Thermal Properties of Matter — NEET High-Yield Notes
Heat is the thermal energy in transit between systems due to temperature gradient. Temperature measures the average translational kinetic energy per molecule.
1. Temperature Scales & Thermometry
The relation between Celsius (°C), Fahrenheit (°F), and Kelvin (K) temperature scales is governed by:
C100 = F - 32180 = K - 273.15100
Triple Point of Water: Ttr = 273.16 K (0.01°C) at 4.58 mm of Hg. Absolute zero is 0 K = -273.15°C.
2. Thermal Expansion of Solids & Liquids
When heated, the separation between atoms increases due to the asymmetric nature of potential energy curves.
- Linear Expansion: ΔL = L0 α ΔT → L = L0 (1 + α ΔT)
- Superficial (Areal) Expansion: ΔA = A0 β ΔT → β = 2α
- Cubical (Volumetric) Expansion: ΔV = V0 γ ΔT → γ = 3α
- Relation between coefficients: α : β : γ = 1 : 2 : 3
Fig 1: Thermal Expansion Relations & Anomalous Density of Water (Peak at 4°C)
NEET Trap Alert (Anomalous Expansion of Water): Between 0°C and 4°C, water contracts on heating (γ is negative!). Density is maximum at 4°C (ρ = 1000 kg/m3). This enables aquatic life to survive at the bottom of frozen lakes in winter.
3. Specific Heat Capacity & Calorimetry
The amount of heat required to raise the temperature of unit mass by 1°C:
Q = m ⋅ s ⋅ ΔT | Heat Capacity S = m ⋅ s = QΔT
For water, specific heat capacity s = 1 cal/g°C = 4186 J/kg°C. Water has exceptionally high specific heat, making it an excellent coolant.
Fig 2: Heating Curve of Water showing Latent Heat Plateaus
Latent Heat & Phase Transition
- Latent Heat of Fusion (Lf): Heat required to melt 1g ice at 0°C → Lf = 80 cal/g = 3.33 × 105 J/kg.
- Latent Heat of Vaporization (Lv): Heat required to boil 1g water to steam at 100°C → Lv = 540 cal/g = 2.26 × 106 J/kg.
- Principle of Calorimetry: Heat Lost by hot bodies = Heat Gained by cold bodies (neglecting external loss).
4. Heat Transfer Modes
Fig 3: Conduction through Composite Slabs & Stefan-Boltzmann Law
A. Conduction
Heat current H through a slab of area A, length L, thermal conductivity K:
H = dQdt = KA(T1 - T2)L = T1 - T2Rth
where Thermal Resistance Rth = LKA.
- Series combination: Req = R1 + R2 → L1 + L2Keq = L1K1 + L2K2
- Parallel combination: 1Req = 1R1 + 1R2 → Keq (A1 + A2) = K1 A1 + K2 A2
B. Radiation
- Stefan-Boltzmann Law: Total power radiated by a blackbody of area A at absolute temperature T:E = σ ⋅ A ⋅ T4 (σ = 5.67 × 10-8 W/m2K4)For a real body with emissivity e (0 < e < 1) in surroundings at T0:Net Radiation Rate Pnet = e σ A (T4 - T04)
Fig 4: Wien's Displacement Law & Newton's Law of Cooling Curve
C. Wien's Displacement Law
The wavelength λmax corresponding to maximum emissive power is inversely proportional to absolute temperature:
λmax ⋅ T = b (Wien constant b = 2.898 × 10-3 m⋅K)
D. Newton's Law of Cooling
For small temperature differences (ΔT = T - T0 ≤ 30°C):
dTdt = -k (T - T0) → T1 - T2t = K T1 + T22 - T0
Thermal Properties — Formula Matrix & 20 Golden Facts
Master Formula Matrix
| Concept | Formula | Key NEET Trap |
|---|---|---|
| Temperature Scale | C5 = F - 329 = K - 273.155 | C = F at -40° |
| Expansion Coeffs | α = β2 = γ3 | For isotropic solid: γ = 3α |
| Thermal Stress | FA = Y α ΔT | Independent of rod length! |
| Calorimetry | Q = m s ΔT + m L | Always check if ice reaches 0°C first |
| Thermal Resistance | Rth = LKA | Analogous to electrical R = ρL/A |
| Stefan-Boltzmann | P = e σ A (T4 - T04) | T must strictly be in Kelvin |
| Wien's Law | λmax T = b | Peak shifts to shorter λ at high T |
| Newton's Cooling | T1 - T2t = K (T1+T22 - T0) | Valid only for small (T - T0) |
20 Golden NEET Must-Know Facts & Traps
1. Celsius and Fahrenheit readings coincide at exactly -40° (-40°C = -40°F).
2. Thermal stress F/A = YαΔT is independent of length L and cross-sectional area A of the rod.
3. When a cavity in a metal sphere is heated, the cavity volume expands exactly as if it were filled with the same metal!
4. Density of water is maximum at 4°C (1000 kg/m3). From 0°C to 4°C, volume decreases on heating.
5. Ratio of expansion coefficients is α : β : γ = 1 : 2 : 3 for isotropic solids. For anisotropic: γ = αx + αy + αz.
6. Bimetallic strip bends towards the metal with smaller α on heating, and towards larger α on cooling.
7. Latent heat of vaporization of water (540 cal/g) is 6.75 times larger than latent heat of fusion (80 cal/g).
8. Steam at 100°C causes much more severe burns than boiling water at 100°C due to 540 cal/g extra latent heat.
9. Mixing equal masses of ice at 0°C and water at 80°C results in all ice melting and final mixture at exactly 0°C!
10. In steady-state conduction, temperature gradient dT/dx is constant across a uniform rod, but heat current H is the same at all cross-sections.
11. Equivalent thermal conductivity for two slabs of equal length in series is Keq = 2K1K2 / (K1 + K2) (Harmonic mean).
12. For parallel slabs of equal area: Keq = (K1 + K2) / 2 (Arithmetic mean).
13. Perfectly black body emissivity e = 1, absorptive power a = 1, reflectivity r = 0, transmissivity t = 0.
14. Kirchhoff's Law: Good absorbers are good emitters (eλ = aλ at any temperature).
15. If absolute temperature of a blackbody is doubled, total radiated energy increases by 24 = 16 times.
16. Wien's constant b = 2.898 × 10-3 m⋅K. Hotter stars look blue/white; cooler stars look red.
17. Solar constant S ≈ 1360 W/m2, used to estimate surface temperature of Sun (≈ 5800 K) via Stefan's law.
18. Cooling rate under Newton's law: A body cools from 80°C to 70°C faster than from 70°C to 60°C because (T - T0) is larger.
19. Thermal capacity = Mass × Specific heat capacity (C = ms), SI unit is J/K. Water equivalent W = m ⋅ s (in grams if s in cal/g°C).
20. Cooking utensils have high thermal conductivity K and low specific heat s for fast heating.
Thermal Properties — 8 Solved NEET Numericals (3-Box Layout)
Problem 1: Thermal Stress in Clamped Steel Wire
Given: Steel wire clamped at both ends at 40°C. Cooled to 20°C (ΔT = 20°C). Young's modulus Y = 2 × 1011 N/m2, α = 1.2 × 10-5 /°C, Area A = 1 mm2 = 10-6 m2. Find tension produced.
Step-by-Step Calculation:
Thermal strain prevented = α ΔT = (1.2 × 10-5)(20) = 2.4 × 10-4.
Thermal stress = Y ⋅ α ΔT = (2 × 1011)(2.4 × 10-4) = 4.8 × 107 N/m2.
Tension T = Stress × Area = (4.8 × 107)(10-6) = 48 N.
Thermal strain prevented = α ΔT = (1.2 × 10-5)(20) = 2.4 × 10-4.
Thermal stress = Y ⋅ α ΔT = (2 × 1011)(2.4 × 10-4) = 4.8 × 107 N/m2.
Tension T = Stress × Area = (4.8 × 107)(10-6) = 48 N.
NEET Shortcut: Tension T = Y ⋅ A ⋅ α ⋅ ΔT. Notice wire length L does NOT matter at all!
Problem 2: Mixture of Ice and Water (Calorimetry Equilibrium)
Given: 50 g of ice at 0°C is mixed with 50 g of water at 40°C. Specific heat of water s = 1 cal/g°C, Latent heat Lf = 80 cal/g. Find final temperature and composition.
Step-by-Step Calculation:
Heat released by 50 g water cooling from 40°C to 0°C: Qlost = m s ΔT = 50 × 1 × 40 = 2000 cal.
Heat required to melt ALL 50 g ice: Qreq = m Lf = 50 × 80 = 4000 cal.
Since Qlost (2000 cal) < Qreq (4000 cal), all ice CANNOT melt!
Mass of ice melted = QlostLf = 200080 = 25 g.
Remaining ice = 50 - 25 = 25 g. Total water = 50 + 25 = 75 g.
Final temperature = 0°C.
Heat released by 50 g water cooling from 40°C to 0°C: Qlost = m s ΔT = 50 × 1 × 40 = 2000 cal.
Heat required to melt ALL 50 g ice: Qreq = m Lf = 50 × 80 = 4000 cal.
Since Qlost (2000 cal) < Qreq (4000 cal), all ice CANNOT melt!
Mass of ice melted = QlostLf = 200080 = 25 g.
Remaining ice = 50 - 25 = 25 g. Total water = 50 + 25 = 75 g.
Final temperature = 0°C.
NEET Shortcut: If heat given by water is less than mice × 80, temperature is ALWAYS 0°C with a mixture of ice and water.
Problem 3: Conduction Interface Temperature of Two Slabs
Given: Two slabs in contact of equal thickness L and equal area A. Slab 1 has K1 = 2K at 100°C. Slab 2 has K2 = K at 0°C. Find interface temperature T.
Step-by-Step Calculation:
In steady state, heat current through slab 1 = heat current through slab 2:
K1 A (100 - T)L = K2 A (T - 0)L
Substitute K1 = 2K, K2 = K:
2K(100 - T) = K(T) → 200 - 2T = T → 3T = 200 → T = 66.7°C (or 200/3 °C).
In steady state, heat current through slab 1 = heat current through slab 2:
K1 A (100 - T)L = K2 A (T - 0)L
Substitute K1 = 2K, K2 = K:
2K(100 - T) = K(T) → 200 - 2T = T → 3T = 200 → T = 66.7°C (or 200/3 °C).
NEET Shortcut: Tinterface = K1 T1 + K2 T2K1 + K2 = 2(100) + 1(0)3 = 2003 = 66.7°C.
Problem 4: Stefan-Boltzmann Radiation Rate Change
Given: A spherical blackbody of radius R at temperature T radiates power P. If radius is halved (R' = R/2) and absolute temperature is doubled (T' = 2T), find new radiated power P'.
Step-by-Step Calculation:
By Stefan's Law: P = σ A T4 = σ (4πR2) T4 ∝ R2 T4.
P' = σ [4π(R/2)2] (2T)4 = σ [4π (R2/4)] (16 T4) = 164 P = 4 P.
By Stefan's Law: P = σ A T4 = σ (4πR2) T4 ∝ R2 T4.
P' = σ [4π(R/2)2] (2T)4 = σ [4π (R2/4)] (16 T4) = 164 P = 4 P.
NEET Shortcut: Ratio P'/P = (R'/R)2 × (T'/T)4 = (1/2)2 × (2)4 = 14 × 16 = 4.
Problem 5: Wien's Displacement Law Peak Ratio
Given: Sun emits max radiation at λ1 = 500 nm (T1 = 5800 K). A distant star emits max radiation at λ2 = 250 nm. Find surface temperature of the star T2.
Step-by-Step Calculation:
By Wien's Law: λmax ⋅ T = constant → T2 = T1 ⋅ λ1λ2.
T2 = 5800 × 500250 = 5800 × 2 = 11,600 K.
By Wien's Law: λmax ⋅ T = constant → T2 = T1 ⋅ λ1λ2.
T2 = 5800 × 500250 = 5800 × 2 = 11,600 K.
NEET Shortcut: Half the wavelength means DOUBLE the absolute temperature!
Problem 6: Newton's Law of Cooling Time Estimate
Given: A cup of coffee cools from 80°C to 60°C in 5 minutes in a room at 20°C. How long will it take to cool from 60°C to 40°C?
Step-by-Step Calculation:
Case 1: (80 - 60)/5 = K [80+602 - 20] → 4 = K [70 - 20] = 50 K → K = 450 = 0.08 min-1.
Case 2: (60 - 40)/t = K [60+402 - 20] → 20t = K [50 - 20] = 30 K.
Substitute K = 4/50: 20t = 30 × 450 = 2.4 → t = 202.4 = 8.33 minutes (8 min 20 s).
Case 1: (80 - 60)/5 = K [80+602 - 20] → 4 = K [70 - 20] = 50 K → K = 450 = 0.08 min-1.
Case 2: (60 - 40)/t = K [60+402 - 20] → 20t = K [50 - 20] = 30 K.
Substitute K = 4/50: 20t = 30 × 450 = 2.4 → t = 202.4 = 8.33 minutes (8 min 20 s).
NEET Trap Alert: Never assume cooling time is constant! As the body approaches surrounding temperature, cooling slows down (8.33 min > 5 min).
Problem 7: Steam and Ice Calorimetry Calculation
Given: 10 g of steam at 100°C is passed into 100 g of water and 20 g of ice at 0°C. Find final temperature of mixture.
Step-by-Step Calculation:
1. Heat needed to melt 20 g ice: Qmelt = 20 × 80 = 1600 cal.
2. Heat released when 10 g steam condenses at 100°C: Qcondense = 10 × 540 = 5400 cal.
Since 5400 > 1600, ALL ice melts! Remaining heat from condensation = 5400 - 1600 = 3800 cal.
Total water now = 100 (initial) + 20 (melted ice) + 10 (condensed steam) = 130 g.
3. Heat released if 10 g water from steam cools from 100 to T, and heat absorbed by 120 g water from 0 to T:
Heat balance: 1600 + 120(1)(T - 0) = 5400 + 10(1)(100 - T)
1600 + 120 T = 5400 + 1000 - 10 T → 130 T = 6400 - 1600 = 4800 → T = 4800130 = 36.9°C.
1. Heat needed to melt 20 g ice: Qmelt = 20 × 80 = 1600 cal.
2. Heat released when 10 g steam condenses at 100°C: Qcondense = 10 × 540 = 5400 cal.
Since 5400 > 1600, ALL ice melts! Remaining heat from condensation = 5400 - 1600 = 3800 cal.
Total water now = 100 (initial) + 20 (melted ice) + 10 (condensed steam) = 130 g.
3. Heat released if 10 g water from steam cools from 100 to T, and heat absorbed by 120 g water from 0 to T:
Heat balance: 1600 + 120(1)(T - 0) = 5400 + 10(1)(100 - T)
1600 + 120 T = 5400 + 1000 - 10 T → 130 T = 6400 - 1600 = 4800 → T = 4800130 = 36.9°C.
NEET Shortcut: Always verify phase changes first before writing heat balance equations!
Problem 8: Apparent Volume Expansion of Liquid in Vessel
Given: A glass flask of volume V0 = 1000 cm3 is completely filled with mercury at 0°C. Real coefficient of mercury γr = 18 × 10-5 /°C, glass linear expansion αg = 9 × 10-6 /°C. Find volume of mercury overflowing at 100°C.
Step-by-Step Calculation:
Glass volumetric coefficient γg = 3 αg = 3(9 × 10-6) = 2.7 × 10-5 /°C.
Apparent expansion coefficient of mercury γapp = γr - γg:
γapp = (18 × 10-5) - (2.7 × 10-5) = 15.3 × 10-5 /°C.
Overflow volume ΔVapp = V0 ⋅ γapp ⋅ ΔT = 1000 × (15.3 × 10-5) × 100 = 15.3 cm3.
Glass volumetric coefficient γg = 3 αg = 3(9 × 10-6) = 2.7 × 10-5 /°C.
Apparent expansion coefficient of mercury γapp = γr - γg:
γapp = (18 × 10-5) - (2.7 × 10-5) = 15.3 × 10-5 /°C.
Overflow volume ΔVapp = V0 ⋅ γapp ⋅ ΔT = 1000 × (15.3 × 10-5) × 100 = 15.3 cm3.
NEET Shortcut: Overflow volume ΔV = V0 (γliquid - 3αvessel) ΔT. Always convert linear α to volumetric 3α!
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App