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Thermal physics and kinetic theory: QCE Physics Unit 1 Year 11

Syllabus dot point

“Thermal energy, temperature and kinetic theory of matter, methods of heat transfer (conduction, convection, radiation), specific heat capacity Q=mcΔTQ = mc\Delta T, and latent heat”

10 min read

Quick answer

Thermal physics applies kinetic theory (temperature = average kinetic energy of particles) and heat transfer (conduction, convection, radiation) to thermal problems; specific heat capacity formula Q=mcΔTQ = mc\Delta T underpins calorimetry and latent heat (Q=mLQ = mL) accounts for energy absorbed at phase changes.

Jump to a section
  1. What this dot point is asking
  2. Kinetic theory of matter
  3. Heat transfer
  4. Specific heat capacity
  5. Latent heat
  6. Calorimetry
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

QCAA wants Year 11 students to apply kinetic theory and heat transfer concepts to thermal problems, and to use the specific heat capacity formula in calorimetry.

Kinetic theory of matter

All matter is made of particles (atoms or molecules) in constant random motion. Kinetic theory links the microscopic motion of these particles to the macroscopic quantities we measure (temperature, pressure and internal energy), and it is the conceptual spine of the whole topic.

  • Temperature is a measure of the average translational kinetic energy of the particles. It is an intensive property, meaning it does not depend on how much substance is present. Two beakers of water at the same temperature have the same average particle kinetic energy regardless of volume. The absolute (kelvin) scale is built so that 0 K0\ \text{K} is the point of minimum particle motion, and on the kelvin scale average kinetic energy is directly proportional to absolute temperature.
  • Internal energy is the total of all the kinetic and potential energies of the particles in a sample. It is an extensive property, so it scales with the amount of substance. A large mass of warm water can hold far more internal energy than a small mass of very hot water, which is the key distinction QCAA tests.
  • In a solid, strong bonds hold particles in a fixed lattice and they only vibrate about fixed positions. In a liquid, particles have enough energy to slide past one another while staying close together. In a gas, particles move freely in mostly straight-line paths between collisions, with negligible intermolecular forces. Heating a substance raises the average kinetic energy; at a phase boundary the added energy goes into potential energy (breaking bonds) instead of kinetic energy, so the temperature pauses.

Heat transfer

Heat is the transfer of thermal energy from a region of higher temperature to one of lower temperature. There are three mechanisms, and QCAA scenarios usually involve all three acting together.

Conduction
Energy flows through a material by particle vibration and collision, passing kinetic energy from hot particles to neighbouring cooler ones. Solids conduct best because their particles are close-packed; metals are exceptional conductors because free (delocalised) electrons carry energy rapidly through the lattice. Gases conduct poorly, which is why trapped air is a good insulator.
Convection
Heat is carried by the bulk movement of a fluid (liquid or gas). When a fluid is heated it expands, becomes less dense and rises, while cooler denser fluid sinks to replace it, setting up a convection current. Convection drives sea breezes, ocean currents and the circulation in a heated room, and it cannot occur in a solid.
Radiation
Energy is transferred by electromagnetic waves (predominantly infrared) and, uniquely, requires no medium, so it crosses a vacuum. Every object above 0 K0\ \text{K} radiates. The Stefan-Boltzmann law gives the power radiated:

P=eσAT4,P = e\sigma A T^4,

where ee is emissivity (00 to 11), σ=5.67×10−8 W m−2K−4\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}, AA is surface area and TT is the absolute temperature in kelvin. Because of the fourth-power dependence, a modest rise in absolute temperature produces a large rise in radiated power.

Specific heat capacity

The specific heat capacity cc of a substance is the energy needed to raise the temperature of 1 kg1\ \text{kg} by 1 K1\ \text{K} (or 1∘C1^\circ\text{C}, since a temperature interval is identical on both scales):

Q=mcΔT,Q = m c \Delta T,

where QQ is energy in joules, mm is mass in kilograms, cc is specific heat capacity in J kg−1K−1\text{J kg}^{-1}\text{K}^{-1} and ΔT\Delta T is the temperature change. Water has an unusually high value, c=4186 J kg−1K−1c = 4186\ \text{J kg}^{-1}\text{K}^{-1}, which is why it stores large amounts of energy, moderates coastal climates and is used as a coolant. Typical comparison values are cice=2100c_{\text{ice}} = 2100, caluminium=900c_{\text{aluminium}} = 900 and ccopper=386 J kg−1K−1c_{\text{copper}} = 386\ \text{J kg}^{-1}\text{K}^{-1}. A positive QQ means energy absorbed and temperature rises; a negative value means energy released and temperature falls.

Latent heat

During a phase change, energy is absorbed or released without any temperature change, because the energy goes into changing the potential energy of the particles (breaking or forming intermolecular bonds) rather than their kinetic energy.

  • Latent heat of fusion LfL_f is the energy per kilogram to melt a solid (or release on freezing). For water, Lf=3.34×105 J kg−1L_f = 3.34 \times 10^5\ \text{J kg}^{-1}.
  • Latent heat of vaporisation LvL_v is the energy per kilogram to boil a liquid (or release on condensing). For water, Lv=2.26×106 J kg−1L_v = 2.26 \times 10^6\ \text{J kg}^{-1}.

Q=mL.Q = m L.

Vaporisation requires far more energy than fusion (about seven times for water) because all the intermolecular bonds must be fully broken to separate the particles into a gas, not merely loosened.

Calorimetry

In an insulated system no energy escapes, so energy is conserved and heat lost by the hotter bodies equals heat gained by the cooler bodies:

m1c1(T1,i−Tf)=m2c2(Tf−T2,i).m_1 c_1 (T_{1,i} - T_f) = m_2 c_2 (T_f - T_{2,i}).

This is the principle behind calorimetry and the standard mixing problem. Write the conservation equation, substitute mcΔTmc\Delta T for each body, then solve for the unknown (usually the final temperature TfT_f or an unknown specific heat). If a phase change occurs during mixing, an extra mLmL term must be added on the appropriate side.

Calorimetry: final temperature of a mixture

Set up the energy balance

A 0.40 kg0.40\ \text{kg} aluminium block (c=900 J kg−1K−1c = 900\ \text{J kg}^{-1}\text{K}^{-1}) at 95∘C95^\circ\text{C} is dropped into 0.60 kg0.60\ \text{kg} of water (c=4186 J kg−1K−1c = 4186\ \text{J kg}^{-1}\text{K}^{-1}) at 18∘C18^\circ\text{C} in an insulated cup. In an insulated system, heat lost by aluminium equals heat gained by water:

mAlcAl(95−Tf)=mwcw(Tf−18).m_{Al} c_{Al}(95 - T_f) = m_w c_w (T_f - 18).

Substitute the values

(0.40)(900)(95−Tf)=(0.60)(4186)(Tf−18)(0.40)(900)(95 - T_f) = (0.60)(4186)(T_f - 18)

360(95−Tf)=2511.6(Tf−18).360(95 - T_f) = 2511.6(T_f - 18).

Expand and collect terms

34200−360Tf=2511.6Tf−45208.834200 - 360 T_f = 2511.6 T_f - 45208.8

79408.8=2871.6 Tf.79408.8 = 2871.6\, T_f.

Solve and sense-check

Tf=79408.82871.6=27.7∘C.T_f = \frac{79408.8}{2871.6} = 27.7^\circ\text{C}.

The final temperature lies between the two starting values and sits close to the water's value, which is the expected result because water has by far the larger heat capacity here.

Common errors
Confusing temperature with internal energy
A large mass of cold water can have more total internal energy than a small hot cup.
Forgetting latent heat at phase changes
No temperature change during melting/boiling but energy is absorbed.
Using degrees C in Kelvin formulas
For Stefan-Boltzmann (T4T^4) you need absolute temperature (K).

Examples in context

Example 1. A Townsville solar hot-water system raises 300 L300 \text{ L} of water from 20∘C20^\circ \text{C} to 65∘C65^\circ \text{C} on a sunny day. Sensible heat is Q=mcΔT=300×4186×45=5.65×107 JQ = mc\Delta T = 300 \times 4186 \times 45 = 5.65 \times 10^7 \text{ J}. Convection circulates water through the rooftop collector while conduction transfers solar-heated copper to the working fluid and radiation contributes about 3030 per cent of the collector input. The kinetic theory underpins all three: hotter water molecules vibrate and translate more vigorously, raising bulk temperature.

Example 2. ANSTO Mt Cotton's calibration lab uses a temperature-stable copper block of mass 5.0 kg5.0 \text{ kg} as a thermal anchor. Heating it through 1.0 K1.0 \text{ K} requires Q=5.0×390×1=1950 JQ = 5.0 \times 390 \times 1 = 1950 \text{ J}, providing slow thermal response that smooths satellite-tracker electronics through diurnal swings. The QCAA Unit 1 dot point binds the kinetic theory (high cc from many vibrational modes per mole) to the engineering choice of material.

Try this

Q1. State the kinetic theory definition of temperature. [2 marks]

  • Cue. Measure of the average translational kinetic energy of the constituent particles.

Q2. A 0.40 kg0.40 \text{ kg} aluminium block (c=900 J kg−1 K−1c = 900 \text{ J kg}^{-1} \text{ K}^{-1}) absorbs 7200 J7200 \text{ J}. Calculate the temperature rise. [2 marks]

  • Cue. ΔT=Q/(mc)=7200/(0.40×900)=20 K\Delta T = Q/(mc) = 7200/(0.40 \times 900) = 20 \text{ K}.

Q3. Identify and analyse the three heat-transfer mechanisms in a Queensland slate-roofed home in summer. (a) Describe each mechanism in context. (b) Calculate the radiative loss from a 30 m230 \text{ m}^2 roof at 60∘C60^\circ \text{C} to a sky at 20∘C20^\circ \text{C} (emissivity 0.90.9, σ=5.67×10−8 W m−2 K−4\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}). (c) Recommend two design measures. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

  • Cue. (a) Conduction through tiles, convection in cavity, radiation to sky; (b) net ≈8.2 kW\approx 8.2 \text{ kW}; (c) reflective foil, ridge venting.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
A copper block of 0.50 kg0.50\ \text{kg} at 80∘C80^\circ\text{C} is placed in 0.30 kg0.30\ \text{kg} of water at 20∘C20^\circ\text{C} in an insulated container. Using cCu=386 J kg−1K−1c_{Cu} = 386\ \text{J kg}^{-1}\text{K}^{-1} and cw=4186 J kg−1K−1c_w = 4186\ \text{J kg}^{-1}\text{K}^{-1}, determine the final temperature of the system.
Show worked answer →

Apply conservation of energy in an insulated container: heat lost by copper equals heat gained by water.

mCucCu(80−T)=mwcw(T−20)m_{Cu} c_{Cu} (80 - T) = m_w c_w (T - 20)

0.50×386×(80−T)=0.30×4186×(T−20)0.50 \times 386 \times (80 - T) = 0.30 \times 4186 \times (T - 20)

193(80−T)=1255.8(T−20)193(80 - T) = 1255.8(T - 20)

15440−193T=1255.8T−2511615440 - 193T = 1255.8T - 25116

40556=1448.8T40556 = 1448.8T

T≈28.0∘CT \approx 28.0^\circ\text{C}.

Markers reward the correct conservation equation, valid algebraic rearrangement, and a final temperature that lies between the two starting temperatures (closer to the water because of its much higher heat capacity and comparable mass).

Practice6 marks
Determine the total energy required to convert 0.25 kg0.25\ \text{kg} of ice at −15∘C-15^\circ\text{C} into steam at 100∘C100^\circ\text{C}. Use cice=2100c_{\text{ice}} = 2100, cwater=4186 J kg−1K−1c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}, Lf=3.34×105L_f = 3.34 \times 10^5 and Lv=2.26×106 J kg−1L_v = 2.26 \times 10^6\ \text{J kg}^{-1}.
Show worked answer →

Split the process at every phase boundary, because temperature only changes within a single phase and stays constant during a phase change.

Stage 1: warm ice −15∘C→0∘C-15^\circ\text{C} \to 0^\circ\text{C}
Q1=mcΔT=0.25×2100×15=7875 JQ_1 = mc\Delta T = 0.25 \times 2100 \times 15 = 7875\ \text{J}.
Stage 2: melt ice at 0∘C0^\circ\text{C}
Q2=mLf=0.25×3.34×105=8.35×104 JQ_2 = mL_f = 0.25 \times 3.34 \times 10^5 = 8.35 \times 10^4\ \text{J}.
Stage 3: warm water 0∘C→100∘C0^\circ\text{C} \to 100^\circ\text{C}
Q3=mcΔT=0.25×4186×100=1.0465×105 JQ_3 = mc\Delta T = 0.25 \times 4186 \times 100 = 1.0465 \times 10^5\ \text{J}.
Stage 4: boil water at 100∘C100^\circ\text{C}
Q4=mLv=0.25×2.26×106=5.65×105 JQ_4 = mL_v = 0.25 \times 2.26 \times 10^6 = 5.65 \times 10^5\ \text{J}.
Total
Q=7875+83500+104650+565000=7.61×105 JQ = 7875 + 83500 + 104650 + 565000 = 7.61 \times 10^5\ \text{J}.

Markers reward four separate stages, correct use of mass in kilograms, and recognition that vaporisation dominates because LvL_v is roughly seven times LfL_f.

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Sources & how we know this

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