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Thermal Properties of Matter

NCERT Class 11 Physics Chapter 10

Thermal Properties of Matter for Class 11 Physics, JEE and NEET: temperature, thermal expansion, specific heat, calorimetry, latent heat and heat transfer.

10 topics Updated 8 Nov 2026

For: Class 11, JEE and NEET, unless a topic says otherwise

In the syllabus: JEE Class 11 syllabus, NEET Class 11 syllabus

With your textbook

NCERT Physics Part II (Class 11), Chapter 10 “Thermal Properties of Matter”, reprint 2026-27.

Open the official NCERT chapter (PDF, 2.3 MB, ncert.nic.in)

Checked on 19 Oct 2026.

Learn in this order

  1. Temperature and heat

    The difference between them, their SI units, thermal equilibrium, the calorie and solved examples.

    NCERT Section 10.2 10 min read 4 solved examples 6 practice questions

  2. Measuring temperature

    Thermometers, fixed points, the Celsius to Fahrenheit formula, faulty thermometers and solved examples.

    NCERT Section 10.3 10 min read 4 solved examples 8 practice questions

  3. Ideal-gas equation and absolute temperature

    Boyle’s and Charles’ laws, PV = μRT, the gas thermometer, absolute zero and the kelvin scale, with examples.

    NCERT Section 10.4 11 min read 5 solved examples 6 practice questions

  4. Thermal expansion

    Linear, area and volume expansion and their coefficients, thermal stress and water’s anomalous expansion.

    NCERT Section 10.5 14 min read 4 solved examples 6 practice questions

  5. Specific heat capacity

    Heat capacity, specific and molar specific heat, ΔQ = msΔT, units, Cp and Cv, and why water matters.

    NCERT Section 10.6 12 min read 5 solved examples 6 practice questions

  6. Calorimetry

    Heat lost equals heat gained, the calorimeter, the method of mixtures, water equivalent and solved examples.

    NCERT Section 10.7 14 min read 5 solved examples 6 practice questions

  7. Experiment: specific heat by the method of mixtures

    Finding the specific heat capacity of a solid and a liquid, with theory, procedure, precautions and errors.

    JEE and NEET only 13 min read 4 solved examples 6 practice questions

  8. Change of state

    Melting, boiling, sublimation, regelation, the triple point and Q = mL, with solved examples.

    NCERT Section 10.8 18 min read 5 solved examples 7 practice questions

  9. Heat transfer

    Conduction, thermal conductivity, convection, radiation, Wien’s law and the Stefan–Boltzmann law, with examples.

    NCERT Section 10.9 19 min read 5 solved examples 7 practice questions

  10. Newton’s law of cooling

    The formula and its derivation, rate of cooling, the exponential cooling curve, the log graph and solved examples.

    Class 11 only NCERT Section 10.10 14 min read 4 solved examples 7 practice questions

Key formulas in this chapter

Temperature and heat

  • T (K) = t (°C) + 273.15
  • ΔT (K) = Δt (°C)
  • 1 cal ≈ 4.186 J (NCERT convention); 1 thermochemical cal = 4.184 J exactly

Measuring temperature

  • (tF − 32)/180 = tC/100
  • tF = 1.8tC + 32
  • ΔtF = 1.8ΔtC
  • (X − Xice)/(Xsteam − Xice) = tC/100 (any linear scale or thermometric property)

Ideal-gas equation and absolute temperature

  • PV = constant (Boyle’s law, T constant)
  • V/T = constant (Charles’ law, P constant)
  • PV = μRT, R = 8.31 J mol K
  • PV/T = PV/T (fixed quantity)
  • T = tC + 273.15

Thermal expansion

  • Δl/l = αl ΔT
  • ΔA/A = 2αl ΔT
  • ΔV/V = αV ΔT, αV = 3αl
  • αV = 1/T (ideal gas, constant pressure, T in kelvin)
  • Thermal stress = Yαl ΔT

Specific heat capacity

  • S = ΔQ/ΔT (heat capacity, J K)
  • s = ΔQ/(mΔT) (specific heat capacity, J kg K)
  • C = ΔQ/(μΔT) (molar specific heat capacity, J mol K)
  • ΔQ = msΔT
  • C = Ms (M = molar mass in kg mol)
  • Cp − Cv = R (ideal gas; ≈ R for the real gases in NCERT Table 10.4)

Calorimetry

  • heat lost = heat gained (isolated system)
  • ms(T − T) = (ms + ms)(T − T)
  • T = (msT + msT) / (ms + ms) (two bodies, no container)
  • W = m3s3/sw (water equivalent)

Experiment: specific heat by the method of mixtures

  • s = (mw sw + mc sc)(T − T2)/[m(T1 − T)] (solid)
  • sl = [ms(T1 − T)/(T − T2) − mc sc]/ml (liquid)
  • Δs/s ≈ Δ(T − T)/(T − T) + Δ(T − T)/(T − T) (first-order worst-case temperature bounds)

Change of state

  • Q = mL (heat for a change of state)
  • Lf = 3.33 × 105 J kg−1 (water)
  • Lv = 22.6 × 105 J kg−1 (water)
  • Triple point of water: 273.16 K, about 611 Pa

Heat transfer

  • H = KA(TC − TD)/L (steady conduction)
  • R = L/(KA); R = R + R in series
  • K′ = 2KK/(K + K) (two equal bars in series)
  • 1/R = 1/R + 1/R (bars in parallel)
  • λm T = 2.9 × 10−3 m K (Wien)
  • H = AeσT4; Hnet = eσA(T4 − Ts^4), σ = 5.67 × 10−8 W m−2 K−4

Newton’s law of cooling

  • −dQ/dt = k(T − T)
  • dT/dt = −K(T − T), K = k/(ms)
  • T2 = T1 + C′e−Kt
  • t = ln(ΔTstart/ΔTend)/K
  • (Ti − Tf)/t ≈ K[(Ti + Tf)/2 − T1] (short interval, average method)

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