Physics for UPSC – Chapter 10: Thermal Properties of Matter, Calorimetry and Heat Transfer

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Physics for UPSC · Chapter 10

Thermal Properties of Matter

Detailed Physics theory on temperature, thermal expansion, calorimetry, latent heat, conduction, convection, radiation and Newton’s law of cooling, including UPSC PYQ analysis and 20 MCQs.

12 theory modulesIn-depth explanation
7 numericalsFully explained
20 MCQsAnswer and reasoning

Heat, temperature and thermal equilibrium

Heat is energy transferred between systems because of a temperature difference; it is not a substance stored inside an object. Temperature is a property that determines the direction of spontaneous heat transfer when systems are in thermal contact. If a hot metal block touches colder water, energy normally flows into the water until thermal equilibrium is approached. Internal energy, by contrast, is a state property related to microscopic motion and interactions.

The zeroth law states that if bodies A and B are each in thermal equilibrium with C, they are in thermal equilibrium with each other. This makes reliable thermometers possible. A large cold reservoir can contain more internal energy than a tiny hot object: temperature and total internal energy must not be treated as synonyms.

  • Heat Q is measured in joules; temperature is measured in kelvin (SI).
  • At equilibrium there is no net heat flow between two bodies at the same temperature.

Temperature scales, measurement and conversion

The kelvin scale is an absolute thermodynamic temperature scale: T(K) = t(°C) + 273.15. Celsius and kelvin increments have the same magnitude: a change of 10 °C equals a change of 10 K. Fahrenheit is related by F = (9/5)C + 32. Converting a temperature is different from converting a temperature interval; adding 273.15 to an interval is an error.

At standard atmospheric pressure pure water freezes around 0 °C and boils around 100 °C; these temperatures are not immutable at every altitude or pressure. A thermometer works by exploiting a physical property that varies with temperature, such as resistance, expansion or emitted thermal radiation. The measurement must be calibrated to an accepted scale.

  • 0 °C corresponds to 273.15 K, and 100 °C to 373.15 K.
  • Absolute temperature in gas-law and blackbody formulas must be kelvin.

Linear, area and volume thermal expansion

Solids usually expand when heated because average separation between constituent particles increases. For a uniform rod under a moderate temperature change, ΔL = αL₀ΔT, where α is the coefficient of linear expansion. For an isotropic material with small changes, the area coefficient is approximately β = 2α and the volume coefficient γ = 3α. These are approximate first-order relations.

A rod containing a circular hole expands the size of the hole when the whole plate is heated uniformly: imagine the missing region made from the same material. Heating rails or bridges without room to expand can create significant thermal stress. The coefficient α is expressed in K⁻¹, so the product αΔT has no dimensions.

  • Thermal expansion coefficients depend on material and conditions.
  • The approximate β and γ relationships require small expansion and isotropic behaviour.
Examination approach: Identify the model assumptions, define variables with SI units and check limiting cases before answering.

Anomalous expansion of water and environmental importance

Unlike most liquids across their ordinary temperature ranges, water behaves anomalously near its freezing point. At atmospheric pressure liquid water reaches its maximum density near 4 °C. As water cools from 4 °C toward 0 °C, its volume generally increases; ice is less dense than liquid water, which is why it floats.

In cold lakes, surface water cools and becomes denser until about 4 °C, then colder surface water tends to remain above the denser water. Ice often forms at the surface, creating an insulating layer and allowing liquid water beneath to support aquatic life. Do not claim all water everywhere is densest at exactly 4 °C irrespective of pressure and dissolved substances.

  • Near 4 °C, fresh water has maximum density at roughly atmospheric pressure.
  • Water’s freezing expansion is a well-known cause of weathering and pipe damage.

Specific heat capacity, heat capacity and sensible heat

The specific heat capacity c is the heat per unit mass required for a temperature change of one kelvin under specified conditions: Q = mcΔT when c is roughly constant and no phase change occurs. Heat capacity C = mc applies to the specified object. A high specific heat means a substance requires relatively more energy per kilogram for the same temperature increase.

Water has a relatively high specific heat capacity, contributing to the moderated temperature swings near large water bodies. Heat capacity for gases depends on process conditions, giving distinct Cp and Cv. In problems that mix substances, identify which portions gain and lose heat; do not set the heat gained by each piece equal unless energy balance demands it.

  • Specific heat capacity SI unit: J·kg⁻¹·K⁻¹.
  • The equation Q = mcΔT does not account for latent heat during a phase change.

Calorimetry and the method of mixtures

An insulated calorimeter allows estimating heat transfer by using energy conservation. In an ideal mixing question, heat lost by the initially hot substances equals heat gained by the initially cold substances, including the calorimeter itself if its heat capacity is given. The final common temperature generally lies between initial temperatures when no phase transition or chemical reaction occurs.

Consider masses m₁ and m₂ of the same liquid at temperatures T₁ and T₂ with no environmental losses: final T = (m₁T₁ + m₂T₂)/(m₁ + m₂). With different heat capacities, use a weighted heat-capacity sum rather than an ordinary average. For ice-water or steam-water mixtures, first calculate latent heat before treating temperature changes.

  • Include the vessel’s heat capacity if it is non-negligible.
  • Check whether complete melting or condensation actually occurs before applying the final-temperature equation.
Examination approach: Identify the model assumptions, define variables with SI units and check limiting cases before answering.

Latent heat and changes of state

When a pure substance changes phase at approximately constant pressure, it can absorb or release latent heat without a temperature change during the transition. For a mass m, the heat associated with a phase change is Q = mL, where L is specific latent heat. Melting ice at 0 °C requires energy even though its temperature remains 0 °C until the melting process is completed.

Latent heat of vaporisation is generally much greater than latent heat of fusion for water at ordinary pressure. Steam condensing onto skin can release significant latent heat, explaining why burns from steam can be severe. Evaporation can occur below boiling temperature because faster molecules escape from a liquid’s surface; boiling involves vapour formation throughout the liquid under appropriate pressure.

  • L has unit J/kg.
  • The temperature plateau in an ideal heating curve represents a phase change, not zero energy input.

Conduction and thermal resistance

Conduction transfers energy through a material from hotter to colder regions without bulk transport of the material. In a uniform slab in steady one-dimensional conduction, the magnitude of heat current is H = kAΔT/L, where k is thermal conductivity, A area, L thickness and ΔT temperature difference. Good thermal conductors typically have larger k than good insulators.

Increasing insulation thickness reduces conduction in this ideal slab model; doubling area doubles conduction under otherwise fixed conditions. Series thermal layers behave like resistances: Rth = L/(kA), and heat current H = ΔT/Rth. Real buildings also lose heat by air infiltration, convection and radiation, so conduction alone is not a complete efficiency analysis.

  • Thermal conductivity uses W·m⁻¹·K⁻¹.
  • Conduction is not the same as bulk circulation of hot fluid.

Convection and heat transfer in fluids

Convection involves heat transfer between a surface and moving fluid, combining bulk fluid motion with conduction in the boundary layer. Natural convection occurs when buoyancy caused by temperature-dependent density differences drives motion; forced convection uses a fan, pump or wind. Heated air above a radiator generally rises in ordinary room conditions due to lower density.

A standard engineering approximation is H = hA(Ts−T∞), where h depends on fluid properties, geometry and flow. Convection cannot occur through a perfect vacuum because no material fluid is present. It is a major component of cooling electronics, atmospheric motion and ocean circulation.

  • Convection requires a fluid capable of motion.
  • Air-conditioning and ventilation design rely on fluid flow and thermal exchange.
Examination approach: Identify the model assumptions, define variables with SI units and check limiting cases before answering.

Thermal radiation and blackbody ideas

All bodies above absolute zero emit electromagnetic radiation due to their thermal state. Radiation can transfer energy through a vacuum, unlike conduction and convection. An ideal blackbody absorbs all incident radiation and emits thermal radiation characterised by temperature. For a perfect blackbody, emitted power P = σAT⁴, where σ is the Stefan–Boltzmann constant.

Real surfaces have emissivity ε, giving approximately P = εσAT⁴ for their own thermal emission; net radiation exchange with a large surrounding at temperature T₀ is Pnet = εσA(T⁴−T₀⁴) under simple assumptions. Wien’s displacement law λmaxT = b means hotter blackbodies peak at shorter wavelengths. Radiation is crucial for solar energy, space heat transfer and climate.

  • Radiation can travel through vacuum.
  • Use kelvin, not Celsius, in any fourth-power temperature expression.

Newton’s law of cooling and real limits

Newton’s empirical law of cooling states that, under suitable conditions, the rate of temperature change is proportional to the temperature difference between an object and its surroundings: dT/dt = −k(T−Ts) for lumped cooling with constant parameters. Its solution is an exponential approach T(t)−Ts = [T(0)−Ts]e−kt.

This model is useful for relatively small temperature differences when convection is dominant and the object remains approximately at a uniform internal temperature. It is not a universal law of all cooling because radiative losses vary with T⁴ and fluid heat-transfer coefficients may change. A larger instantaneous temperature excess generally means faster cooling in the valid regime.

  • An object approaches ambient temperature asymptotically in the simple exponential model.
  • Radiative heat loss cannot always be represented by a constant cooling coefficient.

Applications, comparisons and exam reasoning

Thermal concepts connect diverse UPSC science questions: railway expansion gaps, thermos flasks, cold-weather aquatic life, evaporation cooling, ice-versus-water energy balances, and why silvered surfaces suppress radiative exchange. A thermos reduces conduction through its structure, convection by vacuum insulation, and radiation through reflective surfaces; no single mechanism explains its entire design.

Solve combined heat problems in stages: determine warming/cooling to the phase boundary, add latent heat, then determine any further temperature change. At every stage conserve energy and check the available heat supply. If assumed all ice melts yet the energy is insufficient even for melting, revisit that assumption before computing a final temperature.

  • Separate phase-change energy from sensible heating.
  • Classify each heat-transfer process: conduction, convection or radiation.
Examination approach: Identify the model assumptions, define variables with SI units and check limiting cases before answering.

Formula sheet and physical assumptions

A formula is useful only under the conditions stated in the theory. Distinguish absolute temperature in kelvin from Celsius temperature differences.

Kelvin–CelsiusT(K) = t(°C)+273.15
Fahrenheit–CelsiusF = 9C/5 + 32
Linear expansionΔL = αLΔT
Area/volume coefficientsβ ≈ 2α; γ ≈ 3α
Sensible heatQ = mcΔT
Latent heatQ = mL
Steady conductionH = kAΔT/L
Thermal resistanceRth = L/(kA)
Blackbody radiationP = σAT⁴
Net radiationPnet = εσA(T⁴−T₀⁴)
Wien’s lawλmaxT = b
Newton coolingT−Ts = (T0−Ts)e⁻ᵏᵗ

Detailed worked numericals

Each example identifies the model, substitutes quantities consistently and interprets the result rather than merely reporting a number.

Example 1. Convert 113 °F to Celsius and kelvin

C = (113−32)×5/9 = 45 °C; T = 45+273.15 = 318.15 K.

Example 2. Wire expansion

For L = 2 m, α = 1.2×10⁻⁵ K⁻¹ and ΔT = 50 K, ΔL = 2×1.2×10⁻⁵×50 = 1.2×10⁻³ m = 1.2 mm.

Example 3. Heating water

For 0.5 kg of water, c = 4200 J kg⁻¹ K⁻¹ and a 20 K rise, Q = 0.5×4200×20 = 42,000 J.

Example 4. Ice melting

To melt 0.10 kg ice at 0 °C using Lf = 3.34×10⁵ J/kg, Q = 0.10×3.34×10⁵ = 33.4 kJ.

Example 5. Mixing identical liquids

100 g water at 80 °C mixed with 100 g water at 20 °C in an insulated negligible-capacity container gives 50 °C.

Example 6. Rate of conduction

A slab has k=0.5 W/(m K), A=2 m², ΔT=20 K and thickness 0.1 m. H=0.5×2×20/0.1=200 W.

Example 7. Cooling over one half-life

If excess temperature falls to half in 10 min, initial excess 40 K becomes 20 K after 10 min and 10 K after 20 min, assuming unchanged exponential cooling.

UPSC PYQ analysis: concept-based past-paper references

These are paraphrased topic descriptions indexed under historical UPSC NDA General Ability Test papers, not verbatim question statements or official answer keys. Session/question labels come from a third-party index and should be checked against the UPSC original question-paper archive before quoting the question. The reasoning below is independently prepared by CrackTarget.

PAST-PAPER CONCEPT 1

UPSC NDA (I) 2022 · indexed Heat concept

Question focus: determine the mass of a material from heat supplied, specific heat and temperature rise.

How to solve: Rearrange Q=mcΔT as m=Q/(cΔT). Convert all temperatures to differences; a 10 °C change is also a 10 K change.

Indexed NDA Physics questions and source references
PAST-PAPER CONCEPT 2

UPSC NDA (I) 2022 · indexed Heat concept

Question focus: identify the meaning of specific latent heat of vaporisation.

How to solve: Specific latent heat is heat per unit mass required for liquid-to-vapour conversion at the relevant phase-transition conditions, without sensible temperature rise.

Indexed NDA Physics questions and source references
PAST-PAPER CONCEPT 3

UPSC NDA (II) 2022 · indexed Heat Transfer concept

Question focus: identify the heat-transfer mechanism that can propagate at the speed of light.

How to solve: Electromagnetic thermal radiation travels through vacuum at light speed; ordinary conduction and convection require material media.

Indexed NDA Physics questions and source references

UPSC-style MCQs: 20 original questions with explanations

These are new examination-style exercises, not previously asked UPSC questions. Open the correct answer only after attempting each question.

Q1. The SI unit of specific heat capacity is:
  1. J/K
  2. J/(kg·K)
  3. W/m
  4. J/kg
Show answer and explanation

Answer B. c is energy per mass per temperature interval.

Q2. Zero degree Celsius equals:
  1. 0 K
  2. 100 K
  3. 273.15 K
  4. 373.15 K
Show answer and explanation

Answer C. T(K)=t(°C)+273.15.

Q3. A 20 °C rise corresponds to a temperature difference of:
  1. 20 K
  2. 253 K
  3. 293 K
  4. 36 K
Show answer and explanation

Answer A. Kelvin and Celsius intervals have equal size.

Q4. The zeroth law primarily establishes:
  1. charge conservation
  2. thermal equilibrium transitivity
  3. mass-energy relation
  4. refractive index
Show answer and explanation

Answer B. It provides the basis for comparing temperatures using thermometers.

Q5. A uniformly heated metal plate containing a circular hole will normally show a hole that:
  1. shrinks
  2. expands
  3. closes completely
  4. remains always identical
Show answer and explanation

Answer B. All lengths scale under uniform expansion, including the empty hole.

Q6. For small isotropic expansion, volumetric coefficient is approximately:
  1. α/3
  2. α
  3. 2α
  4. 3α
Show answer and explanation

Answer D. First-order volume expansion coefficient γ≈3α.

Q7. Fresh water at normal pressure has greatest density near:
  1. 0 °C
  2. 4 °C
  3. 20 °C
  4. 100 °C
Show answer and explanation

Answer B. The familiar anomalous expansion gives maximum density near 4 °C.

Q8. The equation Q=mcΔT applies directly when:
  1. phase change occurs
  2. no phase change and c approximately constant
  3. mass becomes zero
  4. temperature stays constant during melting
Show answer and explanation

Answer B. Sensible heating changes temperature without a latent-heat transition.

Q9. Latent heat of fusion is measured in:
  1. J/kg
  2. J/(kg·K)
  3. N/m
  4. W/m
Show answer and explanation

Answer A. Specific latent heat is energy per unit mass.

Q10. Which explains cooling by evaporation?
  1. fast molecules escape and take energy
  2. evaporation adds energy to the remaining liquid
  3. water cannot evaporate below boiling
  4. molecules lose mass
Show answer and explanation

Answer A. Escape of energetic molecules tends to lower remaining average kinetic energy.

Q11. Heat transferred by electromagnetic waves through vacuum is:
  1. conduction
  2. convection
  3. radiation
  4. fluid diffusion
Show answer and explanation

Answer C. Radiation requires no medium.

Q12. Steady conduction through a slab becomes twice as fast when area:
  1. halves
  2. doubles
  3. becomes one-fourth
  4. does not matter
Show answer and explanation

Answer B. H=kAΔT/L.

Q13. The unit of thermal conductivity is:
  1. W/(m·K)
  2. J/kg
  3. kg/m³
  4. N·m
Show answer and explanation

Answer A. From H=kAΔT/L, k has W/(m K).

Q14. A blackbody’s ideal emission power varies as:
  1. T
  2. T²
  3. T³
  4. T⁴
Show answer and explanation

Answer D. Stefan–Boltzmann gives P∝T⁴.

Q15. If an ideal blackbody’s absolute temperature doubles, emission becomes:
  1. 2 times
  2. 4 times
  3. 8 times
  4. 16 times
Show answer and explanation

Answer D. (2T)⁴=16T⁴.

Q16. Wien’s law predicts hotter thermal bodies have peak wavelengths:
  1. longer
  2. shorter
  3. unchanged
  4. undefined
Show answer and explanation

Answer B. λmax=b/T.

Q17. Which property measures heat per degree for an entire object?
  1. heat capacity
  2. specific latent heat
  3. conductivity
  4. density
Show answer and explanation

Answer A. Heat capacity C=Q/ΔT.

Q18. For equal masses of identical liquid mixed at 10°C and 30°C with no losses, final T:
  1. 10°C
  2. 15°C
  3. 20°C
  4. 30°C
Show answer and explanation

Answer C. The equal-heat-capacity mean is 20°C.

Q19. Newton’s cooling law in its simple model predicts excess temperature changes:
  1. exponentially
  2. linearly in all situations
  3. never
  4. as T⁴ exactly
Show answer and explanation

Answer A. With constant cooling coefficient, excess temperature decays exponentially.

Q20. Ice floating on a lake can help aquatic life survive because:
  1. ice is denser than water
  2. surface ice insulates liquid below
  3. ice eliminates gravity
  4. water has zero heat capacity
Show answer and explanation

Answer B. Ice floats and reduces further heat loss, while water near 4°C can remain below.

Last-minute revision checklist

  • Heat is energy transferred; temperature is not heat stored in a system.
  • Never add 273.15 to a temperature difference.
  • Use latent heat when phase changes; use mcΔT for sensible heating.
  • The 4 °C density maximum refers to ordinary fresh-water conditions.
  • Thermal radiation works in vacuum; conduction and convection require matter.
  • T⁴ laws require absolute temperature in kelvin.
Study routine: Revise the formulas for 10 minutes, solve two numericals without notes and reattempt missed MCQs the following day.