Heat and Thermodynamics

Measurement and Mechanical Equivalent of Heat

Heat is a form of energy, measured in energy units. SI unit is Joule (J), C.G.S. unit is erg, commonly used unit is calorie.

Definition: A calorie is the amount of heat required to raise the temperature of 1gm of water through 1°C

Conversion: 1 Calorie = 4.18 Joules

Mechanical Equivalent: J = work done / Heat produced = 4.18 J/cal

In C.G.S.: J = 4.2 × 10⁷ erg/cal

Illustration: A bullet of mass 10 gm moving with a speed of 20 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost Kinetic energy goes to ice? (Temperature of ice block = 0°C).

Solution:

Velocity of bullet + ice block system:

V = (10 gm × 20 m/s) / 1000 gm = 0.2 m/s

Loss of K.E. = ½mv² - ½(m + M)V²

= ½ × 0.01 × (20)² - ½ × 1 × (0.2)²

= ½ × (4 - 0.04) = 1.98 J

Heat generated = 1.98/4.2 Cal

Heat received by ice block = (1.98/4.2) × ½ Cal = 0.24 Cal

Mass of ice melted = 0.24 Cal / (80 Cal/gm) = 0.003 gm

Calorimetry and Thermometry

When two objects having different temperatures are brought in contact, heat flows from the hot object to the cold object. If we neglect heat exchange with the surroundings, the heat lost by the hot object = the heat gained by the cold object. This is the basic principle of calorimetry which follows from the principle of conservation of energy.

Zeroth Law of Thermodynamics and Temperature

If a system A is in thermal equilibrium with system B and the system B is in thermal equilibrium with system C, then systems A and C are in thermal equilibrium with each other.

The common property of these systems in thermal equilibrium is the temperature.

Important Values

• Latent Heat of Fusion of ice: 80 cal/gm
• Latent Heat of Vaporization for water: 536 cal/gm
• Triple point of water: Ttr = 273.16K

Temperature Scale Relations

Scale Name	Symbol	Lower Fixed Point	Upper Fixed Point	No. of Divisions
Reaumur	°R	0°R	80°R	80
Celsius	°C	0°C	100°C	100
Fahrenheit	°F	32°F	212°F	180
Kelvin	K	273K	373K	100

Temperature Scale Relationship:

R/80 = C/100 = (F-32)/180 = (K-273)/100

Specific Heat (Specific Heat Capacity)

Amount of heat (Q) required to raise the temperature of unit mass of the substance through a unit degree.

Q = m.s.ΔT

S = Q/(m.ΔT)

Units: cal g^-1 °C^-1 or J kg^-1 K^-1 (S.I)

Properties of Specific Heat

• S depends on nature of body
• S of a substance when it melts or boils at constant temperature is infinite
• S = 0, if temperature of a substance changes without the transfer of heat (Q = 0)
• S of saturated water is negative

Molar Specific Heat (C)

Molar specific heat of a substance is defined as amount of heat required to raise the temperature of 1 mole of the substance by 1°C or 1K.

Molar specific heat = molecular weight × specific heat

C = Q/(n.ΔT); where n = no. of moles

S = C/m; where m = no. of kilograms per mole

S.I. unit: J mol^-1 K^-1

Heat Capacity

Heat capacity of a substance is defined as the amount of heat required to raise the temperature of the substance through 1°C.

Unit: cal °C^-1 or JK^-1

Illustration: 5 gm of water at 30°C and 5 gm of ice at -20°C are mixed together in a calorimeter. Find the final temperature of mixture and also the final masses of ice and water. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/gm°C and latent heat of ice = 80 cal/gm.

Solution: In this case heat is given by water and taken by ice

Heat available with water to cool from 30°C to 0°C
= m.s.Δθ = 5×1×30 = 150 cal

Heat required by 5 gm ice to increase its temperature up to 0°C
= m.s.Δθ = 5×0.5×20 = 50 cal

Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from -20°C to 0°C. The remaining heat 100 cal is used for melting the ice.

If mass of ice melted is m gm then:
m × 80 = 100 ⟹ m = 1.25 gm

Thus 1.25 gm ice out of 5 gm melts and mixture of ice and water is at 0°C.

Thermal Expansion

Increase in the temperature of a body is generally accompanied by an increase in its size. This is known as Thermal Expansion.

Consider a rod of length L at temperature T. If its temperature is changed to T + ΔT, the average linear coefficient of thermal expansion of the material of the rod in the temperature range T to T + ΔT is defined as:

α = (1/L) × (ΔL/ΔT)

The value of linear coefficient of thermal expansion at temperature T:

α = (1/L) × (dL/dT)

For most solids: L = L₀(1 + αΔT)

Illustration: A certain clock with an iron pendulum is made so as to keep correct time at 10°C. Given α_iron = 12 × 10^-6 per °C. How fast or slow does the clock move per day if the temperature rises to 25°C?

Solution: When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes (24×60×60)/2 = 43200 vibrations/day

If length of pendulum at 10°C is ℓ₁₀ and at 25°C is ℓ₂₅

ℓ₂₅ = ℓ₁₀[1 + α(25-10)] = ℓ₁₀[1+15α]

As T = 2π√(ℓ/g) i.e. T ∝ √ℓ i.e. n ∝ 1/√ℓ

n₂₅/n₁₀ = √(ℓ₁₀/ℓ₂₅) = √(1/(1+15α)) ≈ (1-15α/2)

n₂₅ = n₁₀(1-15α/2) = 43200[1-15×12×10⁻⁶/2] = 43200[1-0.00009] = 43196.12

The clock makes (43200-43196.12) = 3.88 vibrations less per day

Clock loses 3.88×2 = 7.76 sec per day

Area and Volume Expansion

If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by

Area Expansion: A = A₀(1 + βΔT), where β ≈ 2α

Volume Expansion: V = V₀(1 + γΔT), where γ ≈ 3α

Thermal Expansion in Liquids

The experimental measurement of γ for a liquid becomes slightly difficult due to expansion of the container.

The actual increase in volume of liquid = apparent increase in volume of liquid + increase in volume of container

γ_liquid = γ_apparent + γ_container

Anomalous Expansion of Water

If the temperature of water is increased from 0°C, it contracts up to 4°C and expands thereafter. In other words, the density of water is highest at 4°C.

Thermal Stress

When a rod of length L is held between two rigid supports and the temperature of rod is increased by ΔT, the rigid support prevents the rod from expanding. This causes a stress in the rod.

Thermal expansion = Mechanical compression

αLΔT = FL/(AY)

Thermal stress F/A = YαΔT (compressive nature)

What is the strain developed in the rod?

Longitudinal Strain by definition is fractional change in length. One may argue that as the length of the rod has not changed the strain is zero. But, it is nor correct. A more exact definition of strain is fractional change in length of the object at that temperature due to stress.

Variation of Density with Temperature

The density at temperature t°C is given by:

d_t = d₀/(1 + γ.t)

where d₀ = density at 0°C

In general: d₁/d₂ = (1 + γt₂)/(1 + γt₁)

Kinetic Theory of Gases

When the molecules have thermal kinetic energy high enough to overcome their binding forces, the matter is said to be in gaseous state. In this state, the matter can expand to any extent to which it is allowed i.e. it fills the entire volume of the container.

Ideal Gas Equation

Equation of State: PV = nRT

Where n is number of moles and R is universal gas constant

In S.I. units: R = 8.31 J mol^-1 K^-1

Pressure and Kinetic Energy

The pressure exerted by an ideal gas on the wall of its container is:

P = (1/3)ρV_rms²

Where V_rms = root-mean-square velocity and ρ = density of the gas

V_rms = √(3RT/M)

Note: M should be taken as mass of one mole of gas in kg

Illustration: Find r.m.s speed of Hydrogen molecules at room temperature (=300 K).

Solution: Mass of 1 mole of Hydrogen gas = 2 gm = 2 × 10^-3 kg

V_rms = √(3RT/M) = √((3×8.3×300)/(2×10^-3)) = 1.93 × 10³ m/s

Dalton's Law of Partial Pressure

When the container contains more than one gas, total pressure exerted by all the gases on the wall is sum of pressures exerted by each gas as it would while filling the container alone.

P = P₁ + P₂ + P₃ + ...

Standard Temperature and Pressure

One mole of any gas occupies a volume of 22.4 litres at standard temperature and pressure which are 273.15K (=0°C) and 1.013×10⁵ Pa (=1atm) respectively.

Internal Energy

Internal energy, in present context, for any body is sum total of kinetic energies and potential energies of its constituents (at molecular level). In case of an ideal gas, as there is no intermolecular forces, except during collision the possibility of potential energy is ruled out, so it is only kinetic energy.

Nature of Motion	Atomicity	Degree of Freedom (f)			Total
Translational	Rotational	Vibrational*
Monoatomic	1	3	0	0	3
Diatomic	2	3	2	0	5
Poly Linear	>2	3	2	0	5
Non-linear	>2	3	3	0	6

*At room temperature the energy associated with vibrational motion is negligibly small in comparison to translational and rotational K.E.

Equipartition of Energy

According to the Law of equipartition of energy, the total K.E. of molecules is equally shared by the various degrees of freedom. The average energy per degree of freedom of a molecule is ½kT, where k is the Boltzmann constant and T is the absolute temperature.

For a monoatomic molecule: U = (3/2)kT

For one mole: U = (3/2)RT

Work Done in Different Processes

Total work done over a large displacement by the enclosed gas can be given as:

W = ∫PdV

Different Thermodynamic Processes

(a) Isochoric Process

Here, the volume is constant throughout the process.

W_isochoric = 0

(b) Isobaric Process

In this case, pressure of the gas remains constant throughout the process.

W = P∆V = nR∆T (where n = number of moles, ∆T = change in absolute temperature)

(c) Isothermal Process

For such a process, temperature remains constant throughout the process.

Using ideal gas equation: P = nRT/V

W = ∫PdV = nRT∫(dV/V) = nRT ln(V₂/V₁)

(d) Adiabatic Process

PVᵞ = Constant = C

P = CV^-γ

W = ∫PdV = ∫CV^-γdV = (P₁V₁ - P₂V₂)/(γ-1) = -Cv(T₂ - T₁)

The work done by a gas can also be evaluated from the P-V diagram of the process. Area enclosed by the curve in a P-V diagram = work done by the gas

First Law of Thermodynamics

First law of thermodynamics is simply a re-statement of principle of conservation of energy.

Statement: Q = ∆U + W

Where:

• Q = heat given to the system
• ∆U = change in internal energy
• W = work done by the system

Illustration: 3000 J of heat is given to a gas at constant pressure of 2 × 10⁵ N/m². If its volume increases by 10 litres during the process, find the change in the internal energy of the gas.

Solution: Q = 3000 J

W = P∆V = (2×10⁵ N/m²)(10×10⁻³m³) = 2×10³ J

∆U = Q - W = 3000 - 2000 = 1000 J

Specific Heat Capacities of Gases

The amount of heat required for certain temperature change of a given gaseous system depends on the nature of process and hence the work. Therefore, specific heat capacity of gases would be different for different processes.

Specific heat capacity: S = (1/m) × (∆Q/∆T)

Molar heat capacity: C = (1/n) × (∆Q/∆T)

Two Special Cases

(i) Constant Volume Process

Cv = (1/n) × (∆Q/∆T)|_V=constant

This is the molar specific heat at constant volume

(ii) Constant Pressure Process

Cp = (1/n) × (∆Q/∆T)|_P=constant

This is the molar specific heat at constant pressure

Relation Between Cp and Cv (Mayer's Relation)

Cp - Cv = R

Since ∆U is independent of the process: ∆U = nCv∆T is true for all processes

The Values of Cp and Cv

If f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as:

U = (f/2)nRT

∆U = (f/2)nR∆T = nCv∆T

Therefore: Cv = (f/2)R

From Mayer's Relation: Cp = Cv + R = (f/2 + 1)R

And the ratio of specific heats: γ = Cp/Cv = (f+2)/f

Second Law of Thermodynamics

(i) Kelvin Statement

It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.

(ii) Clausius Statement

It is impossible for a self-acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature.

Reversible Process

A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible process.

Irreversible Process

A process which cannot be made to proceed in the reverse direction is called an irreversible process.

Heat Engine

It is a device which converts continuously heat energy into mechanical energy in a cyclic process.

Efficiency of Heat Engine

η = work output/heat input = W/Q₁ = (Q₁-Q₂)/Q₁ = 1 - Q₂/Q₁

Where Q₁ is the heat supplied by the source and Q₂ is the heat rejected to the sink

Carnot Engine

It is an ideal heat engine which is based on Carnot's reversible cycle. Its working consists of four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression.

The efficiency of Carnot engine: η = 1 - Q₂/Q₁ = 1 - T₂/T₁

where T₁ and T₂ are the temperatures of source and sink respectively

Illustration: The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 65°C, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working.

Solution:

Given: η₁ = 1/6, η₂ = 1/3

For first case: η₁ = 1 - T₂/T₁ = 1/6

For second case: η₂ = 1 - (T₂-65)/T₁ = 1/3

Solving: T₁ = 390 K and T₂ = 325 K

Heat Transfer

Heat can be transferred from one place to another by three methods – conduction, convection and radiation.

Conduction

If a spoon is put in a cup of hot tea after some time its other end also becomes warm. Heat is transferred from one end of the spoon to the other end due to molecular vibrations within the spoon. This process is called thermal conduction. In this process there is no movement of mass.

Thermal Conductivity

Consider a disk of cross-sectional area A and thickness x. The temperature of the two ends is maintained at T₁ and T₂ (T₂ > T₁) and the curved surface is kept insulated from surroundings.

∆Q/∆t = -kA(dT/dx)

where k is called the thermal conductivity of the material

The minus sign indicates that temperature decreases along the direction of heat flow.

Ingen Hausz Experiment

Ingen Hausz provided a method to compare the thermal conductivities of different materials. He took wax coated rods of different materials but of same area.

If ℓ₁, ℓ₂, ℓ₃... represent the lengths up to which the wax is melted and K₁, K₂, K₃... are their thermal conductivities, then:

K₁/ℓ₁² = K₂/ℓ₂² = ... = constant

Convection

Convection is that method of heat transfer in which the heat is carried from one place to another by actual movement of heated matter. This process can occur only in fluids. If the heated matter is forced to move by some agent (like a fan) then the process is known as forced convection. If the matter flows on its own due to difference of pressure it is known as natural or free convection.

Radiation

Radiation is that method of heat transfer which does not require any material medium. In this process heat energy is transferred in the form of electromagnetic waves.

Properties of Heat Radiation

• Heat radiation consists of electromagnetic waves of the infra-red region
• It travels with velocity of light (3 × 10⁸ m/s in vacuum)
• It has all qualities of light such as reflection, refraction, interference, polarization etc.
• It obeys inverse square law

Absorption, Reflection and Transmission

If Q is the total incident energy on a body, Q₁ is the part absorbed, Q₂ is the part reflected and Q₃ is the part transmitted then:

Q = Q₁ + Q₂ + Q₃

Absorption coefficient: a = Q₁/Q

Reflection coefficient: r = Q₂/Q

Transmission coefficient: t = Q₃/Q

Thus: a + r + t = 1

If for a body r = t = 0 and a = 1, i.e. it absorbs all the energy falling on it, it is called a black body.

Kirchoff's Law

At a given temperature the ratio of emissive power to absorption power of any body is equal to the emissive power of a black body.

E₁/a₁ = E₂/a₂ = E_{Black body}

Wien's Displacement Law

The wavelength corresponding to highest intensity λₘ is inversely proportional to the absolute temperature.

λₘ = b/T

where b (= 2.89 × 10⁻³ meter Kelvin) is known as the Wien's constant

Stefan's Law

The area under a curve in the energy distribution graph is proportional to the fourth power of corresponding absolute temperature.

Q = σAT⁴

where σ = 5.67 × 10⁻⁸ W/m²-K⁴ is Stefan's constant

For a non-black body: Q = eσAT⁴ where e is the emissivity

For a body in surroundings: Q = eσA(T⁴-T₀⁴)

Newton's Law of Cooling

It states that the rate of loss of heat by a body is directly proportional to difference in temperatures of the body and the surroundings provided the temperature difference is small.

-dT/dt = KA∆T

where K depends on the nature of surface, specific heats and mass of the body

Illustration: A body cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. Find the temperature of surroundings.

Solution:

For the first ten minutes:

dT/dt = (62°-50°)/10 = -1.2°C/min and T̄ = (62+50)/2 = 56°C

⟹ -1.2°C/min = -KA(56-T₀)°C ...(1)

For the next ten minutes:

dT/dt = (50°-42°)/10 = -0.8°C/min and T̄ = (50+42)/2 = 46°C

⟹ -0.8°C/min = -KA(46-T₀)°C ...(2)

Dividing (1) by (2): 3/2 = (56-T₀)/(46-T₀)

⟹ T₀ = 26°C

Concepts and Formulae at a Glance

1. J = work done / Heat produced
2. Q = m·s·ΔT
3. 1 cal g⁻¹ °C⁻¹ = 4.18J g⁻¹ °C⁻¹ = 4180 JKg⁻¹K⁻¹
4. C = Q/(n·ΔT)
5. R/80 = C/100 = (F-32)/180 = (K-273)/100
6. Variation of density: dₜ = d₀/(1 + γ·t)
7. L = L₀(1 + αΔT) : α = Linear coefficient of thermal expansion
8. A = A₀(1 + βΔT) : β = Area coefficient of thermal expansion
9. V = V₀(1 + γΔT) : γ = Coefficient of volume expansion
10. β ≈ 2α ; γ ≈ 3α
11. γₗᵢqᵤᵢd = γₐₚₚₐᵣₑₙₜ + γcₒₙₜₐᵢₙₑᵣ
12. thermal stress = F/A = γ·α·ΔT
13. PV = nRT
14. P = (1/3)ρVᵣₘₛ²
15. Vᵣₘₛ = √(3RT/M)
16. Average energy per degree of freedom = (1/2)KT
17. First law: ΔQ = ΔU + W
18. Cₚ - Cᵥ = R
19. γ = Cₚ/Cᵥ = (f+2)/f
20. η = work output/Heat input = W/Q₁ = 1 - Q₂/Q₁
21. Carnot engine: η = 1 - T₂/T₁
22. ∂Q/∂t = -k·A·dT/dX
23. Ingen Hausz experiment: k₁/ℓ₁² = k₂/ℓ₂² = constant
24. Wien's law: λₘ = b/T
25. Newton's cooling: dT/dt = -KA·ΔT

Solved Examples

Problem 1: Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is:

(A) 30K    

(B) 18K

(C) 50K    

(D) 42K

Solution:

For cylinder A (Isobaric Process): dQ = nC_PdT₁

For cylinder B (Isochoric Process): dQ = nC_vdT₂

Since the same amount of heat is given to both cylinders:

nC_PdT₁ = nC_vdT₂

C_vdT₂ = (C_v + R) × 30

For diatomic gas: C_v = (5/2)R

dT₂ = [(5/2)R + R] × 30 / [(5/2)R] = (7/2)R × 30 / (5/2)R = 42K

Answer: (D) 42K

Problem 2: 840 J of heat is required to raise the temperature of 2g moles of an ideal gas from 20°C to 40°C at constant pressure. The amount of heat required to raise the temperature of the same gas from 40°C to 60°C at constant volume is (take R = 8.4 J g.mole^-1 K^-1):

(A) 504 J    

(B) 420 J

(C) 840 J    

(D) 630 J

Solution:

840 = 2 × C_P × (40 - 20)

840 = 2 × C_P × 20

C_P = 840 / 40 = 21 J g.mole^-1 K^-1

C_v = C_P - R = 21 - 8.4 = 12.6 J g.mole^-1 K^-1

Q = nC_vΔT = 2 × 12.6 × (60 - 40) = 2 × 12.6 × 20 = 504 J

Answer: (A) 504 J

Problem 3: A vessel contains a mixture of 1 mole of oxygen and 2 moles of nitrogen at 300K. The ratio of average rotational kinetic energy per O₂ molecule to that of per N₂ molecule is:

(A) 1 : 1    

(B) 1 : 2

(C) 2 : 1    

(D) Depends on moment of Inertia of two molecules

Solution: Rotational kinetic energy depends only on degree of freedom associated with rotation, which is same for both the diatomic molecules. Both O₂ and N₂ have 2 rotational degrees of freedom.

Answer: (A) 1 : 1

Problem 4: The average translational kinetic energy and the rms speed of molecules in a sample of oxygen at 300K are 6.21 × 10^-21J and 484 m/s respectively. The corresponding values at 600K are nearly (assuming ideal gas behavior):

(A) 12.42 × 10^-21J, 968 m/s    

(B) 8.78 × 10^-21J, 684 m/s

(C) 6.21 × 10^-21J, 968 m/s    

(D) 12.42 × 10^-21J, 684 m/s

Solution:

KE = (3/2)kT and V_rms = √(3RT/M)

Since KE ∝ T:

KE₂/KE₁ = T₂/T₁ = 600/300 = 2

KE₂ = 2 × KE₁ = 2 × 6.21 × 10^-21 = 12.42 × 10^-21 J

Since V_rms ∝ √T:

V_rms,2/V_rms,1 = √(T₂/T₁) = √2

V_rms,2 = √2 × V_rms,1 = √2 × 484 = 684 m/s

Answer: (D) 12.42 × 10^-21J, 684 m/s

Problem 5: At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is:

(A) H₂    

(B) F₂

(C) O₂    

(D) Cl₂

Solution: V_rms = √(3RT/M)

Rearranging: M = 3RT/V_rms²

M = (3 × 8.31 × 300)/(1930)²

M = 7479.3/3,724,900 = 2.0078 × 10^-3 kg = 2.00 gm

This is the molecular weight of hydrogen (H₂)

Answer: (A) H₂

Problem 6: Three closed vessels A, B and C are at the same temperature T and contain gases which obey maxwellian distribution of velocities. Vessel A contains O₂, vessel B only N₂ and C contains mixture of equal quantities of O₂ and N₂. If the average speed of the O₂ molecules in vessel A is V₁, that of N₂ in vessel B is V₂. The average speed of the O₂ molecules in vessel C is:

(A) (V₁+V₂)/2   

(B) V₁

(C) √(V₁² + V₂²)/2    

(D) √(3kT/m)

Solution: The average speed V = √(2kT/m) depends only on temperature and molecular weight. Since the temperature is the same and we're asking about O₂ molecules specifically, the average speed of O₂ molecules in vessel C will be the same as in vessel A.

Answer: (B) V₁

Problem 7: A black body is at a temperature of 2880K. The energy of radiation emitted by this object with wave length between 499 nm and 500 nm is U₁, between 999 nm and 1000nm is U₂ and between 1499nm and 1500nm is U₃. Wien constant b = 2.88 × 10⁶ nmK. Then:

(A) U₁ = 0    

(B) U₃ = 0

(C) U₁ = U₂    

(D) U₂ > U₁

Solution: From Wien's Law: λ_mT = constant, where T is temperature of black body and λ_m is wavelength corresponding to maximum energy of emission.

λ_m = b/T = (2.88 × 10⁶)/(2880) = 1000 nm

Energy distribution analysis:

1. U₁ and U₃ are not zero because a blackbody emits radiation at all wavelengths.
2. Since U₁ corresponds to lower wavelength (499.5 nm average), U₃ corresponds to higher wavelength (1499.5 nm average), and U₂ corresponds to wavelength near maximum (999.5 nm average).
3. Hence U₂ > U₁

Answer: (D) U₂ > U₁

Problem 8: In a room where temperature is 30°C a body cools from 61°C to 59°C in 4 minutes. The time taken by the body to cool from 51°C to 49°C will be:

(A) 4 minutes    

(B) 6 minutes

(C) 5 minutes    

(D) 8 minutes

Solution: Rate of cooling ∝ difference in temperature

-dT/dt ∝ ΔΘ ⟹ -dT/dt = K·ΔΘ

In First Case:

dT = 61° - 59° = 2°

Θ̄ = (60 + 30)/2 = 30°

dt = 4 minutes

∴ K = dT/(dt × Θ̄) = 2/(30 × 4) = 1/60

For second case:

dT = 2°

Θ̄ = (50 - 30) = 20°

∴ dt = dT/(K × Θ̄) = 2/(1/60 × 20) = 6 min

Answer: (B) 6 minutes

Problem 9: A monatomic gas (γ = 5/3) is suddenly compressed to (1/8)th of its volume adiabatically then the pressure of the gas will become _____ times.

(A) (24/5)    

(B) 8

(C) (40/3)   

(D) 32

Solution: For adiabatic process: P₁V₁^γ = P₂V₂^γ

P₂/P₁ = (V₁/V₂)^γ

Given: V₂ = V₁/8

P₂/P₁ = (V₁/(V₁/8))^γ = (8)^γ

P₂/P₁ = (8)^5/3 = (2³)^5/3 = 2⁵ = 32

Answer: (D) 32

Problem 10: Two rods of length ℓ₁ and ℓ₂ are made of materials whose coefficients of linear expansion are α₁ and α₂. If the difference between two lengths is independent of temperature then:

(A) ℓ₁/ℓ₂ = α₂/α₁    

(B) ℓ₁/ℓ₂ = α₁/α₂

(C) ℓ₂²α₂ = ℓ₁²α₁   

(D) ℓ₁²α₁ = ℓ₂²α₂

Solution: If change in length is Δℓ

Then: Δℓ₁ = L₀₁α₁ΔT

Δℓ₂ = L₀₂α₁ΔT

difference in length is ℓ₂ - ℓ₁ = (L₀₂ + Δℓ₂) - (L₀₁ + Δℓ₁)

= (L₀₂ - L₀₁) + (Δℓ₂ - Δℓ₁)

L₀₂ - L₀₁ is independent of temperature for ℓ₂ - ℓ₁ to be independent of temperature.

Δℓ₂ - Δℓ₁ must be equal to zero

i.e., L₁α₁ΔT = L₂α₂ΔT

i.e., L₁α₁ = L₂α₂

i.e., L₁/L₂ = α₂/α₁

Answer: (A) ℓ₁/ℓ₂ = α₂/α₁

Assignment

Q1. At what temperature the root-mean square velocity of hydrogen molecules equal to that of oxygen molecule at 47°C.

(A) 20 K    

(B) 80 K

(C) -73 K    

(D) 3 K

Q2. Two vessels having equal volume contain molecular hydrogen at one atmosphere pressure and helium at two atmospheres pressure respectively. If both samples are at same temperature the mean velocity of hydrogen molecule is:

(A) Equal to that of helium    

(B) Twice that of helium

(C) half of the helium    

(D) √2 times that of helium

Q3. The number of molecules per cc of a gas at NTP is:

(A) 2.68 × 10¹⁷    

(B) 2.68 × 10¹⁴

(C) 6 × 10²⁸   

(D) 22,400 × 6 × 10²³

Q4. If one mole of a monatomic gas (γ = 5/3) is mixed with one mole of diatomic gas (γ = 7/5), the value of γ for the mixture is:

(A) 1.40    

(B) 1.9

(C) 1.53    

(D) 3.07

Q5. The first excited state of hydrogen atom is 10.2 eV above its ground state. The temperature to excite hydrogen atom to first excite level is:

(A) 7.88 × 10⁴°C    

(B) 7.88 × 10⁴K

(C) 78.8 × 10⁴°C    

(D) 78.8 × 10⁴K

Q6. A diatomic gas does 100J of work when it is expanded isobarically. The heat given to the gas during this process equals:

(A) 700 J    

(B) 350 J

(C) 175 J    

(D) 1050 J

Q7. One Kg of ice at 0°C is mixed with 9 kg of water at 45°C then the resulting temperature will be:

(A) 0°C    

(B) 45°C

(C) 32.5°C    

(D) 35°C

Q8. A block of ice of mass 1 kg absorbs 80 K cal and is converted into water at 0°C. The change in entropy of substance is:

(A) Zero    

(B) 80 k cal k⁻¹

(C) 80/273 k cal k⁻¹   

(D) 80 × 273 k Cal × b

Q9. A thermoelectric thermometer works on:

(A) Seebeck effect    

(B) Joule effect

(C) Peltier effect    

(D) Thomson effect

Q10. The length of steel rod which would have the same difference in length with a copper rod of length 16 cm at all temperature (αcopper = 18 × 10⁻⁶ k⁻¹, αsteel = 12 × 10⁻⁶ k⁻¹) is:

(A) 20 cm    

(B) 18 cm

(C) 24 cm    

(D) 30 cm

Q11. At 4°C, 0.98 of the volumes of a body is immersed in water. The temperature at which the entire body get immersed in water (γw = 3.3 × 10⁻⁴ k⁻¹) is (neglect the expansion of the body):

(A) 40.8°C    

(B) 64.6°C

(C) 60.6°C    

(D) 58.8°C

Q12. Starting with the same initial conditions an ideal gas expands from volume v₁ to v₂ in three different ways. The work done by the gas is w₁ if the process is purely isothermal, w₂ if purely isobaric and w₃ if purely adiabatic. Then:

(A) w₂ > w₁ > w₃    

(B) w₂ > w₃ > w₁

(C) w₁ > w₂ > w₃    

(D) w₁ > w₃ > w₂

Q13. A solid body is floating in a liquid bath at temperature 30°C with 3/4 of its volume inside liquid. The temperature is increased and it is found that at 60°C the body starts sinking. Find the coefficient of cubical expansion of liquid neglecting the expansion of solid body:

(A) 1/180 °C⁻¹   

(B) 1/120 °C⁻¹

(C) 1/60 °C⁻¹    

(D) 1/30 °C⁻¹

Q14. A solid ball of metal has a spherical cavity inside it. If the ball is heated the volume of the cavity will:

(A) increases    

(B) decreases

(C) remain unchanged    

(D) have its shape changed

Q15. A metal ball immersed in alcohol weights W₁ at 0°C and W₂ at 50°C. The coefficient of cubical expansion of the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that:

(A) W₁ > W₂    

(B) W₁ = W₂

(C) W₁ < W₂    

(D) None of these

Answers to Assignment

Question No.	Answer	Question No.	Answer	Question No.	Answer
1	A	6	B	11	B
2	D	7	C	12	B
3	B	8	C	13	B
4	C	9	A	14	A
5	B	10	C	15	C