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Class XI · Physics · Unit 11 · Interactive Lecture

Heat & Thermodynamics

The complete lecture — every idea comes alive in the live panel on the right as you read. Scroll down; molecules jiggle, rails grow, the heating curve draws itself and a heat engine shows its energy budget live.

Hold a cup of hot chai — energy flows from cup to hand. Energy in transit because of a temperature difference is heat (Q), measured in joules. A body never "contains" heat.

  • Temperature (T) — degree of hotness; microscopically, a measure of the average translational KE of one molecule.
  • Internal energy (U) — the total KE + mutual PE of all the molecules. A warm bucket beats a boiling cup: lower T, far more molecules, more U.
Exam point: heat and work are energy in transit; temperature and internal energy are state properties of the body.

Thermometers are calibrated between two fixed points — ice point and steam point: 0–100 °C, 32–212 °F, 273–373 K.

ConversionsT(K) = T(°C) + 273 · T(°F) = (9/5)T(°C) + 32 · T(°C) = (5/9)[T(°F) − 32]

Kelvin is the absolute scale: at 0 K (−273 °C) molecular translation is minimum — so gas-law sums must use kelvin.

worked — fever on three scales
98.6 °F = ?
(5/9)(98.6 − 32) = 37 °C = 37 + 273 = 310 K

Heated molecules vibrate over larger distances — the solid expands. Tiny, but strong enough to buckle a track.

Expansion formulaslinear: ΔL = α L ΔT · volumetric: ΔV = β V ΔT, with β ≈ 3α
steel α ≈ 1.2 × 10⁻⁵ K⁻¹
  • Railway gaps — a 30 m rail heated 30 °C grows ≈ 11 mm; the gap absorbs it.
  • Bridge rollers — one end of a steel bridge rides on rollers, free to creep with the seasons.
  • Bimetallic strip — brass + iron bend on heating; the thermostat inside an iron.
Anomalous water: 0→4 °C water contracts; ponds freeze top-down and fish survive below.

A gas is mostly empty space: vast numbers of molecules in ceaseless random motion, with negligible volume, no forces except in collisions, and perfectly elastic collisions.

The two key resultsP = (1/3) ρ <v²> — pressure from molecular bombardment
<½mv²> = (3/2) k T — temperature ∝ average KE

Heating a gas literally means speeding its molecules up. At one temperature all gases share the same average KE per molecule — light H₂ just moves faster than heavy O₂.

  • Boyle's law (T constant) — P ∝ 1/V → P₁V₁ = P₂V₂. Squeezed gas hits the walls more often.
  • Charles's law (P constant) — V ∝ T → V₁/T₁ = V₂/T₂ (kelvin!). Faster molecules need more room.
Ideal gas equationPV = nRT · R = 8.314 J·mol⁻¹·K⁻¹ · combined: P₁V₁/T₁ = P₂V₂/T₂
worked — bicycle pump
200 kPa, 1.5 L → 0.5 L at constant T?
P₂ = 200 × 1.5/0.5 = 600 kPa
Pressure cooker: fixed V, rising T → rising P → water boils above 100 °C → daal in minutes.
  • Specific heat c — heat to raise 1 kg by 1 K (J·kg⁻¹·K⁻¹). Water 4200, sand ≈ 800, iron 450.
  • Heat capacity C — for the whole body: C = mc (J·K⁻¹).
The heat equationQ = m c ΔT

Karachi's sea breeze: by afternoon the land (small c) is far hotter than the Arabian Sea (huge c); air rises over the land and cool sea air flows in. At night the flow reverses — the land breeze.

worked — chai water
2 kg water, 25→100 °C?
Q = 2 × 4200 × 75 = 630 kJ

On the plateaus, heat flows in but T refuses to rise — the latent heat is breaking intermolecular bonds (raising PE), not speeding molecules up.

Latent heatmelting: Q = m L_f · L_f(ice) = 3.36 × 10⁵ J/kg
boiling: Q = m L_v · L_v(water) = 2.26 × 10⁶ J/kg (≈ 7× bigger!)

The huge L_v is why steam scalds worse than boiling water, and why sweating cools you — each gram of evaporating sweat carries off ~2260 J.

worked — ice in your drink
Melt 0.5 kg of ice at 0 °C?
Q = 0.5 × 3.36 × 10⁵ = 168 kJ absorbed from the drink
First lawΔU = Q − W
Q = heat added to the gas · W = work done by the gas (PΔV)
  • Isothermal — T constant → ΔU = 0 → Q = W: all heat becomes expansion work (slow, conducting walls).
  • Adiabatic — Q = 0 → ΔU = −W: an expanding gas cools (clouds form as moist air rises); rapid compression heats (a pump's barrel, diesel ignition).
worked — bookkeeping
Gas absorbs 800 J, does 300 J of work?
ΔU = 800 − 300 = +500 J stays inside
  • Kelvin statement — no cyclic engine converts all its heat into work; some must be rejected to a colder body.
  • Clausius statement — heat never flows cold → hot by itself; a refrigerator must be driven by work.
Engine efficiencyW = Q₁ − Q₂ · η = W/Q₁ = 1 − Q₂/Q₁
Carnot limit: η = 1 − T₂/T₁ (kelvin) — always < 100%

A motorbike's petrol engine manages only 25–30%: most fuel energy leaves as hot exhaust and radiator heat — that is the second law, not bad engineering.

worked — engine cycle
Q₁ = 1000 J, Q₂ = 600 J?
W = 400 J → η = 400/1000 = 40%
numerical — ice at 0 °C to steam at 100 °C (0.1 kg)
mL_f + mcΔT + mL_v = 33.6 + 42 + 226 kJ = ≈ 302 kJ — boiling costs the most.
numerical — combined gas law
600 cm³ at 27 °C, 100 kPa → 127 °C, 150 kPa?
V₂ = (100×600×400)/(300×150) = 533 cm³
numerical — engine at 28%
Q₁ = 2500 J?
W = 700 J · Q₂ = 1800 J rejected
  1. Heat = energy in transit; T ∝ average KE of one molecule; U = total KE + PE of all.
  2. K = °C + 273; °F = (9/5)°C + 32; gas laws in kelvin only.
  3. ΔL = αLΔT, β ≈ 3α — railway gaps, bridge rollers, bimetal strips.
  4. P = ⅓ρ<v²>; Boyle P₁V₁ = P₂V₂; Charles V/T constant; PV = nRT.
  5. Q = mcΔT (water c = 4200 → sea breeze); Q = mL on the plateaus.
  6. ΔU = Q − W; isothermal Q = W; adiabatic ΔU = −W.
  7. η = 1 − Q₂/Q₁ ≤ 1 − T₂/T₁ — no engine reaches 100%.
🔥 Live panelHeat & Thermodynamics
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