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EAPCET Physics · Intermediate first year

Thermal Properties of Matter

Thermal Properties of Matter covers thermal expansion, calorimetry, change of state, and heat transfer by conduction, convection and radiation.

Subject
Physics
Taught in
Intermediate first year
Physics in the paper
40 of 160 questions

Key formulas and results

  • Δl=lα ΔT\Delta l = l\alpha\,\Delta T; β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha
  • Q=mc ΔTQ = mc\,\Delta T; latent heat Q=mLQ = mL
  • Conduction: Qt=kA ΔTl\frac{Q}{t} = \frac{kA\,\Delta T}{l}
  • Stefan–Boltzmann law: E=σT4E = \sigma T^{4}
  • Wien’s law: λmT=\lambda_{m}T = constant
  • Newton’s law of cooling: rate of cooling ∝(T−T0)\propto (T - T_{0})
  • Thermal stress in a rod clamped at both ends and heated by Δθ\Delta\theta: stress =Yα Δθ= Y\alpha\,\Delta\theta, so the force on the supports F=YAα ΔθF = YA\alpha\,\Delta\theta
  • Time lost or gained per day by a pendulum clock when temperature changes by Δθ\Delta\theta: Δt=12α Δθ×86400 s\Delta t = \tfrac{1}{2}\alpha\,\Delta\theta \times 86400\ \text{s}
  • Real and apparent expansion of a liquid: γreal=γapparent+γcontainer\gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{container}}
  • Density after heating: ρ=ρ01+γ Δθ≈ρ0(1−γ Δθ)\rho = \frac{\rho_{0}}{1 + \gamma\,\Delta\theta} \approx \rho_{0}(1 - \gamma\,\Delta\theta)
  • Thermal resistance of a slab: R=lKAR = \frac{l}{KA}; heat current H=ΔTRH = \frac{\Delta T}{R}, with R=R1+R2R = R_{1} + R_{2} in series and 1R=1R1+1R2\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} in parallel
  • Equivalent conductivity of two equal-length slabs in series: K=2K1K2K1+K2K = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}; of two equal-area slabs in parallel: K=K1+K22K = \frac{K_{1} + K_{2}}{2}
  • Radial conduction: cylindrical shell H=2πKl(T1−T2)ln⁡(b/a)H = \frac{2\pi Kl(T_{1} - T_{2})}{\ln(b/a)}; spherical shell H=4πKab(T1−T2)b−aH = \frac{4\pi Kab(T_{1} - T_{2})}{b - a}
  • Power radiated by a body of emissivity ee: P=σeAT4P = \sigma eAT^{4}, with σ=5.67×10−8 W m−2 K−4\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\,\text{K}^{-4} and e=1e = 1 for a perfect black body

What this chapter covers

  • Thermal expansion
  • Specific heat and calorimetry
  • Latent heat
  • Conduction
  • Radiation laws
  • Newton’s law of cooling

How it is usually asked

  • Calorimetry and mixtures
  • Thermal expansion
  • Conduction through composite slabs
  • Radiation laws

Understanding Thermal Properties of Matter

Most solids expand when heated. The change in length is proportional to the original length and the temperature rise, with the coefficient of linear expansion α; area and volume expand with coefficients about 2α and 3α. When bodies at different temperatures are mixed, calorimetry says the heat lost by the hot body equals the heat gained by the cold one. During melting or boiling the temperature stays constant while latent heat mL is absorbed.

Heat travels by conduction, convection and radiation. Conduction through a slab is proportional to area and temperature difference and inversely proportional to thickness, so slabs in series behave like resistors in series. A black body radiates energy per unit area proportional to T⁴, and its peak wavelength shifts inversely with T (Wien’s law). Newton’s law of cooling applies to small temperature differences.

EAPCET questions typically involve ice–water mixtures, expansion of rods and liquids, composite slabs, radiation ratios and cooling-time comparisons. For slabs in series the heat current is the same through each, so the temperature drop across each slab is proportional to its thermal resistance l/(kA). For radiation, power scales as AT⁴, so doubling the absolute temperature multiplies the emitted power sixteen times.

Worked example

EAPCET-style question

10 g of ice at 0 °C is added to 50 g of water at 40 °C. Taking the latent heat of fusion of ice as 3.36×105 J/kg3.36 \times 10^{5}\ \text{J/kg} and the specific heat of water as 4200 J/(kg K), and ignoring heat losses, the final temperature of the mixture is

  1. (1)20 °C
  2. (2)24 °C
  3. (3)33.3 °C
  4. (4)0 °C

Solution

  1. Heat to melt the ice: 0.01×3.36×105=3360 J0.01 \times 3.36 \times 10^{5} = 3360\ \text{J}. The warm water can give up to 0.05×4200×40=8400 J0.05 \times 4200 \times 40 = 8400\ \text{J}, so all the ice melts.
  2. Let the final temperature be TT. Heat lost by warm water =0.05×4200×(40−T)=210(40−T)= 0.05 \times 4200 \times (40 - T) = 210(40 - T).
  3. Heat gained = 3360 (melting) + 0.01×4200×T=3360+42T0.01 \times 4200 \times T = 3360 + 42T.
  4. 8400−210T=3360+42T8400 - 210T = 3360 + 42T, so 252T=5040252T = 5040.
  5. T=20 ∘CT = 20\ ^\circ\text{C}.

Answer: 20 °C

Shortcuts and results to remember

  • α:β:γ=1:2:3\alpha : \beta : \gamma = 1 : 2 : 3 for an isotropic solid, so β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha.
  • Water contracts when heated from 0 °C to 4 °C and has its maximum density at 4 °C.
  • Remember the calorimetry constants: specific heat of ice and steam 0.5 cal/g °C, of water 1 cal/g °C, latent heat of fusion 80 cal/g, of vaporisation 540 cal/g, and 1 cal ≈ 4.2 J.
  • On a temperature versus heat-supplied graph the slope is 1ms\frac{1}{ms}, and the flat portions are phase changes at constant temperature.
  • Newton’s law of cooling in average form: T1−T2t=k[T1+T22−T0]\frac{T_{1} - T_{2}}{t} = k\left[\frac{T_{1} + T_{2}}{2} - T_{0}\right], which avoids logarithms in exam problems.

More practice questions

Practice question 1

50 g of ice at −20 °C is kept in a container of water equivalent 80 g, also at −20 °C. The heat needed to convert the ice into water at 50 °C (and bring the container to 50 °C) is

  1. (1)7000 cal
  2. (2)9600 cal
  3. (3)12600 cal
  4. (4)5600 cal

Solution

  1. Ice from −20 °C to 0 °C: 50×0.5×20=500 cal50 \times 0.5 \times 20 = 500\ \text{cal}
  2. Melting: 50×80=4000 cal50 \times 80 = 4000\ \text{cal}
  3. Water from 0 °C to 50 °C: 50×1×50=2500 cal50 \times 1 \times 50 = 2500\ \text{cal}
  4. Container from −20 °C to 50 °C: 80×1×70=5600 cal80 \times 1 \times 70 = 5600\ \text{cal}
  5. Total =500+4000+2500+5600=12600 cal= 500 + 4000 + 2500 + 5600 = 12600\ \text{cal}

Answer: 12600 cal

Practice question 2

Two rods of equal length and area, with conductivities 3K3K and KK, are joined end to end. The free end of the 3K3K rod is at 100 °C and the free end of the KK rod is at 20 °C. The junction temperature in the steady state is

  1. (1)40 °C
  2. (2)60 °C
  3. (3)80 °C
  4. (4)90 °C

Solution

  1. In steady state the heat current is the same through both rods
  2. 3KA(100−T)l=KA(T−20)l\frac{3KA(100 - T)}{l} = \frac{KA(T - 20)}{l}
  3. 300−3T=T−20300 - 3T = T - 20, so 4T=3204T = 320
  4. T=80 ∘CT = 80\ ^\circ\text{C}

Answer: 80 °C

Common mistakes to avoid

  • Ignoring the latent heat needed to melt ice before its water can warm up — check first whether all of it melts.
  • Forgetting that the melted ice must then be heated from 0 °C to the final temperature.
  • Using Celsius temperatures in the Stefan–Boltzmann law or Wien’s law — both need kelvin.
  • Taking the area expansion coefficient as 3α — it is 2α, and the volume coefficient is 3α.

How to prepare it

Slabs in series add like resistors in series — use thermal resistance l/(kA).

Once the formulas feel familiar, test them the way the exam does: practise Thermal Properties of Matter questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Thermal Properties of Matter chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Thermal Properties of Matter part of the EAPCET Physics syllabus?
Yes. Thermal Properties of Matter is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Thermal Properties of Matter for EAPCET?
Slabs in series add like resistors in series — use thermal resistance l/(kA).
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Thermal Properties of Matter?
Ignoring the latent heat needed to melt ice before its water can warm up — check first whether all of it melts. Forgetting that the melted ice must then be heated from 0 °C to the final temperature. Using Celsius temperatures in the Stefan–Boltzmann law or Wien’s law — both need kelvin. Taking the area expansion coefficient as 3α — it is 2α, and the volume coefficient is 3α.