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Non-Rationalised Science NCERT Notes and Solutions (Class 6th to 10th)
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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
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Physics Chemistry Biology

Class 11th (Physics) Chapters
1. Physical World 2. Units And Measurements 3. Motion In A Straight Line
4. Motion In A Plane 5. Laws Of Motion 6. Work, Energy And Power
7. System Of Particles And Rotational Motion 8. Gravitation 9. Mechanical Properties Of Solids
10. Mechanical Properties Of Fluids 11. Thermal Properties Of Matter 12. Thermodynamics
13. Kinetic Theory 14. Oscillations 15. Waves



Chapter 11 Thermal Properties Of Matter



Introduction

The concepts of heat and temperature are fundamental to our understanding of the physical world. While we intuitively understand temperature as a measure of the 'hotness' or 'coldness' of an object, physics requires a more precise definition. This chapter delves into the thermal properties of matter, defining heat and temperature, exploring how they are measured, and examining the processes by which heat is transferred.

We will study why materials expand when heated, a principle used by blacksmiths to fit iron rings onto wooden wheels. We will also investigate the phenomena of phase changes, such as water boiling or freezing, where temperature remains constant despite the continuous flow of heat. Understanding these concepts is essential for numerous applications, from engineering design to explaining everyday phenomena like the changing direction of sea breezes.



Temperature and Heat

Temperature is a physical quantity that provides a relative measure of how hot or cold an object is. An object with a higher temperature is considered hotter than an object with a lower temperature. Our sense of touch gives us a qualitative idea of temperature, but it is unreliable and limited for scientific purposes.

When two systems at different temperatures are brought into contact, energy is transferred from the hotter system to the colder one. This transfer of energy continues until both systems reach the same temperature, a state known as thermal equilibrium.

Heat (Q) is defined as the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. Heat always flows from a region of higher temperature to a region of lower temperature.

When a body is heated, it can undergo several changes, such as a rise in temperature, thermal expansion, or a change of state (e.g., melting or boiling).



Measurement of Temperature

Temperature is measured using a device called a thermometer. Thermometers work based on a physical property of a substance that changes predictably with temperature. This property is called the thermometric property.

The most common thermometric property is the thermal expansion of a liquid. Liquid-in-glass thermometers (using mercury or alcohol) rely on the principle that the volume of the liquid expands or contracts linearly with temperature over a wide range.

Temperature Scales

To define a temperature scale, two fixed reference points are needed. The most common reference points are:

Celsius Scale (°C)

On the Celsius scale, the ice point is defined as 0 °C and the steam point as 100 °C. The interval between these points is divided into 100 equal parts.

Fahrenheit Scale (°F)

On the Fahrenheit scale, the ice point is 32 °F and the steam point is 212 °F. The interval is divided into 180 equal parts.

Conversion Formula

The relationship between the Celsius temperature ($t_C$) and Fahrenheit temperature ($t_F$) is linear and given by the formula:

$\frac{t_F - 32}{180} = \frac{t_C}{100}$

This simplifies to:

$t_F = \frac{9}{5}t_C + 32$



Ideal-Gas Equation and Absolute Temperature

Thermometers using different liquids can give slightly different readings between the fixed points due to non-uniform expansion. A gas thermometer, especially using a low-density gas, provides a more fundamental and consistent measure of temperature.

The state of a gas is described by its pressure (P), volume (V), and temperature (T). For low-density gases, these variables are related by the ideal-gas law:

$PV = \mu RT$

where $\mu$ is the number of moles of the gas and $R$ is the universal gas constant ($R \approx 8.31 \text{ J mol}^{-1} \text{ K}^{-1}$).

Absolute Temperature Scale (Kelvin Scale)

In a constant-volume gas thermometer, the volume of the gas is kept fixed, so the ideal gas law simplifies to $P \propto T$. This means the pressure of the gas can be used as a direct measure of temperature.

If we plot the pressure of a low-density gas versus its temperature in Celsius and extrapolate the line to zero pressure, all gases are found to intersect the temperature axis at the same point: -273.15 °C. This temperature is called absolute zero. It is the theoretical temperature at which a gas would exert no pressure.

A plot of pressure versus temperature for different low-density gases. All lines extrapolate to the same point at -273.15 °C, which is absolute zero.

The Kelvin temperature scale (K) is based on this absolute zero. On this scale, 0 K is defined as -273.15 °C.

The size of a kelvin is the same as the size of a degree Celsius. The relationship between the Kelvin temperature (T) and the Celsius temperature ($t_C$) is:

$T(\text{K}) = t_C(^\circ\text{C}) + 273.15$



Thermal Expansion

Most substances expand when their temperature increases and contract when their temperature decreases. This phenomenon is called thermal expansion. The change in dimensions is a result of the increased kinetic energy of the atoms, which causes them to vibrate with greater amplitude and increase the average interatomic distance.

1. Linear Expansion

This refers to the expansion in one dimension, such as the length of a rod.

The fractional change in length ($\Delta l/l$) is directly proportional to the change in temperature ($\Delta T$).

$\frac{\Delta l}{l} = \alpha_l \Delta T$

where $\alpha_l$ is the coefficient of linear expansion. Its unit is K⁻¹ or °C⁻¹.


2. Area Expansion

This refers to the expansion in two dimensions, such as the area of a sheet.

$\frac{\Delta A}{A} = \alpha_A \Delta T$

where $\alpha_A$ is the coefficient of area expansion. For an isotropic material, $\alpha_A = 2\alpha_l$.


3. Volume Expansion

This refers to the expansion in three dimensions, such as the volume of a cube or a liquid.

$\frac{\Delta V}{V} = \alpha_V \Delta T$

where $\alpha_V$ is the coefficient of volume expansion. For an isotropic solid, $\alpha_V = 3\alpha_l$.


Anomalous Expansion of Water

Water exhibits an unusual behaviour. Instead of expanding continuously on heating, it contracts when heated from 0 °C to 4 °C. This means that water has its maximum density at 4 °C. Above 4 °C, it expands normally.

This anomalous property is crucial for aquatic life in cold climates. As a lake cools, the surface water at 4 °C is the densest and sinks. Water cooler than 4 °C is less dense and stays at the surface, where it eventually freezes. This allows the water at the bottom of the lake to remain at 4 °C, enabling fish and other organisms to survive.

Graphs showing the variation of volume and density of water with temperature, illustrating the minimum volume and maximum density at 4°C.

Example 1. A blacksmith fixes an iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim is 5.243 m and the diameter of the iron ring is 5.231 m, both at 27 °C. To what temperature should the ring be heated so as to fit the rim of the wheel? (Coefficient of linear expansion of iron, $\alpha_l = 1.20 \times 10^{-5} \text{ K}^{-1}$).

Answer:

Let the initial temperature be $T_1 = 27$ °C. Let the final temperature be $T_2$.

The initial length (diameter) of the ring is $L_1 = 5.231$ m.

The ring must be heated so that its final length (diameter) becomes $L_2 = 5.243$ m to fit the rim.

We use the formula for linear expansion: $L_2 = L_1(1 + \alpha_l (T_2 - T_1))$.

$5.243 = 5.231 [1 + (1.20 \times 10^{-5})(T_2 - 27)]$

$\frac{5.243}{5.231} = 1 + (1.20 \times 10^{-5})(T_2 - 27)$

$1.00229 = 1 + (1.20 \times 10^{-5})(T_2 - 27)$

$0.00229 = (1.20 \times 10^{-5})(T_2 - 27)$

$T_2 - 27 = \frac{0.00229}{1.20 \times 10^{-5}} \approx 191$

$T_2 = 191 + 27 = 218$ °C.

The ring should be heated to approximately 218 °C.



Specific Heat Capacity

The amount of heat required to change the temperature of a substance depends on its mass, the desired temperature change, and the nature of the substance itself.

The heat capacity (S) of a substance is the amount of heat ($\Delta Q$) required to change its temperature by $\Delta T$.

$S = \frac{\Delta Q}{\Delta T}$ (Unit: J/K)

A more useful property is the specific heat capacity (s), which is the heat capacity per unit mass.

Specific Heat Capacity (s): It is the amount of heat required to raise the temperature of a unit mass of a substance by one degree (1 K or 1 °C).

$s = \frac{1}{m} \frac{\Delta Q}{\Delta T}$

From this, the heat absorbed or released by a mass $m$ during a temperature change $\Delta T$ is:

$\Delta Q = ms \Delta T$

Molar Specific Heat Capacity (C)

If the amount of substance is given in moles ($\mu$), we define the molar specific heat capacity (C).

$C = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$

The SI unit is J mol⁻¹ K⁻¹.

For gases, the specific heat capacity depends on the process. The two important types are:



Calorimetry

Calorimetry is the measurement of heat. It is based on the principle of conservation of energy applied to heat transfer.

Principle of Calorimetry: In an isolated system, when a hot body is brought into contact with a cold body, the heat lost by the hot body is equal to the heat gained by the cold body.

Heat Lost = Heat Gained

This principle is used to determine quantities like specific heat capacity or latent heat. The device used for these measurements is a calorimeter. A simple calorimeter consists of an insulated metallic vessel (often copper) with a stirrer and a lid for a thermometer. The insulation minimizes heat exchange with the surroundings.


Example 2. A sphere of 0.047 kg aluminum at 100 °C is placed in a 0.14 kg copper calorimeter containing 0.25 kg of water at 20 °C. The final steady temperature is 23 °C. Calculate the specific heat capacity of aluminum ($s_{Al}$).

(Given: $s_{water} = 4186 \text{ J kg}^{-1} \text{K}^{-1}$, $s_{copper} = 386 \text{ J kg}^{-1} \text{K}^{-1}$)

Answer:

According to the principle of calorimetry, Heat Lost by aluminum = Heat Gained by water + Heat Gained by calorimeter.

Heat Lost by Aluminum Sphere:

Mass $m_{Al} = 0.047$ kg. Temperature change $\Delta T_{Al} = 100^\circ\text{C} - 23^\circ\text{C} = 77^\circ\text{C}$.

Heat Lost = $m_{Al} s_{Al} \Delta T_{Al} = (0.047) s_{Al} (77)$.

Heat Gained by Water:

Mass $m_{w} = 0.25$ kg. Temperature change $\Delta T_{w} = 23^\circ\text{C} - 20^\circ\text{C} = 3^\circ\text{C}$.

Heat Gained = $m_{w} s_{w} \Delta T_{w} = (0.25)(4186)(3) = 3139.5$ J.

Heat Gained by Copper Calorimeter:

Mass $m_{cu} = 0.14$ kg. Temperature change $\Delta T_{cu} = 23^\circ\text{C} - 20^\circ\text{C} = 3^\circ\text{C}$.

Heat Gained = $m_{cu} s_{cu} \Delta T_{cu} = (0.14)(386)(3) = 162.12$ J.

Equating Heat Lost and Gained:

$(0.047) s_{Al} (77) = 3139.5 + 162.12$

$3.619 s_{Al} = 3301.62$

$s_{Al} = \frac{3301.62}{3.619} \approx 912.3 \text{ J kg}^{-1} \text{K}^{-1}$.

The specific heat capacity of aluminum is approximately 912 J kg⁻¹ K⁻¹.



Change of State

Matter can exist in three primary states: solid, liquid, and gas. A transition from one state to another is called a change of state or phase transition. These transitions occur at a constant temperature when heat is either supplied to or removed from the substance.

A graph of temperature versus heat added to a sample of water, starting as ice at -10°C and ending as steam. Plateaus are seen at 0°C (melting) and 100°C (boiling).

Melting and Fusion

Vaporization and Condensation

Sublimation

Some substances can change directly from the solid state to the vapor state without passing through the liquid phase. This process is called sublimation. Examples include dry ice (solid CO₂) and iodine.

Latent Heat

During a phase change, the heat supplied does not change the temperature but is used to change the state. This "hidden" heat is called latent heat (L).

The heat Q required to change the state of a mass $m$ of a substance is given by:

$Q = mL$

where L is the specific latent heat of the substance for that transition.

The very high latent heat of vaporization of water is why steam at 100 °C can cause much more severe burns than boiling water at 100 °C.



Heat Transfer

Heat is transferred from a region of higher temperature to a region of lower temperature through three distinct modes: conduction, convection, and radiation.

1. Conduction

Conduction is the transfer of heat through a material without any bulk movement of the material itself. It occurs primarily through molecular collisions. Metals are good conductors of heat because their free electrons can transport energy quickly. Gases are poor conductors.

The rate of heat flow (or heat current), $H$, through a uniform bar of length $L$ and cross-sectional area $A$, with its ends at temperatures $T_C$ and $T_D$ ($T_C > T_D$) is given by:

$H = \frac{dQ}{dt} = KA \frac{T_C - T_D}{L}$

where $K$ is the thermal conductivity of the material. Its SI unit is W m⁻¹ K⁻¹.

A rod with ends at different temperatures, showing heat flow H by conduction from the hot end to the cold end.

2. Convection

Convection is the transfer of heat through the actual bulk movement of a fluid (liquid or gas).


3. Radiation

Radiation is the transfer of heat in the form of electromagnetic waves. This mode of heat transfer does not require any material medium and can travel through a vacuum. This is how the Earth receives heat from the Sun.

All objects with a temperature above absolute zero emit thermal radiation. The rate at which an object radiates energy depends on its temperature, surface area, and the nature of its surface.

Stefan-Boltzmann Law

The total energy radiated per unit time (Power, $H$) by a perfect blackbody (a perfect emitter and absorber) is proportional to the fourth power of its absolute temperature ($T$) and its surface area ($A$).

$H = \sigma A T^4$

where $\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}$ is the Stefan-Boltzmann constant.

For a real body, the law is modified by its emissivity ($e$), a number between 0 and 1:

$H = e \sigma A T^4$

If a body at temperature $T$ is in surroundings at temperature $T_s$, the net rate of heat loss by radiation is:

$H_{net} = e \sigma A (T^4 - T_s^4)$

Wien's Displacement Law

The wavelength ($\lambda_m$) at which the radiated energy is maximum is inversely proportional to the absolute temperature ($T$) of the body.

$\lambda_m T = \text{constant}$ (Wien's constant $\approx 2.9 \times 10^{-3} \text{ m K}$)

This law explains why the color of a heated object changes from red to yellow to white as its temperature increases.



Newton’s Law of Cooling

Newton's law of cooling provides a simple model for the rate at which a hot body cools down when the temperature difference between the body and its surroundings is small.

Newton's Law of Cooling: The rate of loss of heat ($-dQ/dt$) of a body is directly proportional to the temperature difference ($\Delta T$) between the body ($T_2$) and its surroundings ($T_1$).

$-\frac{dQ}{dt} = k(T_2 - T_1)$

where $k$ is a constant that depends on the surface area and nature of the body.

Since the heat lost is also related to the temperature drop of the body ($dQ = ms dT_2$), the law can be written in terms of the rate of cooling:

$-ms \frac{dT_2}{dt} = k(T_2 - T_1) \implies \frac{dT_2}{dt} = -K(T_2 - T_1)$

where $K=k/ms$ is another constant. This differential equation shows that the rate of cooling is greatest when the body is hottest and decreases as the body's temperature approaches the temperature of the surroundings.

The solution to this equation is an exponential decay function:

$T_2(t) = T_1 + (T_{2, initial} - T_1)e^{-Kt}$


Example 3. A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?

Answer:

We use the approximate form of Newton's law of cooling: $\frac{\Delta T}{\Delta t} = -K(T_{avg} - T_{surroundings})$.

Case 1: Cooling from 94 °C to 86 °C

Change in temperature, $\Delta T = 86 - 94 = -8$ °C.

Time interval, $\Delta t = 2$ min.

Average temperature of the body, $T_{avg} = \frac{94+86}{2} = 90$ °C.

Temperature of surroundings, $T_{surroundings} = 20$ °C.

Temperature difference, $T_{avg} - T_{surroundings} = 90 - 20 = 70$ °C.

Applying the law: $\frac{8^\circ\text{C}}{2 \text{ min}} = K(70^\circ\text{C})$. (We use the magnitude of temperature drop).

This gives $4 = 70K \implies K = \frac{4}{70} = \frac{2}{35}$ min⁻¹.

Case 2: Cooling from 71 °C to 69 °C

Change in temperature, $\Delta T = 71 - 69 = 2$ °C.

Time interval, $\Delta t = ?$

Average temperature of the body, $T_{avg} = \frac{71+69}{2} = 70$ °C.

Temperature difference, $T_{avg} - T_{surroundings} = 70 - 20 = 50$ °C.

Applying the law: $\frac{2^\circ\text{C}}{\Delta t} = K(50^\circ\text{C})$.

Substitute the value of K from Case 1:

$\frac{2}{\Delta t} = \left(\frac{2}{35}\right)(50)$

$\frac{1}{\Delta t} = \frac{50}{35} = \frac{10}{7}$

$\Delta t = \frac{7}{10} = 0.7$ minutes.

$0.7 \text{ min} \times 60 \text{ s/min} = 42$ seconds.



Exercises



Question 11.1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

Question 11.2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between $T_A$ and $T_B$ ?

Answer:

Question 11.3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :

$R = R_o [1 + \alpha (T – T_o )]$

The resistance is 101.6 $\Omega$ at the triple-point of water 273.16 K, and 165.5 $\Omega$ at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 $\Omega$ ?

Answer:

Question 11.4. Answer the following :

(a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?

(c) The absolute temperature (Kelvin scale) T is related to the temperature $t_c$ on the Celsius scale by

$t_c = T – 273.15$

Why do we have 273.15 in this relation, and not 273.16 ?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ?

Answer:

Question 11.5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

Temperature Pressure thermometer A Pressure thermometer B
Triple-point of water $1.250 \times 10^5$ Pa $0.200 \times 10^5$ Pa
Normal melting point of sulphur $1.797 \times 10^5$ Pa $0.287 \times 10^5$ Pa

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?

Answer:

Question 11.6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = $1.20 \times 10^{–5} \text{ K}^{–1}$.

Answer:

Question 11.7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : $\alpha_{steel} = 1.20 \times 10^{–5} \text{ K}^{–1}$.

Answer:

Question 11.8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = $1.70 \times 10^{–5} \text{ K}^{–1}$.

Answer:

Question 11.9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = $2.0 \times 10^{–5} \text{ K}^{–1}$; Young’s modulus of brass = $0.91 \times 10^{11}$ Pa.

Answer:

Question 11.10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = $2.0 \times 10^{–5} \text{ K}^{–1}$, steel = $1.2 \times 10^{–5} \text{ K}^{–1}$).

Answer:

Question 11.11. The coefficient of volume expansion of glycerine is $49 \times 10^{–5} \text{ K}^{–1}$. What is the fractional change in its density for a 30 °C rise in temperature ?

Answer:

Question 11.12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g$^{–1}$ K$^{–1}$.

Answer:

Question 11.13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g$^{–1}$ K$^{–1}$; heat of fusion of water = 335 J g$^{–1}$).

Answer:

Question 11.14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm$^3$ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?

Answer:

Question 11.15. Given below are observations on molar specific heats at room temperature of some common gases.

Gas Molar specific heat ($C_v$) (cal mol$^{–1}$ K$^{–1}$)
Hydrogen4.87
Nitrogen4.97
Oxygen5.02
Nitric oxide4.99
Carbon monoxide5.01
Chlorine6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?

Answer:

Question 11.16. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g$^{–1}$.

Answer:

Question 11.17. A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s$^{–1}$ m$^{–1}$ K$^{–1}$. [Heat of fusion of water = $335 \times 10^3$ J kg$^{–1}$]

Answer:

Question 11.18. A brass boiler has a base area of 0.15 m$^2$ and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s$^{–1}$ m$^{–1}$ K$^{–1}$ ; Heat of vaporisation of water = $2256 \times 10^3$ J kg$^{–1}$.

Answer:

Question 11.19. Explain why :

(a) a body with large reflectivity is a poor emitter

(b) a brass tumbler feels much colder than a wooden tray on a chilly day

(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

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

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

Answer:

Question 11.20. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answer:

ADDITIONAL EXERCISES

Question 11.21. Answer the following questions based on the P-T phase diagram of carbon dioxide:

(a) At what temperature and pressure can the solid, liquid and vapour phases of CO$_2$ co-exist in equilibrium ?

(b) What is the effect of decrease of pressure on the fusion and boiling point of CO$_2$ ?

(c) What are the critical temperature and pressure for CO$_2$ ? What is their significance ?

(d) Is CO$_2$ solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm ?

Answer:

Question 11.22. Answer the following questions based on the P – T phase diagram of CO$_2$:

(a) CO$_2$ at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase ?

(b) What happens when CO$_2$ at 4 atm pressure is cooled from room temperature at constant pressure ?

(c) Describe qualitatively the changes in a given mass of solid CO$_2$ at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.

(d) CO$_2$ is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe ?

Answer: