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Thermal Physics and Heat

Subject: Physics
Topic: 9
Cambridge Code: 0625

Temperature and Heat​

Temperature (T) - Measure of average kinetic energy of particles

  • Unit: Kelvin (K) or Celsius (°C)
  • K = °C + 273.15

Heat (Q) - Energy transferred between objects at different temperatures

  • Unit: Joules (J)
  • Flows from hot to cold (second law of thermodynamics)

Thermal Equilibrium​

Thermal equilibrium - No net heat flow

  • Objects same temperature
  • No further temperature change

Specific Heat Capacity​

Specific heat capacity (c) - Energy per unit mass per degree

Q=mcΔTQ = mc\Delta T

Where:

  • Q = energy transferred
  • m = mass
  • c = specific heat capacity
  • ΔT = temperature change

Unit: J/(kg·K) or J/(kg·°C)

Interpretation​

High specific heat capacity:

  • Lots of energy needed to change temperature
  • Example: Water (4200 J/(kg·K))
  • Slow to heat, slow to cool

Low specific heat capacity:

  • Little energy needed to change temperature
  • Example: Metals (~500 J/(kg·K))
  • Quick to heat, quick to cool

Latent Heat​

Latent heat - Energy needed for change of state (no T change)

Q=mLQ = mL

Where:

  • Q = energy
  • m = mass
  • L = latent heat capacity

Types​

Latent heat of fusion (melting/freezing):

  • Energy to/from heat during phase change solid ↔ liquid
  • Example: Ice water at 0°C

Latent heat of vaporization (boiling/condensing):

  • Energy to/from heat during phase change liquid ↔ gas
  • Example: Steam-water mixture at 100°C
  • Usually larger than fusion

Heating Curves​

Three stages:

  1. Solid heating: Q=mcΔTQ = mc\Delta T
  2. Melting: Q=mLfQ = mL_f (constant T)
  3. Liquid heating: Q=mcΔTQ = mc\Delta T
  4. Boiling: Q=mLvQ = mL_v (constant T)
  5. Gas heating: Q=mcΔTQ = mc\Delta T

Thermodynamic Laws​

First Law of Thermodynamics​

Energy conservation for heat and work:

ΔU=Q−W\Delta U = Q - W

Where:

  • ΔU = change in internal energy
  • Q = heat supplied to system
  • W = work done by system

Interpretation:

  • Heat in increases internal energy
  • Work out decreases internal energy

Second Law of Thermodynamics​

Entropy (disorder) increases: ΔS≥0\Delta S ≥ 0

Consequences:

  • Heat flows from hot to cold (not reverse)
  • Some energy always "wastes" as heat
  • Perfect efficiency impossible
  • Process direction always one way (irreversible)

Internal Energy​

Internal energy (U) - Total kinetic + potential energy of particles

ΔU=mcΔT\Delta U = mc\Delta T

(For temperature change without phase change)

Work Done​

Work done by system expanding: W=PΔVW = P\Delta V

For constant pressure (isobaric process)

Heat and Work​

Q = positive: Heat into system Q = negative: Heat out of system W = positive: System does work (expands) W = negative: Work done on system (compresses)

Ideal Gas​

Ideal gas - Gas molecules with:\

  • Negligible volume
  • Elastic collisions
  • No intermolecular forces

Ideal Gas Law​

PV=nRTPV = nRT

Where:

  • P = pressure (Pa)
  • V = volume (m³)
  • n = number of moles
  • R = gas constant = 8.31 J/(mol·K)
  • T = temperature (K)

For fixed amount of gas: PVT=constant\frac{PV}{T} = \text{constant}

Kinetic Theory​

Average kinetic energy per molecule: Ekˉ=32kBT\bar{E_k} = \frac{3}{2}k_BT

Where kBk_B = Boltzmann's constant = 1.38 × 10⁻²³ J/K

Pressure from particle collisions: P=13ρcˉ2P = \frac{1}{3}ρ\bar{c}^2

Where:

  • ρ = density
  • cˉ\bar{c} = mean molecular speed

Thermodynamic Processes​

Isothermal (Constant Temperature)​

T = constant, so ΔU = 0 Q=WQ = W PV=constantPV = \text{constant}

Heat in equals work out

Adiabatic (No Heat Exchange)​

Q = 0 ΔU=−W\Delta U = -W TVγ−1=constantTV^{γ-1} = \text{constant}

Internal energy change equals work

Isobaric (Constant Pressure)​

P = constant VT=constant\frac{V}{T} = \text{constant} W=PΔVW = P\Delta V

Isochoric (Constant Volume)​

V = constant PT=constant\frac{P}{T} = \text{constant} W=0W = 0

All heat goes to internal energy

Efficiency​

Heat Engine Efficiency​

η=WQin=Qin−QoutQinη = \frac{W}{Q_{\text{in}}} = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}}

Or: η=1−QoutQinη = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}}

Always less than 100% (can't convert all heat to work)

Carnot efficiency (maximum possible): ηCarnot=1−TcoldThotη_{\text{Carnot}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}

Heat Transfer​

Conduction​

Heat through solid: Q=kAΔTd×tQ = \frac{kA\Delta T}{d} \times t

Where:

  • k = thermal conductivity
  • A = area
  • ΔT = temperature difference
  • d = thickness

Metals: Good conductors (high k) Insulators: Poor conductors (low k)

Convection​

Heat transfer by fluid circulation

  • Warmer fluid less dense, rises
  • Cooler fluid denser, sinks
  • Repeat causes circulation

Radiation​

Heat transfer as electromagnetic waves P=σAT4P = σAT^4

(Stefan-Boltzmann law)

Key Points​

  1. Q = mcΔT for temperature changes
  2. Q = mL for phase changes (constant T)
  3. First law: ΔU = Q - W
  4. Second law: Entropy increases
  5. Ideal gas: PV = nRT
  6. Temperature = average kinetic energy
  7. Work from expansion: W = PΔV
  8. Internal energy from temperature
  9. Heat engine: η = 1 - Q_out/Q_in
  10. Heat transfers by conduction, convection, radiation

Practice Questions​

  1. Calculate energy from Q = mcΔT
  2. Calculate latent heat energy
  3. Analyze heating curves
  4. Apply ideal gas law
  5. Calculate work done
  6. Apply first law
  7. Solve thermodynamic processes
  8. Calculate efficiency
  9. Analyze heat transfer
  10. Complex thermal scenarios

Revision Tips​

  • Know specific values (c, L for common substances)
  • Understand phase changes
  • Apply ideal gas law correctly
  • Know thermodynamic laws
  • Practice P-V diagrams
  • Understand entropy concept
  • Remember K in temperatures
  • Calculate work from P-V graphs
  • Know heat transfer methods
  • Connect to particle theory