Build complete energy balances, decide the final phase and separate free from constrained expansion.
JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections
BY THE END, YOU CAN
Calculate sensible and latent energy with container heat capacity included.
Determine whether ice or steam fully changes phase before solving for temperature.
Apply linear, area and volume expansion with correct small-change assumptions.
Analyse apparent liquid expansion, thermal stress and bimetallic bending.
BEFORE YOU START
Energy conservation, mass and density.
Linear elasticity, temperature scales and fractional changes.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Temperature is a state variable; heat is energy in transit
Temperature determines the direction of spontaneous thermal energy transfer between systems in contact: net heat flows from higher to lower temperature until equilibrium, unless another process intervenes. Two bodies at the same temperature need not contain the same internal energy, because their masses, materials and phases can differ.
Heat is energy transferred because of a temperature difference; it is not a substance stored inside an object. Internal energy is a property of the state, while heat describes a transfer during a process. A body can also gain internal energy through mechanical or electrical work, so a temperature rise alone does not prove that the transfer occurred as heat.
For temperature differences, a change of 1 K equals a change of 1°C. For absolute-temperature ratios or formulas such as ideal-gas laws and radiation, use kelvin. The conversion is T(K) = t(°C) + 273.15. A sample at 40°C is not twice as hot in the thermodynamic ratio sense as one at 20°C.
Thermal equilibrium means no net thermal transfer between the bodies under the stated contact conditions. It does not mean molecular motion stops. In a calorimetry calculation, the final common temperature is usually assumed uniform, which requires enough time and mixing; a temperature measured immediately after contact may not represent the whole sample.
02 / UNDERSTAND & APPLY
2. Heat capacity relates an energy change to temperature under a specified process
For a small temperature change without phase transition under a specified heating condition, heat capacity C relates heat input to temperature change, Q ≈ CΔT. Specific heat capacity c = C/m is per unit mass, and molar heat capacity is per mole. Their units are J K⁻¹, J kg⁻¹ K⁻¹ and J mol⁻¹ K⁻¹ respectively.
If c is approximately constant over the interval, Q = mc(T_final − T_initial). If it varies appreciably, use Q = m∫c(T)dT along the specified process. A large specific heat means more energy is required per kilogram for the same temperature rise; it does not imply the sample begins with a larger temperature.
For gases, the process matters strongly: heating at constant pressure can include expansion work, so C_P differs from C_V. For ordinary solid and liquid calorimetry, volume-change work is often small and the supplied c is used under the stated pressure conditions. Treating Q = mcΔT as a universal identity for every thermodynamic path would be incorrect.
An electrical heater delivering constant power P supplies energy Pt. If a known fraction η enters the sample as useful thermal energy, ηPt = mcΔT when there is no phase change and other losses are accounted for by η. If the container also warms, its heat capacity belongs in the balance. Heater rating alone is not necessarily the absorbed sample power.
ORIGINAL WORKED EXAMPLE
Heating a sample and its container
A 0.20 kg liquid with c = 4000 J kg⁻¹ K⁻¹ is in a container of heat capacity 100 J K⁻¹. A heater supplies 180 W to the combined system with negligible losses. Find the time for a 20 K rise, with no phase change.
Liquid heat capacity is mc = 0.20(4000) = 800 J K⁻¹.
Combined capacity is 800 + 100 = 900 J K⁻¹, so required energy is 900(20) = 18000 J.
Time is energy/power = 18000/180 = 100 s. Ignoring the container would underestimate it.
Answer 100 s.
03 / UNDERSTAND & APPLY
3. Calorimetry is a conservation-of-energy equation for the chosen system
For an insulated collection of bodies exchanging only thermal energy internally, the sum of their energy changes is zero. With constant heat capacities and no phase changes, Σmᵢcᵢ(T_f − Tᵢ) + C_container(T_f − T_container,initial) = 0. Hot parts contribute negative changes and cold parts positive changes when the signs are kept consistently.
The familiar statement “heat lost equals heat gained” is a convenient rearrangement, not a separate conservation law. Include every part that changes temperature: water, metal, vessel, stirrer or thermometer if their heat capacities matter. If external heating, leakage, evaporation or chemical reaction is present, those energy transfers require additional terms.
Without phase changes, reactions or external work, the final equilibrium temperature must lie between the initial temperatures for positive heat capacities. An answer outside that range often reveals a sign or unit error. A simple arithmetic average is correct only when the combined heat capacities of the two parts are equal, not merely when their masses match.
A calorimeter’s water equivalent is a mass of water having the same heat capacity, W = C_calorimeter/c_water. It is a convenient parameter, not the actual mass of water hidden inside the apparatus. If W is given, replace container heat capacity by Wc_water once; do not add both the water-equivalent term and the same container capacity again.
ORIGINAL WORKED EXAMPLE
Mixing water with a hot metal
A 0.10 kg metal sample with specific heat 400 J kg⁻¹ K⁻¹ at 100°C is placed in 0.20 kg water at 20°C. Take c_water = 4200 J kg⁻¹ K⁻¹, neglect container capacity and losses, and assume no phase change. Find final temperature.
Metal heat capacity is 40 J K⁻¹; water heat capacity is 840 J K⁻¹.
Energy balance is 40(T_f − 100) + 840(T_f − 20) = 0.
The result lies close to water’s initial temperature because water has much greater total heat capacity.
Answer Approximately 23.6°C under the stated model.
04 / UNDERSTAND & APPLY
4. A phase transition can absorb energy without raising temperature
At a specified equilibrium transition pressure, energy supplied during melting or boiling can change phase while temperature remains at the transition value. The latent heat per mass L gives Q = mL for the amount that changes phase. Fusion and vaporisation have different latent heats; neither should be replaced by a specific heat times an assumed temperature interval.
An ideal heating curve consists of sloped single-phase segments and flat coexistence segments when pressure is fixed and the transition is sharp. Along a slope, added energy changes temperature; on a plateau, it changes phase fraction. The plateau is not evidence that energy input has stopped. Real mixtures, finite-rate gradients and changing pressure can make the curve more complicated.
To warm ice below melting temperature and obtain warm liquid water, use separate stages: warm solid to melting temperature, supply fusion energy, then warm liquid. To make steam, add liquid heating to the boiling point and vaporisation, then any superheating. Use the appropriate heat capacity and latent heat for each stage.
At the usual atmospheric-pressure idealization, ice–water coexistence is near 0°C and water–steam coexistence near 100°C. These are pressure-dependent transition conditions, not fixed universal temperatures under every circumstance. A closed heated vessel can change pressure, so its path need not follow an open-pan constant-pressure curve.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative model using cice 2100 J/(kg K), fusion latent heat 334000 J/kg and cwater 4200 J/(kg K), at the specified atmospheric-pressure transition. The plateau absorbs energy without a temperature rise.
ORIGINAL WORKED EXAMPLE
Energy to melt and warm ice
Find energy needed to turn 0.050 kg ice at −10°C into water at 20°C. Use c_ice = 2100 J kg⁻¹ K⁻¹, fusion latent heat 3.34 × 10⁵ J kg⁻¹ and c_water = 4200 J kg⁻¹ K⁻¹ at the stated atmospheric-pressure model.
Warm the ice to 0°C: Q₁ = 0.050(2100)(10) = 1050 J.
Melt it: Q₂ = 0.050(3.34 × 10⁵) = 16700 J.
Warm the resulting water: Q₃ = 0.050(4200)(20) = 4200 J.
Total is 21950 J, approximately 22 kJ. Most of the energy in this example goes into melting.
Answer Approximately 22 kJ.
05 / UNDERSTAND & APPLY
5. Check whether enough energy exists to complete the phase change
When ice meets warmer water, do not assume immediately that all ice melts. First calculate the energy the warm parts can release by cooling to the melting temperature. Compare it with the energy needed to warm any subzero ice and then melt all the ice. The comparison decides which final-state equation is physically valid.
If there is insufficient energy to melt all the ice after reaching 0°C in the atmospheric-pressure model, the final mixture can contain both ice and water at 0°C. Solve for the melted mass using available energy divided by latent heat. Assigning a positive final temperature while solid ice remains in ordinary equilibrium would contradict the assumed phase conditions.
If all ice melts and surplus energy remains, distribute that surplus to heat the total final liquid mass and any container. If the initial cold material absorbs more energy than the warm water can release before freezing begins, some water may freeze; a separate stage and phase balance are then needed. Energy conservation is the same, but the final phase assumption changes the algebra.
Steam condensation can transfer much more energy than merely cooling the same mass of hot liquid over a moderate temperature range. A steam-mixing problem must include latent energy released on condensation, and then cooling of the condensate if appropriate. If not all steam condenses, the final phase mixture must again be determined rather than assumed.
ORIGINAL WORKED EXAMPLE
Not all the ice melts
An insulated vessel contains 0.20 kg water at 20°C and receives 0.10 kg ice at 0°C. Neglect container capacity. Use c_water = 4200 J kg⁻¹ K⁻¹ and L_f = 3.36 × 10⁵ J kg⁻¹. Find the final state.
Water can release 0.20(4200)(20) = 16800 J by cooling to 0°C.
Melting all ice would require 0.10(3.36 × 10⁵) = 33600 J, which is twice the available energy.
Melted ice mass is 16800/(3.36 × 10⁵) = 0.050 kg.
Final state is 0.050 kg ice plus 0.25 kg liquid water at 0°C. No energy remains for a temperature rise.
Answer A 0°C mixture: 0.050 kg ice and 0.25 kg water.
06 / UNDERSTAND & APPLY
6. Free thermal expansion changes length, area and volume
For a small uniform temperature change with approximately constant coefficient of linear expansion α, a freely expanding length changes by ΔL = αL₀ΔT. The coefficient has units K⁻¹. It is often positive, but its sign and temperature dependence are material properties rather than definitions. The approximation assumes fractional changes remain small.
For an isotropic solid, each linear dimension has the same fractional change. Area therefore changes approximately by ΔA = 2αA₀ΔT and volume by ΔV = 3αV₀ΔT, neglecting higher-order products. More generally use an area or volume expansion coefficient appropriate to the material; anisotropic crystals need direction-dependent information.
A hole in a uniformly heated free plate expands along with the surrounding material. Imagine filling it with an identical piece: that piece would expand consistently with the rest, so the boundary of the hole moves outward. This reasoning assumes uniform free expansion; constrained or uneven heating can change the geometry.
For fixed mass, density varies inversely with volume. With volume coefficient β, ρ(T) = ρ₀/[1 + βΔT] in the first-order volume model, or approximately ρ₀(1 − βΔT) for a small fractional change. Do not change mass merely because the material expands. Comparing buoyancy after heating may require both object and fluid density changes.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Schematic expansion is strongly exaggerated for visibility; the drawn factor is not the numerical strain in the worked example. Uniform isotropic unconstrained expansion is assumed.
ORIGINAL WORKED EXAMPLE
A heated plate with a circular hole
A freely expanding isotropic plate has α = 2.0 × 10⁻⁵ K⁻¹. A circular hole initially has diameter 10.0 mm. The plate is heated uniformly by 100 K. Find the first-order diameter change and fractional area change.
The hole’s diameter expands like any length: Δd = αd₀ΔT = 2.0 × 10⁻⁵(10.0 mm)(100) = 0.020 mm.
The fractional diameter increase is 0.0020.
Area fractional increase is approximately twice that, 0.0040 or 0.40%. The hole grows rather than closes.
Answer Diameter increases by 0.020 mm; area increases by approximately 0.40%.
07 / UNDERSTAND & APPLY
7. Containers and supports change what expansion is observed
When a liquid expands inside a container, the visible rise or overflow depends on both expansions. For small changes from the same initial volume V₀, apparent volume increase relative to the container is approximately V₀(β_liquid − β_container)ΔT. A solid container’s capacity expands with its volumetric coefficient, approximately 3α for an isotropic material.
If a rod is prevented from expanding between perfectly rigid supports, its free thermal strain αΔT is cancelled by elastic mechanical strain. Under the small linear uniaxial model, heating produces compressive stress of magnitude YαΔT. Cooling from an initially unstressed constrained state produces tensile stress if the supports maintain contact and the rod cannot separate.
The fully prevented-expansion formula assumes rigid supports, uniform temperature, no yielding, no buckling and no initial gap. If a gap exists, the rod expands freely until the gap closes, and stress develops only for further temperature rise. If the supports themselves expand or deform, their compatibility must be included.
A bonded bimetallic strip bends because the two layers would freely expand by different amounts. On heating, the higher-expansion layer tends to occupy the longer outer arc and the strip bends toward the lower-expansion side. The actual curvature depends on thicknesses and elastic properties, so the simple qualitative direction is not a complete curvature formula.
Water near ordinary pressure contracts when heated from about 0°C to about 4°C and expands above that range. Its density is greatest near 4°C. A single positive constant β cannot model both sides of this anomaly. Ice is less dense than liquid water near melting, which also matters for floating and volume-change questions.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative piecewise linear model with alpha 10⁻⁵ K⁻¹ and Y 2×10¹¹ Pa. Rigid supports, uniform temperature, no buckling and no yielding are assumed.
ORIGINAL WORKED EXAMPLE
Expansion first closes a gap
A 1.0 m rod with α = 1.0 × 10⁻⁵ K⁻¹ and Y = 2.0 × 10¹¹ Pa has one end fixed; its other end faces a 0.20 mm gap to a rigid end stop. It is uniformly heated by 50 K from an unstressed state. Find the gap-closing temperature rise and final stress, assuming no buckling or yielding.
Gap closes when αLΔT_gap = 0.00020 m, giving ΔT_gap = 0.00020/(1.0 × 10⁻⁵ × 1.0) = 20 K.
Only the remaining 50 − 20 = 30 K of attempted expansion is prevented.
Answer Gap closes after 20 K; final compressive stress 60 MPa.
08 / UNDERSTAND & APPLY
8. Original practice with phase and expansion checks
Choose the final phase before solving a one-line temperature equation. For expansion, distinguish free dimensions from constrained deformation and actual liquid volume from apparent volume relative to its container.
ORIGINAL PRACTICE · 1
A 0.50 kg sample with c = 800 J kg⁻¹ K⁻¹ warms by 10 K. Find energy absorbed in the stated single-phase model.
Reveal answer & reasoning
4000 J.
Q = mcΔT = 0.50(800)(10).
ORIGINAL PRACTICE · 2
A body has heat capacity 200 J K⁻¹ and mass 0.40 kg. Find specific heat capacity.
Reveal answer & reasoning
500 J kg⁻¹ K⁻¹.
c = C/m.
ORIGINAL PRACTICE · 3
Equal heat-capacity bodies at 10°C and 50°C equilibrate without loss or phase change. Find final temperature.
Reveal answer & reasoning
30°C.
Their equal capacities give the arithmetic mean. Equal mass alone would not guarantee it.
ORIGINAL PRACTICE · 4
A calorimeter has water equivalent 0.020 kg. With c_water = 4200 J kg⁻¹ K⁻¹, find its heat capacity.
Reveal answer & reasoning
84 J K⁻¹.
C = Wc_water.
ORIGINAL PRACTICE · 5
Does a melting plateau mean no energy is entering?
Reveal answer & reasoning
No.
Energy changes phase fraction through latent heat while temperature remains at the transition value.
ORIGINAL PRACTICE · 6
How much energy melts 0.020 kg ice at 0°C if L_f = 3.3 × 10⁵ J kg⁻¹?
Reveal answer & reasoning
6600 J.
Q = mL_f.
ORIGINAL PRACTICE · 7
An ice–water mixture remains at 0°C while heat is added. Must all the ice have melted?
Reveal answer & reasoning
No.
The supplied energy can be melting only part of the ice, with both phases coexisting.
ORIGINAL PRACTICE · 8
A free rod has α = 2 × 10⁻⁵ K⁻¹ and length 1 m. Find extension for 50 K heating.
Reveal answer & reasoning
1.0 mm.
ΔL = αLΔT = 0.001 m.
ORIGINAL PRACTICE · 9
For an isotropic solid with linear coefficient α, what is the first-order volume coefficient?
Reveal answer & reasoning
Approximately 3α.
All three dimensions expand, and products of small strains are neglected.
ORIGINAL PRACTICE · 10
A free uniformly heated metal ring has positive α. Does its hole shrink?
Reveal answer & reasoning
No, it expands.
The entire geometry scales outward under uniform unconstrained expansion.
ORIGINAL PRACTICE · 11
A material’s volume increases by 1% at fixed mass. Is density exactly 1% smaller?
Reveal answer & reasoning
It becomes ρ₀/1.01, approximately 0.9901ρ₀.
Density is inverse volume; a 1% decrease is the first-order approximation.
ORIGINAL PRACTICE · 12
A liquid and container have equal volumetric expansion coefficients. What is their first-order apparent expansion?
Reveal answer & reasoning
Zero.
β_apparent = β_liquid − β_container.
ORIGINAL PRACTICE · 13
A rod is fully prevented from expanding while heated uniformly in the linear elastic regime. Is stress tensile or compressive?
Reveal answer & reasoning
Compressive.
The supports oppose the rod’s attempted expansion.
ORIGINAL PRACTICE · 14
For a heated bimetallic strip, which layer lies on the outside of the bend?
Reveal answer & reasoning
The higher-expansion layer under the standard bonded-layer model.
Its greater preferred length is accommodated by the longer outer arc.
09 / UNDERSTAND & APPLY
9. Write the energy inventory before chasing a final temperature
List each initial mass, temperature, phase and heat capacity. Include the container once and name any external energy transfer. If a transition is possible, compare available energy with the latent-energy threshold first. Only then decide whether the unknown is a final temperature, a melted mass or a remaining phase fraction.
For expansion, identify the relevant coefficient and whether the body is free, contained or constrained. Use temperature differences directly in kelvin or Celsius degrees, but use absolute temperature for ratios in other laws. Check that the resulting strain is small enough for the approximation and that any claimed stress has not silently ignored a gap or deformable support.
Heat is a transfer; internal energy and temperature describe different aspects of state.
Specific heat formulas require a process and no unaccounted phase transition.
Final-temperature averages use heat capacities, not merely masses.
Latent-energy thresholds determine whether a phase change completes.
Uniform free heating expands holes as well as outer dimensions.
Constrained expansion creates stress only after any gap is closed.
KEEP THE BIG IDEAS
Quick revision
A calorimetry equation is an energy balance for a clearly chosen system.
The final phase must be checked rather than assumed.
Expansion depends on geometry and constraints, not only on temperature rise.
Temperature differences and absolute-temperature ratios use different conversions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.