is applied to the rod, its moment of inertia increases, and a higher moment of inertia equals a lower angular velocity. As a result, we can say that the rotation's speed will slow.


  1. (A)

First, define the term "simple pendulum." A simple pendulum is an ideal pendulum composed of a point mass hung by a weightless inextensible perfectly flexible thread and then free to oscillate without friction. This is what distinguishes it from a physical pendulum.

A basic pendulum's time period is defined as the time for one complete cycle, a left swing and a right swing, which is referred to as a period. The time period of a pendulum is determined by its length and, to a lesser extent, by its amplitude, or the width of its swing.

The following formula defines the time period of a basic pendulum:

$T=2 \pi \sqrt{\dfrac{L}{g}}$......(i)

L denotes the effective length of the pendulum in the preceding expression, and g is the acceleration due to gravity.

We know that as the temperature rises, so does the effective length or L of the simple pendulum. This occurs even when the centre of mass remains at the centre of the bob.

From the equation (1) we can say that $T \alpha \sqrt{L}$

Thus, the $\mathrm{T}$ rises as the temperature increases.


  1. (A)

Heating a substance stimulates the molecules of that substance, causing the molecules to move vigorously, giving the molecules velocity, and we know kinetic energy of a body is by virtue of its motion.


  1. (C)

The masses of the provided forms are all the same. The surface area of the round plate will be the greatest of the three. This will allow the plate to cool down more quickly. The sphere will have the smallest surface area for a given volume of the three shapes. As a result, the sphere will cool the slowest. As a result, the plate will cool the fastest and the sphere will cool the slowest.

(C) is the correct answer: the plate will cool the fastest and the spherical will cool the slowest.


  1. (B and C)

As in an oven, when we set any thing to be cooked in it and turn it on, the electromagnetic radiation created inside it begins to change in thermal energy, which is the cause of the cooked substance that is placed in the oven. Since the heating time of a substance is determined by its surface area, the larger the surface area, the longer it will take to cook, and the smaller the surface area, the sooner it will cook.

From the specified sizes of Gulab Jamuns, the smaller one will have a smaller surface area than the larger one, and among the pizzas, the pizza with the smaller radius will be heated and baked before the larger one. As a result, smaller gulab jamuns are prepared before larger ones, and smaller pizzas are heated before larger ones.

As a result, B and C are the correct answers.


Importance of NEET Physics Thermal Properties of Matter

Thermal Properties of Matter is an essential chapter for NEET students as it has significant weightage in the examination. This chapter introduces the students to different laws of heat and temperature. They can familiarise themselves with the concepts of calorimetry, thermal expansion, transfer of heat and much more. These scientific processes and concepts will be used throughout the chapter. 


The introduction to the chapter talks about the measurement of temperature and the Celsius-Fahrenheit scale. Students learn the SI unit of temperature, which is Kelvin (K). They are also taught how to convert Celsius into Fahrenheit and vice-versa. Delving deeper into the chapter, students are taught about the Absolute Temperature Scale, Thermal Expansion and its types, Volume Expansion Coefficient, and much more. 


The chapter also delves into the concepts of heat and teaches students about the process of calorimetry. They can learn about the properties of heat, Specific Heat Capacity, Molar Heat Capacity, Latent Heat of Fusion and Vaporisation, and Specific Latent Heat. Furthermore, the chapter teaches students about different modes of heat transfer such as Conduction, Radiation, and Convection. They also learn about Newton’s Law of Cooling and its mathematical representation. 


The details provided in the chapter along with the Thermal Properties of Matter Class 11 important questions and answers will help students tackle different questions in the competitive examinations. They can learn from the solutions and practice to be prepared for their entrance test. 


Benefits of Thermal Properties of Matter Important Questions

  • The subject matter experts at Vedantu have explained all the concepts and principles relevant to the chapter in the solutions for the important questions. Students will be able to build a strong foundation for the chapter and implement them to solve calculative and explanatory questions efficiently. 

  • Students can understand the meaning of different thermal properties which the states of matter have. They can imbibe the concepts of Calorimetry, Heat Transfer, Temperature Scales, and much more without any difficulty with the help of the explanations provided for Thermal Properties of Matter NEET Questions

  • Preparing for the NEET examination will be much easier by solving the important questions that are related to this topic. Students can figure out the answers to numerical problems, MCQs, and explanatory answers and cross-check the given solutions to rectify any mistakes. 


Download Thermal Properties of Matter Important Questions Free PDF

Get your hands on the PDF version of the Thermal Properties of Matter Class 11 important questions and complete your study material. Learn how different states of matter react to temperature and heat. Find out how to solve numerical questions step-by-step with the help of Vedantu’s resources. This is your chance to download the files and start your preparation for NEET now.

FAQs on Thermal Properties of Matter Class 11 Important Questions NEET Physics [Free PDF Download]

1. What are the different types of thermal properties?

There are basically three different types of thermal properties for different states of matter. These properties are thermal conductivity, thermal diffusivity, and specific heat. 

2. Mention the 4 different properties of heat.

The four major properties of heat are thermal expansion, thermal stress, heat capacity, and thermal conductivity.

3. What is thermal conduction?

The process that results in the transfer of thermal energy from a hotter part of a body to a colder part without transferring any material particles is called thermal conduction.

4. Why is it impossible to have a negative reading on the Kelvin temperature scale?

Since the minimum possible temperature in the Kelvin scale is absolute zero, it is not possible to have a negative reading.