1. Chapter at a Glance
Thermal properties of matter deals with temperature, heat and their effects. You will learn temperature scales, thermal expansion, specific heat capacity, calorimetry, change of state and latent heat, and the three ways heat travels — conduction, convection and radiation — along with Newton's law of cooling and the laws of black-body radiation. NEET usually asks one or two questions, often on calorimetry, expansion or radiation.
- Conversion between Celsius, Fahrenheit and Kelvin.
- Linear, area and volume expansion; relation β = 2α, γ = 3α.
- Calorimetry (mixing, including ice and steam).
- Heat conduction through rods; Stefan's and Wien's laws.
- Newton's law of cooling.
2. Temperature and Thermal Expansion
- Temperature tells how hot a body is and decides the direction of heat flow. Heat is energy transferred because of a temperature difference.
- Scales: C/100 = (F − 32)/180 = (K − 273.15)/100. −40° is the same on the Celsius and Fahrenheit scales.
- Ideal-gas (absolute) temperature: PV ∝ T; absolute zero is 0 K = −273.15 °C.
- Expansion: ΔL = LαΔT; ΔA = AβΔT; ΔV = VγΔT, with β = 2α and γ = 3α for isotropic solids.
- Anomalous expansion of water: water contracts on heating from 0 °C to 4 °C and has maximum density at 4 °C. That is why lakes freeze from the top and aquatic life survives below.
- Applications: gaps in railway tracks, bimetallic strips in thermostats, loosening tight lids by heating.
3. Specific Heat, Calorimetry and Latent Heat
- Heat needed: Q = mcΔT. Specific heat of water c = 4200 J kg⁻¹ K⁻¹ (1 cal g⁻¹ °C⁻¹) — very large, so water is used as a coolant.
- Heat capacity C = mc; molar specific heat = heat per mole per kelvin.
- Principle of calorimetry: in an insulated system, heat lost by hot bodies = heat gained by cold bodies.
- Latent heat L: heat per unit mass for a change of state at constant temperature: Q = mL. Lfusion (ice) = 3.34 × 10⁵ J kg⁻¹ (80 cal g⁻¹); Lvaporisation (water) = 2.26 × 10⁶ J kg⁻¹ (540 cal g⁻¹).
- Steam burns are worse than boiling-water burns because steam gives up its large latent heat on condensing.
- Boiling point rises with pressure (pressure cooker); melting point of ice falls with pressure (regelation).
4. Heat Transfer and Cooling
- Conduction (solids): H = dQ/dt = KA(T₁ − T₂)/L. Thermal resistance R = L/(KA). Rods in series add resistances; in parallel add conductances.
- Convection (fluids): transfer by bulk movement — sea breeze, land breeze, trade winds.
- Radiation needs no medium and travels at the speed of light.
- Stefan–Boltzmann law: P = eσAT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴); net loss = eσA(T⁴ − Ts⁴).
- Wien's displacement law: λmT = b ≈ 2.9 × 10⁻³ m K — hotter bodies glow bluer.
- A perfect black body absorbs all radiation falling on it (e = 1); good absorbers are good emitters (Kirchhoff).
- Newton's law of cooling: for small temperature differences, rate of cooling ∝ (T − Ts). Average form: (T₁ − T₂)/t = k[(T₁ + T₂)/2 − Ts].
5. Formula Master Table
| Quantity | Formula |
|---|---|
| Temperature scales | C/100 = (F − 32)/180 = (K − 273)/100 |
| Expansion | ΔL = LαΔT; γ = 3α; β = 2α |
| Heat (no state change) | Q = mcΔT |
| Latent heat | Q = mL |
| Conduction | H = KAΔT/L |
| Stefan's law | P = eσAT⁴ |
| Wien's law | λmT = 2.9 × 10⁻³ m K |
| Newton's cooling (average form) | (T₁ − T₂)/t = k[(T₁ + T₂)/2 − Ts] |
6. Master Summary Table
| Concept | Key point | NEET Note |
|---|---|---|
| Temperature scales | Equal at −40 | Kelvin never negative |
| Expansion | γ = 3α | Water densest at 4 °C |
| Specific heat | Water has a very high c | Used as coolant, moderates climate |
| Latent heat | No temperature change | Steam burns are more severe |
| Conduction | H = KAΔT/L | Series rods like resistors |
| Radiation | P ∝ T⁴; λm ∝ 1/T | Always use kelvin |
7. Mind Map
- Scales
- Absolute zero
- α, β, γ
- Anomaly of water
- Q = mcΔT
- Latent heat
- Mixtures
- Conduction
- Convection
- Radiation, cooling
8. Tips and Tricks
- Mixing water samples only: final T = (m₁T₁ + m₂T₂)/(m₁ + m₂).
- With ice, first check whether the hot water can melt all the ice (compare mL with mcΔT).
- Doubling absolute temperature raises radiated power 16 times and halves λm.
- For two rods of equal length in series, Keq = 2K₁K₂/(K₁ + K₂).
- Using °C inside T⁴ — always convert to kelvin first.
- Forgetting latent heat when a change of state occurs.
- Thinking a hole in a heated plate becomes smaller — it expands like the plate.
9. Important One-liners
- Water has maximum density at 4 °C.
- Temperature remains constant during a change of state.
- Radiation is the only mode of heat transfer that works through vacuum.
- A good absorber of radiation is a good emitter.
- The colour of a hot star tells its surface temperature (Wien's law).
- Sea breeze during the day is due to convection.
- Specific heat of water is higher than that of most substances.
10. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take cwater = 4200 J kg⁻¹ K⁻¹ (or 1 cal g⁻¹ °C⁻¹). Click to reveal each answer and working.
Average level
Normal body temperature is 98.6 °F. Express it in °C and K.
Show answer & solution
Concept: C/100 = (F − 32)/180.
- C = (98.6 − 32) × 5/9 = 66.6 × 5/9 = 37 °C.
- K = 37 + 273 = 310 K.
Answer: 37 °C; 310 K
A 2 m brass rod (α = 2 × 10⁻⁵ K⁻¹) is heated through 50 °C. Find the increase in its length.
Show answer & solution
Concept: ΔL = LαΔT.
- ΔL = 2 × 2 × 10⁻⁵ × 50.
- = 2 × 10⁻³ m = 2 mm.
Answer: 2 mm
How much heat is needed to warm 2 kg of water from 20 °C to 70 °C?
Show answer & solution
Concept: Q = mcΔT.
- Q = 2 × 4200 × 50.
- = 420 000 J = 420 kJ.
Answer: 420 kJ
How much heat is needed to melt 500 g of ice at 0 °C (L = 3.34 × 10⁵ J kg⁻¹)?
Show answer & solution
Concept: Q = mL; temperature stays at 0 °C.
- Q = 0.5 × 3.34 × 10⁵.
- = 1.67 × 10⁵ J.
Answer: 1.67 × 10⁵ J
The Sun's spectrum peaks at about 500 nm. Estimate its surface temperature (b = 2.9 × 10⁻³ m K).
Show answer & solution
Concept: Wien's law λmT = b.
- T = 2.9 × 10⁻³/(500 × 10⁻⁹).
- T = 5800 K.
Answer: about 5800 K
Advanced level
200 g of water at 80 °C is mixed with 300 g of water at 20 °C. Find the final temperature (no heat loss).
Show answer & solution
Concept: heat lost = heat gained.
- 200(80 − T) = 300(T − 20).
- 16 000 − 200T = 300T − 6000 ⇒ 500T = 22 000.
- T = 44 °C.
Answer: 44 °C
10 g of steam at 100 °C is passed into 190 g of water at 20 °C. Find the final temperature (Lsteam = 540 cal g⁻¹, c = 1 cal g⁻¹ °C⁻¹).
Show answer & solution
Concept: steam first condenses (releasing latent heat), then the condensed water cools to T.
- Heat given by steam = 10 × 540 + 10 × (100 − T) = 6400 − 10T.
- Heat taken by water = 190 × (T − 20) = 190T − 3800.
- 6400 − 10T = 190T − 3800 ⇒ 200T = 10 200.
- T = 51 °C.
Answer: 51 °C
A copper rod 0.5 m long with cross-section 4 cm² (K = 400 W m⁻¹ K⁻¹) has its ends at 100 °C and 0 °C. Find the rate of heat flow.
Show answer & solution
Concept: H = KAΔT/L.
- A = 4 × 10⁻⁴ m², ΔT = 100 K.
- H = 400 × 4 × 10⁻⁴ × 100/0.5.
- H = 32 W.
Answer: 32 W
A black sphere of radius 10 cm is at 1000 K. Find the power it radiates (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴). How much would it radiate at 2000 K?
Show answer & solution
Concept: P = σAT⁴ with A = 4πr², e = 1.
- A = 4π × 0.01 ≈ 0.1257 m².
- P = 5.67 × 10⁻⁸ × 0.1257 × 10¹² ≈ 7.1 × 10³ W.
- At 2000 K, P × 2⁴ = 16 × 7.1 kW ≈ 1.1 × 10⁵ W.
Answer: about 7.1 kW; about 114 kW
A cup of tea cools from 90 °C to 70 °C in 4 minutes in a room at 30 °C. How long will it take to cool from 70 °C to 50 °C?
Show answer & solution
Concept: average form of Newton's law of cooling.
- First interval: 20/4 = k(80 − 30) ⇒ 5 = 50k ⇒ k = 0.1 min⁻¹.
- Second interval: 20/t = 0.1 × (60 − 30) = 3.
- t = 20/3 ≈ 6.7 min — slower, because the tea is closer to room temperature.
Answer: about 6.7 minutes
11. NEET-pattern Questions with Detailed Answers
These questions are original and modelled on concepts repeatedly asked in NEET UG.
The temperature at which the Celsius and Fahrenheit scales show the same reading is:
- 0°
- −40°
- 40°
- −273°
Show solution
Put C = F = x: x/100 = (x − 32)/180 ⇒ 180x = 100x − 3200 ⇒ x = −40.
Answer: (B)
The coefficient of volume expansion of an isotropic solid is related to its coefficient of linear expansion as:
- γ = α
- γ = 2α
- γ = 3α
- γ = α/3
Show solution
Volume changes in three dimensions, so γ = 3α (area: β = 2α).
Answer: (C)
If the absolute temperature of a black body is doubled, the wavelength at which it emits most strongly becomes:
- Double
- Half
- One-sixteenth
- Four times
Show solution
λmT = constant, so λm halves. The emitted power, however, becomes 16 times.
Answer: (B)
Two bodies of the same material, shape and size are at 400 K and 800 K. The ratio of the energies they radiate per second is:
- 1 : 2
- 1 : 4
- 1 : 8
- 1 : 16
Show solution
P ∝ T⁴: (400/800)⁴ = (1/2)⁴ = 1/16.
Answer: (D)
Burns from steam at 100 °C are more severe than from water at 100 °C because steam:
- Is hotter
- Has higher specific heat
- Releases latent heat on condensing
- Is less dense
Show solution
Both are at 100 °C, but each gram of steam releases 540 cal of latent heat as it condenses on the skin, in addition to the heat released while cooling.
Answer: (C)
Lakes in cold countries freeze from the top first because:
- Ice is a good conductor
- Water is densest at 4 °C
- Air is warmer than water
- Water has low specific heat
Show solution
Water at 4 °C is densest and sinks; colder water floats on top and freezes first, while ice (a poor conductor) insulates the water below.
Answer: (B)
Two rods of equal length and area with conductivities K and 3K are joined end to end. The equivalent conductivity is:
- 2K
- 1.5K
- 4K
- 0.75K
Show solution
Series: Keq = 2K₁K₂/(K₁ + K₂) = 2 × 3K²/4K = 1.5K.
Answer: (B)
A metal plate with a circular hole is heated. The diameter of the hole:
- Decreases
- Increases
- Stays the same
- First increases then decreases
Show solution
Thermal expansion enlarges every length of the plate, including the edge of the hole, as if the hole were made of the same material. So the hole grows.
Answer: (B)
12. Quick Revision Summary
- C/100 = (F − 32)/180 = (K − 273)/100; equal at −40.
- ΔL = LαΔT; γ = 3α; water densest at 4 °C.
- Q = mcΔT, Q = mL; heat lost = heat gained.
- H = KAΔT/L; P = eσAT⁴; λmT = b.
- Newton's cooling: rate ∝ (T − Ts).