1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter

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Karnataka 1st PUC Physics Question Bank Chapter 11 Thermal Properties of Matter

Temperature and Heat

Question 1.
What is heat? What is SI units.
Answer:
Heat is the form of energy transferred between two or more systems or system and its surroundings by virtue of temperature difference. SI unit of heat is joule (J).
Note: Quantity of heat depends on the mass of the body.

Question 2.
What is meant by temperature? Give its SI unit.
Answer:
Temperature is an indication of relative measure of hotness or coldness of matter.
The SI unit of temperature is kelvin (K).

Question 3.
What is the main cause for the flow of heat in a medium?
Answer:
The difference of temperature in two regions of medium causes the flow of heat.

Measurement of Temperature

Question 1.
Name the device used to measure the temperature of a body.
Answer:
Thermometer.

Question 2.
What is the commonly used property of material for constructing thermometer?
Answer:
The commonly used property of material is variation of the volume of a liquid with temperature.

Question 3.
Name the common type of thermometer used in temperature measurement.
Answer:
Liquid in glass type

Question 4.
Name the commonly used liquids for making liquid in glass type thermometers?
Answer:
Mercury and alcohol are the liquids used in most liquid-in-glass thermometers.

Question 5.
Name the types of temperature scales used in common measurements.
Answer:

1. Celsius temperature scale
2. Fahrenheit temperature scale
3. Kelvin temperature scale

Question 6.
What is the ice point (freezing point) and steam point (boiling point) of water in degree Celsius and degree Fahrenheit.
Answer:

Question 7.
Write the relationship between degree Celsius (°C) and degree Fahrenheit (°F) temperature.
Answer:

Ideal Gas Equation and Absolute Temperature

Question 1.
State Boyle’s law. Give its mathematical expressions.
Answer:
This law states that “at constant temperature, the pressure of a fixed amount of gas inversely proportional to its volume”.
Mathematically, P ∝ \(\frac{1}{V}\) at constant T
\(P=k_{1} \frac{1}{V}\)
where k₁ is the proportionality constant.
PV = constant ………(1)
It means that at constant temperature, product of pressure and volume of a fixed amount of gas is constant.
Under two conditions, at constant temperature T, equation (1) can be written as
P₁V₁ = constant and P₂V₂ = constant
P₁V₁ = P₂V₂
or
\(\frac{P_{1}}{P_{2}}=\frac{V_{2}}{V_{1}}\)

Question 2.
State Charles’ law. Give is mathematical expression.
Answer:
This law states that “pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature”.
V ∝ T
V = k₂T
\(\frac{V}{T}\) = constant
This relation is called Charles’ law.
Under two different conditions,
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
or
V₁T₂ = V₂T₁

Question 3.
Write the combined gas equation.
Answer:
\(\frac{P V}{T}\) =constan t
or
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

Question 4.
State Gay-Lussac’s law.
Answer:
It states that, “the pressure of a given mass of a gas is directly proportional to the absolute temperature at constant volume.”
Thus, P ∝ T at constant volume for a given mass of gas
or P = k₃T where k₃ is constant.
\(\frac{P}{T}\) = constant T
Therefore, we can write, \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\), where P₁ and P₂ are the pressures of the gas at temperatures T₁ and T₂,
respectively.

Question 5.
What is ideal gas?
Answer:
A gas which obeys combined gas laws i.e., both Boyle’s law and Charles’ law is called ideal gas.

Question 6.
Write ideal gas equation.
Answer:
PV = μRT
Where μ is the number of moles of gas and R is the universal gas constant.

Question 7.
What is the value of R (universal gas constant) in SI units?
Answer:
R = 8.314 JK^-1 mol^-1

Question 8.
Draw the plots of P vs T of a low density gas at constant V for different gases and extrapolate the lines to absolute temperature.
Answer:

Question 9.
What is absolute zero temperature?
Answer:
The temperature -273.15°C at which the pressure of the given mass of the gas becomes zero at constant volume is called absolute zero. Absolute zero is the foundation of Kelvin temperature scale or absolute scale temperature.

Question 10.
At what temperature does extrapolation of P-T curve of all ideal gases at low density meet?
Answer:
-273.15°C

Question 11.
Write the relation between kelvin temperature and degree Celsius temperature.
Answer:
T= t_c + 273.15

Thermal Expansion

Question 1.
What is thermal energy?
Answer:
The total kinetic energy of all the molecules of the body is called thermal energy.

Question 2.
What is thermal expansion?
Answer:
The increase in the dimensions of a body due to the increase in its temperature is called thermal expansion.

Question 3.
Define linear expansion of a body.
Answer:
The expansion of a body in length due to the increase in its temperature is called linear expansion. It is given by the expression
\(\frac{\Delta l}{l}\) = α₁∆T
where \(\frac{\Delta l}{l}\) = the fractional change in length, ∆T = the small change in temperature and α = coefficient of linear expansion.

Question 4.
Define co-efficient of linear expansion?
Answer:
The ratio of increase in length to its original length per degree rise in the temperature is called the coefficient of expansion.
\(\alpha_{l}=\frac{\left(l_{2}-l_{1}\right)}{l\left(t_{1}-t_{1}\right)}=\frac{\Delta l}{l \cdot \Delta T}\)

Question 5.
Define area expansion of a body.
Answer:
It is given by \(\frac{\Delta A}{A}\) = α_a∆T = 2α_a∆T
The expansion in area of a body due to the increase in its temperature is called area expansion.

Question 6.
Define coefficient of area expansion.
Answer:
It is the ratio of increase in the area to its original area per degree rise in the temperature is called coefficient of area expansion. Mathematically it is given by
\(\alpha_{a}=\frac{\Delta A}{A \cdot \Delta T}\)
Where α is the coefficient of area expansion, \(\frac{\Delta A}{A}\) the fractional change in area, ∆T = the small change in temperature.

Question 7.
Define volume expansion of body.
Answer:
The expansion in volume of a body due increase in temperature is called volume expansion. It is given by
\(\frac{\Delta V}{V}=\alpha_{v} \Delta T\)
Where \(\frac{\Delta V}{V}\) = the fractional change in volume, ΔT = the small change in temperature and α = coefficient of volume expansion.

Question 8.
Define co-efficient of volume expansion.
Answer:
It is the ratio of change in volume to the original volume per degree rise in temperature.
\(\alpha_{v}=\frac{\Delta V}{V \cdot \Delta T}\)

Question 9.
Draw the graph of coefficient of volume expansion of copper as a function of temperature.
Answer:

Question 10.
Explain anomalous thermal expansion of water?
Answer:
Water contracts on heating between 0°C and 4°C. The volume of a given amount of water decreases as it is cooled from room temperature till 4°C and then volume increases and density decreases.

Question 11.
At what temperature density of water is maximum?
Answer:
At 4°C

Question 12.
How does coefficient of volume expansion depends on temperature in liquids?
Answer:
The coefficient of volume expansion is relatively constant in liquids.

Question 13.
Explain the variation of coefficient of volume expansion of ideal gas with temperature at constant pressure.
Answer:
For ideal gas, the variation of coefficient of volume expansion with temperature is obtained by ideal gas equation
PV = μRT
At constant pressure,

Thus, coefficient of volume expansion decreases with increase in temperature.

Question 14.
Arrive at the relation between co-efficient of linear and volume expansion OR Prove that α_v = 3α_l
Answer:
Let α and αv be the linear and volume expansion.

Question 15.
Arrive at the relation between co-efficient of linear and area expansion OR Prove that α_A = 2α_l.
Answer:
Consider a rectangular sheet of the solid material of length ‘a’ and breadth ‘b’ as shown in the figure. The area of sheet = A = ab
When the temperature increases by ΔT, the length ‘a’ increases by
A a = α_l aΔT
And breadth ‘b’ increases by
A b = α_l bΔT

From the figure the increase in area is given by
ΔA = ΔA₁ + ΔA₂ + ΔA₃
ΔA = aΔb + bΔa + (Δa)(Δb)
ΔA = aα₁bΔT + bα₁a ΔT + (α)²ab(ΔT)² = 2abα₁ΔT + α₁²abΔT²
ΔA = αabΔT(2 + α₁ΔT)
ΔA = α₁AΔT(2 + α₁ΔT)
Since α is very small( 10^-5 K^-1), the product α₁ΔT for fractional temperature is small in comparison with 2 and may be neglected.
Therefore, ΔA = α₁AΔT x 2
\(\frac{\Delta A}{A \cdot \Delta T}=2 \alpha_{1}\)
α₁ = 2α₁

Question 16.
Give the relation between α, α and αv.
Answer:
α, αA and av are in the ratio α_l: α_A : α_v :: 1 : 2 : 3

Question 17.
Define thermal stress?
Answer:
Thermal stress is equal to the force per unit area developed when thermal expansion or contraction is resisted.

Specific Heat Capacity

Question 1.
Define heat capacity of a substance.
Ans: It is defined as the quantity of heat required to raise the temperature of a given substance by unit degree Celsius. OR The amount of heat per unit mass absorbed or rejected by the substance to change its temperature by one unit.
\(S=\frac{\Delta Q}{\Delta T}\)

Question 2.
Define molar heat capacity. Write its unit.
Answer:
The amount of heat required to raise the temperature of one mole of gas by one kelvin is called molar heat capacity. Its SI unit is J mol^-1K^-1. It is calculated by
\(C=\frac{S}{\mu}=\frac{1}{\mu} \times \frac{\Delta Q}{\Delta T}\)

Question 3.
Define specific heat capacity(s)? Give its SI unit.
Answer:
The amount of heat required to raise the temperature of one kg of gas by one kelvin is called specific heat capacity. Its SI unit is Jkg^-1 K^-1. It is calculated by
\(s=\frac{S}{m}=\frac{1}{m} \times \frac{\Delta Q}{\Delta T}\)
where m is the mass of gas in kg.

specific heat Calculator of liquid water at constant volume or constant pressure at temperatures from 0 to 360 °C.

Question 4.
Define molar heat capacitr at constant pressure.
Answer:
It is defined as the quantity of heat required to raise the temperature of one mole of the gas through one kelvin keeping pressure constant. It is denoted by C_p.

Question 5.
Define molar heat capacity at constant volume.
Answer:
It defined as the quantity of heat required to raise the temperature of one mole of the gas through one kelvin keeping volume constant. It is denoted by C_v.

Calorimetry

Question 1.
What is calorimetry? Write principle of calorimetry.
Answer:
It means measurement of heat.
Principle: Heat lost by hot body = Heat gained by cold body.

Question 2.
What is calorimeter? .
Answer:
A device in which heat measurement can be done is called calorimeter.

Change of State

Question 1.
What is melting?
Answer:
The change of state from solid to liquid is called melting.

Question 2.
What is fusion?
Answer:
The change of state from liquid to solid is called fusion.

Question 3.
Define the terms melting point.
Answer:
The temperature at which the solid and the liquid states of the substances are in thermal equilibrium with each other is called melting point.

Question 4.
What is normal melting point?
Answer:
The melting point of a substance at standard atmospheric pressure is called normal melting point.

Question 5.
What is regelation?
Answer:
The phenomenon of decrease in melting point of ice due to increase in the pressure is called regelation.

Question 6.
What is vapourisation?
Answer:
The change of state from liquid to vapour is called vapourisation.

Question 7.
Define boiling point.
Answer:
The temperature at which the liquid and the vapour states of the substance coexist is called boiling point.

Question 8.
What is normal boiling point?
Answer:
The boiling point of a substance at standard pressure is called normal boiling point.

Question 9.
What is sublimation?
Answer:
The change from solid state to vapour state without passing through the liquid state is called sublimation.

Question 10.
How does boiling point depends on pressure?
Answer:
When pressure is increased, the boiling point increases.

Question 11.
Why cooking is difficult on hills?
Answer:
At high altitudes, atmospheric pressure is lower, which reduces the boiling point.

Question 12.
What is the principle involved in pressure cooker?
Answer:
Boiling point increases with increase in pressure. Hence food gets cooked faster.

Question 13.
What is latent heat? On what factors it depend on? Give its unit.
Answer:
It is the quantity of heat required to change the state of the substance without change in temperature. It depends on the heat of transformation and the mass of the substance undergoing change. If m is the mass of the substance undergoing a change from one state to other, the quantity of heat (Q) required is given by
Q = mL
L = \(\frac{Q}{M}\)
Where L is called latent heat.
Unit of latent heat = J/kg

Question 14.
Define the terms, latent heat of fusion and latent heat of vapourisation.
Answer:
Latent heat of fusion: Quantity of heat required to change the state from solid to liquid at constant temperature.
Latent heat of vaporistaion: Quantity of heat required to change the state from liquid to vapour at constant temperature.

Heat Transfer

Question 1.
Mention the different modes of heat transfer.
Answer:
Conduction, convection and radiation.

Question 2.
What is conduction?
Answer:
It is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature differences. OR It is the mode of transfer of heat in solids without actual movement of the particles of the medium.

Question 3.
Explain the process of heat conduction.
Answer:
Heat conduction is described in terms of rate of heat flow in a material for a given temperature difference. Consider a metallic bar of length L and uniform cross section A with two ends maintained at two different temperatures as shown in the figure.
Let one end of the bar be heated by some means. Initially, the temperature near the heated end of the bar is higher than the other extreme end. Let T_C and T_D be the temperatures of heated end and other end of the bar. There is a gradual decrease in the temperature as we move from the hotter end of the bar to the colder end. When the bar is heated for sufficiently long time, a steady state is reached at which temperature at any point remains constant. At steady state, the rate of flow of heat (or heat current) H is proportional to
(i) the temperature difference (T_C – T_D)
(ii) the area of cross section A and
(iii) is inversely proportional to the length L.
Therefore, the rate of rate of heat flow is given by


Where K is proportionality constant called thermal conductivity of the material.

Question 4.
Give the SI unit and importance of thermal conductivity.
Answer:
SI unit of thermal conductivity is Js^-1K^-1 or Wm^-1K^-1

Question 5.
What is the significance of thermal conductivity for a material?
Answer:
The greater the value of thermal conductivity (K) for a material, the more rapidly will it conduct heat i.e., Large thermal conductivity materials are good conductors.

Question 6.
Why a layer of earth or foam insulation on the cooling of concrete roofs is preferred?
Answer:
Houses made of concretes roofs get very hot during summer, because thermal conductivity of concrete is high. So to keep the rooms cooler it is preferred.

Question 7.
Why cooking pots have copper coating at the bottom?
Answer:
Copper is a good conductor of heat, promotes distribution of heat over the bottom of a pot for uniform cooking.

Question 8.
What is convection?
Answer:
The process of transmission of heat by the actual motion of matter is called convection. Convection is possible only liquids and gases (only in fluids).

Question 9.
What is forced convection? Give examples.
Answer:
A convection process during which material is forced to move by a pump or by some other physical means is called forced convection.
Examples:

• Forced air heating systems in home- transferring heat by forced convection and maintaining it at a uniform temperature.
• Human circulatory system- the heart acts as the pump that circulates the blood through different parts of the body.
• The cooling system of automobile engine.

Question 10.
What is natural convection or free convection? Give examples.
Answer:
The convection process during which material is moved on its own during the transfer heat is called natural convection or free convection. Examples: Sea breeze and land breeze, trade wind etc.

Question 11.
Explain sea breeze.
Answer:
During the day time, the land gets heated up faster than the water body. Warm air from the land rises up and moves towards the sea and cold air from the sea moves towards the land. This is called sea breeze. Therefore, sea breeze will be cooler during the day.

Question 12.
Explain land breeze.
Answer:
During the nights, the land cools faster than the sea. Hence the temperature of the land falls. The warm air from the sea move towards the land and the cold land breeze blows towards the sea during nights. This is called land breeze.

Question 13.
Explain trade wind.
Answer:
The equatorial and polar regions of the earth receive unequal solar heat. Air at the earth’s surface near the equator is hot while the air in the upper atmosphere of the poles is cool. As a result of this, a convection current is set up due to which air at the equator first rises and moves towards the pole and then descends towards the equator. However, this convection current is modified by the rotation of the earth.

Because of this, air close to the equator has an eastward speed of 1600 km/h, while it is zero close to the poles. As a result, air descends not at the poles, but at 30N(north) latitude and returns to the equator. This is called trade wind.

Question 14.
What is radiation?
Answer:
It is a process of transmission of heat from one place to another without the aid of any intervening medium.

Question 15.
What is radiant energy?
Answer:
The energy radiated by electromagnetic waves is called radiant energy.

Question 16.
What is thermal radiation?
Answer:
The electromagnetic radiation emitted by a body by virtue of its temperature is called thermal radiation.

Question 17.
Write the properties of thermal radiation.
Answer:

• They travel in straight line.
• They are electromagnetic in nature.
• They travel with speed of light.
• They undergo reflection, refraction and interference.
• They obey inverse square law.

Question 18.
What are black bodies?
Answer:
Black bodies are those which absorb all the thermal radiations falling on them and emit the same when maintained at suitable higher temperature.

Question 19.
The bottoms of the utensils for cooking food are blackened. Give reason.
Answer:
The surfaces coated with black body absorb complete thermal radiations and conduct the heat. This increases the speed of cooking food.

Question 20.
What is Dewar flask or thermos bottle? What is its use?
Answer:
It is device to minimise heat transfer between the contents of the bottle and outside. It is useful for preventing hot contents from getting cold.

Newton’s Law of Cooling

Question 1.
State Newton’s law of cooling.
Answer:
The rate of loss of heat of the body is directly proportional to the difference of temperature of the body and its surroundings.
\(-\frac{d Q}{d t}\) = k(T₂ – T₁)

Question 2.
Derive the expression T₂ = T₁ + e^-kt+c using Newton’s law of cooling.
Answer:
Let us consider a body of mass m and specific heat capacity S at temperature T₂. Let T₂ is temperature of surroundings.
If the fall in temperature is dT₂ in time dt, then the amount of heat loss, dQ = msdT₂

on integrating,
Iog_e(T₂ – T₁)= -Kt+c
T₂ – T₁ = e^-kt+c
T₂ – T₁ = e^-kt e^c
T₂ = T₁ +C’ e^-kt
where C’ = e^c

Question 3.
Give the experimental verification of Newton’s law of cooling.
Answer:
The following experimental set up is used for the verification of Newton’s law of cooling.

• The set -up consists of a double walled vessel (V) containing water in between the two walls.
• The copper colorimeter (c) containing hot water is placed inside the double walled vessel.
• Two thermometers through the corks are used to note the temperature T₂ of the water in calorimeter and T₁ of hot water in between the double walls respectively.
• The temperature of hot water in the calorimeter is noted after equal intervals of time.
• A graph is plotted between log_e (T₂ – T₁) and time (t).
  The nature of the graph is observed to be a straight line having a negative slope as shown in the figure. This give the verification of Newton’s law of cooling.

Numerical Problems

Question 1.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit, the rim of the wheel?
Answer:
Given, T₁ = 27°C, L_T1 = 5.231 m L_T2 = 5.243 m
L_T2 = L_T1 + α₁(T₂ – T₁)]
5.243 m = 5.231 m [1 + 1.20 x 10^-5 K^-1 (T₂ – 27°C)]
T₂ = 218° C.

Question 2.
A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.
Answer:
Mass of aluminium sphere (m₁) = 0.047 kg
Initial temp, of aluminium sphere = 100°C
Final temp. = 23°C
Change in temp (ΔT) = [100°C – 23°C] = 77°C
Let specific heat capacity of aluminium be S_Al.
The amount of heat lost by the aluminium sphere = msΔT = 0.047 kg x s x 77° C
Mass of water (m₂) = 0.25 kg
Mass of calorimeter (m₃) = 0.14 kg
Initial temperature of water and calorimeter = 20°C
Final temperature of the mixture = 23°C
Change in temperature (ΔT2) = 23°C – 20°C = 3°C
Specific heat capacity of water (s_w) = 4.18 x 10³ J kg^-1 K^-1
Specific heat capacity of copper calorimeter = 0.386 x 10³ J kg^-1 K^-1
The amount of heat gained by water and calorimeter = m₂s_wΔT₂ + m₃s_cuΔT₂
= (m₂s_w+ m₃s_cu )(Δ₂)
= 0.25 kg x 4.18 x 103 J kgK + 0.14 kg x 0.386 x 10 J kgK) (23°C – 20°C)
In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
= 0.047 kg x s_Al x 77°C
= (0.25 kg x 4.18 x 10³ J kg^-1 K^-1 + 0.14 kg x 0.386 x 10³ J kg^-1 K^-1) (3°C)
S_Al = 0.911 kJ kg^-1 K^-1

Question 3.
When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice. (s_water = 4186 J kg^-1 K^-1)
Answer:
Heat lost by water = ms_w(θ_f – θ_l)_w
= (0.30 kg)(4186 Jkg^-1 K^-1)(50.0°C – 6.7°C)
= 54376.14 J
Heat required to melt ice = m₂L_f = (0.15 kg) L_f
Heat required to raise temperature of ice water to final temperature = m₁s_w(θ_f – θ_l)_l
= (0.15 kg) (4186 J kg^-1K^-1 )(6.7°C – 0°C)
=4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 kg) L_f + 4206.93 J
L_f = 3.34 x 10⁵J kg^-1

Question 4.
Calculate the heat required to convert 3 kg of ice at – 12°C kept in a calorimeter to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg^-1 K^-1, specific heat capacity of water = 4186 J kg^-1 K^-1, latent heat of fusion of ice = 3.35 x 10⁵ J kg^-1 and latent heat of steam = 2.256 x 10⁶ Jkg^-1.
Answer:
Mass of the ice, m = 3 kg
specific heat capacity of ice, s_ice = 2100 J kg^-1 K^-1
specific heat capacity of water, s_water = 4186 J kg^-1 K^-1
latent heat of fusion of ice, L_{f ice} = 3.35 x 10⁵ J kg^-1
latent heat of steam, L_steam = 2.256 x 10⁶ J kg^-1
Now, Q = heat required to converts kg of ice at – 12°C to steam at 100°C
Q₁ = heat required to convert ice at – 12°C to ice at 0°C.
= ms_ice ΔT₁ =(3 kg) (2100 J kg^-1 K^-1)[0 – (-12)]°C = 75600 J
Q₂ = heat required to melt ice at 0°C to water at 0°C
mL_{f ice} = (3kg)(3 – 35 x 10⁵ J kg^-1) = 1005000 J
Q₃ = heat required to convert water at 0°C to water at 100°C.
= ms_w ΔT₂ = (3 kg)(4186 J kg^-1K^-1) (100°C) = 1255800 J
Q₄ = heat required to convert water at 100°C to steam at 100°C.
= mL_steam = (3 kg)(2.256 x 10⁶J kg^-1) = 668000 J
So, Q = Q₁ + Q₂ + Q₃ + Q₄ = 75600 J + 1005000 J + 1255800 J + 6768000 J = 9.1 x 10⁶ J

Question 5.
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:
Relation between °C and Kelvin is given by
T_c = T – 273.15
∴T_c for neon = 24.57 – 273.15
∴ T_c for carbon dioxide = 216.55 – 273.15= – 56.60°C
Relation between °C and °F is given by
t_F = \(\frac{9}{5}\) t_c + 32
∴ t_F for neon = \(\frac{9}{5}\) x (-248.58) + 32 = -415.44°C
t_F for carbon dioxide = \(\frac{9}{5}\) x (-56.60) + 32 = -69.88° F

Question 6.
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:
Triple point of water = 273.16 K = 200 A = 350 B

Question 7.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀ [1 + α (T – To)]. The resistance is 101.6 Ω at the triple point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Q?
Answer:
At T₀ = 273.16 K, R0= 101.6 Ω
At T₁ = 600.5 K, R₁ = 165.5 Ω
Let T₂ be the temperature at which the resistance is R₂ = 123.4 Ω

= 293.16 +(600.5 – 273.16)\(\left(\frac{123 \cdot 4-101 \cdot 6}{165 \cdot 5-101 \cdot 6}\right)\)
T₂ = 384.83 K

Question 8.
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 h = T₀ – 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:
(a) The triple point is used as a standard because of the unique conditions of temperature and pressure at this
point. The melting point of ice and boiling point of water depend on pressure. So they cannot be used as reference
points.
(b) The other fixed point is absolute zero, the theoretically lowest possible temperature.
(c) We have 273.15 in the relation and not 273.16 because the triple point of water is 0.01°C and not 0°C
In other words, 0.01°C → 273.16 K
⇒ 0°C → 273.15 K
∴t₀ = T – 173.15
(d) The unit interval size in the Fahrenheit scale = \(\frac{212-32}{100}\) = 1.8
∴The triple point of water =273.16 x 1.8 = 491.69

Question 9.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(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 the 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:
(a) Let P be the pressure in the pressure thermometer and T be the corresponding absolute temperature. Let P₀ and T₀ be the values for the triple point of water.
Then,


(b) The discrepancy is due to the non-ideal behaviour of the gases. To reduce this discrepancy, measurements at lower pressures should be taken.

Question 10.
A steel tape 1 m 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 x 10^-5 K^-1.
Answer:
70 = 63 cm, ΔT = 45°C – 27°C = 18°C = 18 K, α = 1.2 x 10^-5 K^-1
I = I₀ (l + αΔT)
= 63(1 + 1.2 x 10^-5 x 18) = 63.0136 cm
∴Actual length of the steel rod at 45°C = 63.0136 cm
At 27° C, the steel tape is correctly calibrated,
∴the length of the steel rod = 63.0 cm

Question 11.
A hole 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 x 10^-5 K^-1.
Answer:
Given a =1.70 x 10^-5 K^-1, β = 2α = 3.40 x 10^-5 K^-1, d₁ = 4.24 cm
Initial area A₁ at 27°C ( 300 K) is
\(A_{l}=\pi\left(\frac{d_{1}}{2}\right)^{2}=\pi\left(\frac{4 \cdot 24}{2}\right)^{2}\)
= 4.4944 π cm²
Area A₂ at 227°C (500 K) is
A₂ = Aj [1 + β(T₂ – T₁)]
A₂ = 4.49447 π[1 + 3.40 x 10 (500 -300)]= 4.5250 π cm²

∴ Change in diameter = d₂ – d₁
= 4.2544 – 4.25 = 0.0144 cm

Question 12.
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? Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1, Young’s modulus of brass = 0.91 x 10¹¹ Pa.
Answer:
Given, I = 1.8 m, T₁ = 27°C = 300 K, T₂ = -39°C = 234 K,
d = 2mm = 2 x 10^-3, α_brass = 2.0 x 10 K^-1, Y_brass =0.91 x 10¹¹ Pa.

= -377.37 N
Tension developed ΔF = 380 N (Positive value, rounded to 2 significant figures)

Question 13.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter is the change in length of 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 (Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1, steel = 1.2 x 10^-5 K^-1).
Answer:
Given:

(round off to two significant figures)
Since the rods are free to expand, no thermal stress is developed at the junction.

Question 14.
The coefficient of volume expansion of glycerin is 49 x 10^-5 K^-1. What is the fractional change in its density for a 30°C rise in temperature?
Answer:
Given, r_glycerin = 49 x 10^-5 K^-1, ΔT = 30°C = 30 K
Let the volume be V₁ at temperature V₂ at temperature T₂
\(\frac{V_{1}}{V_{2}}=\frac{1}{1+r \Delta T}\)
= \(\frac{1}{1+49 \times 10^{-5} \times 30}\)
= 0.9855
Fractional change in density is given by

(The negative sign denotes a decrease in density)

Question 15.
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 blocks 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:
Given, P = 10 kW, t = 2.5 minutes = 150 s, m = 8 kg,
S_aluminium = 0.91 J g^-1 K^-1 = 910 J kg^-1 K^-1
Energy transferred to the aluminium 50% of P x t
This energy heats up the aluminium.
So, \(\frac{50}{100}\) x pt = ms_aluminium ΔT
\(\frac{50}{100}\) x 10 x 10 x 150 = 8 x 910 x ΔT
ΔT = 103 K or ΔT = 103°C

Question 16.
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 Jg^-1 K^-1, heat of fusion of water = 335 Jg^-1)
Answer:
Given m = 2.5 kg, T₁ = 500°C, T₂ = 0°C, s_copper = 0.39 Jg^-1 K^-1, L_ice = 335 Jg^-1
Let the mass of ice that melts be m_ice
Heat dissipated by the copper block = Heat absorbed by the ice.
m_copper s_copper ΔT = m_ice L_ice
2500 x 0.39 x (500 – 0) = m x 335
m_ice = 1455 g
⇒ m_ice = 1.5 kg
(rounding off to 2 significant figures)

Question 17.
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 cm³ 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 of the specific heat of the metal?
Answer:
Given, mass of metal block, m₁ = 0 – 20 kg
Initial temperature of the metal block, T₁ = 150°C
Initial temperature of the calorimeter and water, T₂ = 27°C
Final temperature of the mixture, T₃ = 40°C
Volume of water in the calorimeter, V = 150 cm³ = 150 x 10⁶ m³
Mass of the water = Density x Volume
= 1000 x 150 x 10 = 150 x 10 kg
Let s be the specific heat of the metal and s₂ be the specific heat of water.
Water equivalent of the calorimeter, W = 0.025 kg
Heat lost by the metal block = Heat gained by the calorimeter and water.
m₁s₁ΔT₁ = (m₂+W)s₂ΔT₂
0.2 x s^-1 x (150 – 40) = (150 x 10^-3 + 0.025) x 4200 x (40 – 27)
s₁ =434.3 J kg^-1 K^-1 = 430 J kg^-1 K^-1 or 0.43 Jg^-1K^-1
If the heat losses to the surroundings are not negligible, the left hand side of the expression would have had an extra term added to it to account for this loss. Upon solving for s₂ we would get a smaller value than the actual value.

Question 18.
Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat (Cv) (cal mol-1 K-1)
Hydrogen	4.87
Nitrogen	4.97
Oxygen	5.02
Nitric oxide	4.99
Carbon monoxide	5.01
Chlorine	6.17

The measured molar specific heats of these gases are markedly different from those for mono atomic gases. Typically, molar specific heat of a monoatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the the rest) value for chlorine?
Answer:
All the gases mentioned above are diatomic, and hence have other degrees of freedom in addition to the translational degrees of freedom. To raise the temperature of the gas, heat has to be supplied to increase the average energy of all the modes. Consequently, the molar specific heats for diatomic gases are higher than that for mono atomic gases.

The higher value of the molar specific heat of chlorine indicates that at room temperature, vibrational modes are also present in addition to the rotational modes of freedom.

Question 19.
Answer the following questions based on the P-T diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂, exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas arbitrary (i) – 70°C under 1 atm, (ii) 60°C under 10 atm, (iii) 15°C under 56 atm?
Answer:
(a) At – 56.6°C and 5.11 atm, which is the triple point of CO₂, all 3 phases can co-exist in equilibrium.
(b) From the fusion and vaporisation curves, we see that reducing the pressure lowers the fusion and boiling point.
(c) The critical temperature and pressure are 31.1 °C and 73.0 atm respectively. Above this temperature, CO₂ will not liquefy even if compressed to high pressures.
(d) (i) Gas (ii) Solid (iii) Liquid

Question 20.
Answer the following questions based on the P-T phase diagram of CO₂:
(a) CO₂ at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
Answer:
(a) When subjected to isothermal compression, CO₂ will condense to solid directly without passing through the liquid phase.
(b) When cooled at constant pressure, CO₂, will condense directly to solid phase without passing through the liquid phase.
(c) As the solid CO₂ is heated up, it first turns to liquid phase and then solid phase. The fusion and boiling points can be obtained from finding out the points of intersection between the horizontal 10 atm line on the P-T diagram with the fusion and vaporisation curves. .
(d) CO₂ will not exhibit any clear transition to the liquid phase, but will deviate more and more from ideal behaviour as the pressure increases.

Question 21.
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 mm, 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 the 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:
Mass of the child, m = 30 kg
Change in temperature ΔT = 101° F – 98°F = 3°F
\(=3 \times \frac{5}{9}=\frac{5}{3}\) °C
Specific heat of water, s = 4.2 x 103 J kg^-1 °C^-1
Latent heat of vaporization, L = 580 cal g^-1 = 580 x 4.2 x 10³ J kg^-1
Time taken for the sweat to evaporate, t = 20 min.
Let m₁be the mass of sweat.
Now, heat lost by the child during evaporation = Heat needed for the sweat to evaporate.
msΔT =m₁ x L
30 x 4.2 x 10³ x \(\frac{5}{3}\) =m₁ x 580 x 4.2 x 10³
⇒ m₁ = 0.0862 kg = 86.2 g
Rate of extra evaporation caused by the drug = Rate of evaporation of sweat

Question 22.
A ‘thermacole’ ice box is a cheap and 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 6h. 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 = 835 x 10 J kg^-1]
Answer:
Total surface area of the cube, A = 6 x (30 cm)² = 0.54 m²
Mass of ice in the box M = 4 kg
Thickness of the box, L = 5 cm = 0.0F
Thermal conductivity of thermacole, k = 0.01 Js^-1 m^-1 K^-1
Outside temperature T₁ = 45 °C
Temperature of ice T₂ = 0°C
Let the mass of ice that melts be m kg.
Heat of fusion of water, L = 335 x 10³ J kg^-1
Time t = 6h = 6 x 3600 s = 21600 s
Heat gained by ice during fusion = Heat conducted through the walls of the box

⇒ m = 0.31 kg
∴ Mass of ice left in the box = M – m
= 4 – 0.31 =3.69 = 3.7 kg.

Question 23.
A brass boiler has a base area of 0.15 m² 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 x 103 J kg^-1.
Answer:
Let the temperature of the flame be T₁
T₂ = boiling point of water = 100°C
Thermal conductivity of brass, k = 109 J s^-1 m^-1 K^-1
Base area of the boiler, A = 0.15 m²
Thickness of the boiler, L = 1 cm = 0.01 m
Heat of vaporisation of water, L_stream = 2256 x 10 J kg^-1
Rate of boiling water = 6 kg min^-1 = \(\frac{6}{60}\)kg s^-1 = 0.1 kg s^-1
60
Power consumed = 0.1 x L_stream
= 0.1 x 2256 x 10³ W
∴Power transmitted through the boiler = 2256 x 10² W

T₁ -100 = 138 ⇒ T₁ =238° C
∴Temperature of the flame = 238° C

Question 24.
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 n circulation of hot water.
Answer:
(a) We know that a + r + t = 1, where ‘a’ is absorbance, ‘r’ is reflectance and ‘t’ is transmittance or emittance. According to Kirchhoff’s law, emittance α absorbance, i.e., good absorbers are also good emitters and so are poor reflectors. Therefore, if reflectivity is large (V is large), then ‘a’ is small and hence emittance is smaller (the object behaves as a poor emitter).
(b) The coefficient of thermal conductivity of brass is higher than that of wood. When a brass tumbler is touched, heat quickly flows from human body to the tumbler and so the tumbler appears colder. But in the case of the wooden tray heat does not flow from the human body to the wooden tray, and so it feels relatively hotter.
(c) Let the temperature of the hot iron in the furnace be ‘T’ and that of the external environment be ‘T₀‘. According to Stefan’s Law, heat energy radiated per second E = σT⁴ in the case of furnace and E’= σ(T⁴ – T₀⁴) in the open. Clearly E > E’. Hence the low readings of the optical pyrometer when the iron piece is in the open.
(d) The earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the earth. If it were absent, the heat would have escaped and the planet would be very cold.
(e) Steam has a higher heat capacity (due to the latent heat) than boiling water. Therefore steam-based water heating systems are more effective.

Question 25.
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:
The average temperature of 80°C and 50°C is 65°C, which is 45° C above the room temperature. Under these
conditions, the body cools 30°C in 5 minutes. Using the equation

The average of 60° C and 30° C is 45° C which is 25°C above the room temperature. K is the same for this situation as the original.
Hence \(\frac{30^{\circ} \mathrm{C}}{t}\) = K(45° C) …….(2)
Divide (1) by (2), we get,

t = 9 min
∴Time taken = 9 minutes.