Thermal properties of matter for JEE Main

Thermal physics questions in JEE Main are rarely hard in concept. They go wrong on bookkeeping: a mixture where not all the ice melts, a composite wall where the slabs are in series rather than parallel, or a temperature in °C plugged into a T⁴ law. This guide gives you the ideas and, just as important, the checks that stop those errors.

The syllabus covers heat and temperature, thermal expansion, specific heat capacity and calorimetry, change of state and latent heat, and heat transfer by conduction, convection and radiation.

Temperature scales

The Celsius, Fahrenheit and Kelvin scales are linked by:

C/100 = (F − 32)/180 = (K − 273.15)/100

A change of 1 °C equals a change of 1 K and of 1.8 °F. The Celsius and Fahrenheit readings are equal at −40.

Thermal expansion

These hold for small ΔT and for isotropic solids. A few consequences come up again and again:

• Liquids in containers: the vessel expands too, so the apparent expansion coefficient is γ_liquid − γ_vessel.
• Pendulum clocks: the period is proportional to √L, so ΔT_period/T_period = ½αΔθ. A clock with a metal pendulum loses time in summer.
• Thermal stress: a rod clamped between rigid walls and heated develops stress = YαΔθ, because the wall prevents the expansion.
• Water contracts on heating from 0 °C to 4 °C and is densest at about 4 °C. That is why lakes freeze from the top down.
• Holes expand as if they were made of the surrounding material.

Calorimetry and change of state

• Heat to change temperature: Q = mcΔT, where c is the specific heat capacity. mc is the heat capacity of the body.
• Heat to change phase at a constant temperature: Q = mL, where L is the latent heat.
• Principle of calorimetry: in an insulated system, heat lost by the hot bodies = heat gained by the cold ones.

Use whatever values the question gives.

Conduction

The rate of heat flow through a slab is H = kAΔT/L, where k is the thermal conductivity.

Define the thermal resistance R = L/(kA). Then H = ΔT/R, exactly like current = voltage/resistance, and the same rules apply.

• Series (heat passes through one slab and then the next): R = R₁ + R₂. The same H flows through each. For two slabs of equal length and area, k_eq = 2k₁k₂/(k₁ + k₂).
• Parallel (slabs side by side between the same two temperatures): 1/R = 1/R₁ + 1/R₂. For equal lengths and areas, k_eq = (k₁ + k₂)/2.

At the junction of slabs in series, the temperature follows from equating H in each slab.

Convection

Convection carries heat by the bulk motion of a fluid. Hot fluid is less dense and rises. Sea breezes and the heating of a room by a heater near the floor are examples. JEE Main questions on convection are usually conceptual.

Radiation

• Stefan–Boltzmann law: a body of emissivity e, area A and absolute temperature T radiates P = eσAT⁴. Its net loss to surroundings at T₀ is eσA(T⁴ − T₀⁴). σ ≈ 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
• Wien's displacement law: λ_max × T = b, with b ≈ 2.9 × 10⁻³ m K. Hotter bodies peak at shorter wavelengths.
• Kirchhoff's law: a good absorber is a good emitter. A perfect black body has e = 1.

Always use kelvin in these laws.

Cooling

For a small temperature difference, a body's rate of cooling is proportional to the difference between its temperature and the surroundings (Newton's law of cooling). In problems, use the average temperature over the interval:

(T₁ − T₂)/t = K[(T₁ + T₂)/2 − T₀]

Worked problems

Problem 1 (numerical answer): ice in water. 100 g of ice at 0 °C is added to 200 g of water at 50 °C in an insulated container. Find the final temperature in °C (c = 4.2 J/g K, L = 336 J/g).

Heat the water can give in cooling to 0 °C: 200 × 4.2 × 50 = 42,000 J.
Heat to melt all the ice: 100 × 336 = 33,600 J. This is less, so all the ice melts.
The remaining 8,400 J warms 300 g of water: ΔT = 8,400/(300 × 4.2) = 6.67 °C.
Final temperature ≈ 6.7 °C.

Problem 2: composite slab. Two slabs of equal thickness and area are joined face to face. The first has twice the conductivity of the second. The outer faces are kept at 100 °C and 0 °C, with the better conductor on the hot side. Find the junction temperature.

In series the same H flows: 2k(100 − T) = k(T − 0), so 200 = 3T and T ≈ 66.7 °C.

Problem 3: pendulum clock. A clock with a brass pendulum (α = 2 × 10⁻⁵ per °C) keeps correct time at 20 °C. How much does it lose per day at 30 °C?

Fractional change in period = ½ × 2 × 10⁻⁵ × 10 = 10⁻⁴.
Time lost per day = 10⁻⁴ × 86,400 = 8.64 s.

Problem 4: cooling. A body cools from 70 °C to 60 °C in 5 minutes in a room at 20 °C. How long does it take to cool from 60 °C to 50 °C?

First interval: 10/5 = K(65 − 20), so K = 2/45 per minute.
Second interval: 10/t = (2/45)(55 − 20) = 70/45, so t = 450/70 ≈ 6.4 minutes. Cooling slows as the body approaches room temperature.

Practice set

1. If the absolute temperature of a black body doubles, what happens to λ_max and to the power radiated?
2. How much heat is needed to melt 10 g of ice at 0 °C (L = 336 J/g)?
3. At what temperature do the Celsius and Fahrenheit scales read the same?
4. A 1 m steel rod (α = 1.2 × 10⁻⁵ per °C) is heated through 50 °C. Find its increase in length.
5. A solid has α = 2 × 10⁻⁵ per K. Find its coefficient of volume expansion.
6. 100 g of ice at 0 °C is added to 100 g of water at 50 °C. Find the final temperature and the mass of ice that melts (c = 4.2 J/g K, L = 336 J/g).
7. Two rods of equal length and area, with conductivities k and 2k, are joined in series. Find the equivalent conductivity.
8. Two black spheres of radii 1 cm and 2 cm are at 4,000 K and 2,000 K. Find the ratio of the power they radiate.

Answers

1. λ_max halves; the power becomes 16 times.
2. 10 × 336 = 3,360 J.
3. −40°.
4. ΔL = 1 × 1.2 × 10⁻⁵ × 50 = 6 × 10⁻⁴ m = 0.6 mm.
5. γ = 3α = 6 × 10⁻⁵ per K.
6. Water can give 100 × 4.2 × 50 = 21,000 J, less than the 33,600 J needed. So the final temperature is 0 °C and 21,000/336 = 62.5 g of ice melts.
7. k_eq = 2 × k × 2k/(3k) = 4k/3.
8. P ∝ r²T⁴. The radius ratio squared is 1 : 4 and the temperature ratio to the fourth power is 16 : 1, so the powers are 16 : 4 = 4 : 1.

What to do next

• Before every mixture problem, check whether all the ice melts (or all the steam condenses).
• Practise ten composite-slab problems using the resistance analogy.
• Convert every temperature to kelvin before using a radiation law.
• Move on to thermodynamics and kinetic theory, and revise elasticity and fluids for thermal stress.