Heat is one of the most fundamental forms of energy transfer. From car engines to refrigerators to the human body, understanding internal energy, specific heat capacity and latent heat underpins all of thermodynamics.
🌡️Define internal energy as the sum of random KE and PE of all particles in a system
⚖️State and apply the first law of thermodynamics: ΔU = Q + W
💧Apply Q = mcΔθ and Q = mL to calculate thermal energy transfers
📊Interpret heating curves including phase changes at constant temperature
🔄Distinguish between isothermal, adiabatic, isobaric and isochoric processes
⚙️Explain why temperature stays constant during a change of state
Internal Energy
The internal energy U of a system is the total random kinetic energy and potential energy of all the particles (atoms, molecules) making up the system:
U = Σ(KE of particles) + Σ(PE of particles)
The kinetic energy arises from the random translational, rotational and vibrational motion of the particles. The potential energy arises from the intermolecular forces between particles.
Ideal gas: No intermolecular forces assumed, so U = KE only. U ∝ T (absolute temperature).
Real substances: Both KE and PE contribute. During a phase change, PE changes at constant KE (constant T).
Internal energy: The sum of the randomly distributed kinetic and potential energies of all particles in a system. It is a state function — its value depends only on the current state of the system, not on how that state was reached.
Internal energy is different from temperature: temperature is a measure of average random KE per particle, while U is the total energy of all particles. A large cold lake has more internal energy than a small hot cup of tea — there are vastly more particles.
The First Law of Thermodynamics
The first law is a statement of conservation of energy applied to thermal systems:
ΔU = Q + W
where ΔU is the increase in internal energy of the system, Q is the heat added to the system (positive if energy flows in), and W is the work done on the system (positive if compressed).
⚠️ Sign convention matters. AQA uses ΔU = Q + W where W is work done ON the system. Some textbooks use ΔU = Q − W where W is work done BY the system. Always check the sign convention in the question.
Work done on a gas during compression: W = −pΔV (for constant pressure), where ΔV is negative (volume decreases) so W is positive (energy added to gas).
Process
What's constant
Consequence
Isothermal
Temperature T
ΔU = 0 for ideal gas; Q = −W
Adiabatic
No heat exchange (Q = 0)
ΔU = W; compression raises T
Isobaric
Pressure p
W = −pΔV; both Q and W non-zero
Isochoric (isovolumetric)
Volume V
W = 0; ΔU = Q
Specific Heat Capacity
The specific heat capacity c of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 K (or 1 °C):
Q = mcΔθ
where Q is heat energy (J), m is mass (kg), c is specific heat capacity (J kg⁻¹ K⁻¹), and Δθ is temperature change (K or °C).
Material
c (J kg⁻¹ K⁻¹)
Notes
Water
4181
Very high — water is an excellent thermal store
Aluminium
897
Light with moderate c — used in cookware
Iron/steel
450
Lower c — heats up faster
Lead
128
Very low c — heats and cools rapidly
Air (constant pressure)
1005
Important in atmospheric physics
The high specific heat capacity of water (4181 J kg⁻¹ K⁻¹) explains why oceans regulate climate, why water is used as a coolant in engines, and why coastal areas have milder temperatures than inland regions.
In calorimetry experiments, heat loss to the surroundings is the main source of error. Insulation, a lid, and quick measurements minimise this systematic error.
Specific Latent Heat
During a change of state (phase transition), a substance absorbs or releases energy without changing temperature. This energy changes the potential energy of the particles (breaking or forming intermolecular bonds) without changing their average kinetic energy.
Q = mL
where L is the specific latent heat (J kg⁻¹): the energy per unit mass needed to change state at constant temperature and pressure.
Specific latent heat of fusion L_f: Energy to melt/freeze 1 kg. For water: L_f = 334 000 J kg⁻¹ = 334 kJ kg⁻¹.
Specific latent heat of vaporisation L_v: Energy to vaporise/condense 1 kg. For water: L_v = 2 260 000 J kg⁻¹ = 2260 kJ kg⁻¹.
L_v > L_f because vaporisation requires breaking nearly all intermolecular bonds (large increase in separation) while fusion only partially disrupts the lattice structure. Also, expansion against atmospheric pressure during vaporisation requires additional work.
Why temperature is constant during phase change: Energy input goes entirely into changing particle PE (breaking intermolecular bonds) rather than increasing KE. Since temperature measures average KE, it remains constant until the phase change is complete.
Heating Curves and Thermodynamic Processes
A heating curve (temperature vs heat added) shows flat regions during phase changes and sloping regions during temperature rise. The gradient of the sloping sections equals 1/(mc) — a steeper gradient means lower heat capacity.
Key features of a water heating curve:
Ice warming from −20°C to 0°C: slope with gradient 1/(m × 2090)
Melting at 0°C: flat region, Q = mL_f = m × 334 000 J
Water warming from 0°C to 100°C: slope with gradient 1/(m × 4181)
Vaporisation at 100°C: flat region, Q = mL_v = m × 2 260 000 J (much longer flat section)
Steam warming above 100°C: slope with gradient 1/(m × 2010)
For adiabatic processes: when a gas is compressed rapidly (no time for heat exchange), its temperature rises because work done on the gas increases its internal energy. This explains why a bicycle pump heats up and why diesel engines ignite fuel by compression alone.
A 1.5 kg aluminium block (c = 897 J kg⁻¹ K⁻¹) is heated from 20°C to 180°C using an electric heater of power 500 W. (a) Calculate the heat energy required. (b) Calculate the minimum time needed, assuming no heat losses.
3Minimum time: t = Q/P = 215 280/500 = 430.6 s ≈ 431 s ≈ 7.2 minutes
Q = 215 kJ; t_min = 431 s (7.2 min). Actual time would be longer due to heat losses.
How much energy is required to convert 200 g of ice at −10°C into steam at 100°C? (c_ice = 2090 J kg⁻¹ K⁻¹; L_f = 334 kJ kg⁻¹; c_water = 4181 J kg⁻¹ K⁻¹; L_v = 2260 kJ kg⁻¹)
Total energy = 607 kJ. Vaporisation alone accounts for 74% of the total energy required.
A gas is compressed adiabatically. 3000 J of work is done on the gas. (a) Calculate the change in internal energy. (b) Explain what happens to the temperature. (c) If the same compression were done isothermally, what would ΔU be?
1Adiabatic means Q = 0. By first law: ΔU = Q + W = 0 + 3000 = +3000 J
2Since ΔU > 0, the internal energy increases. For an ideal gas (U ∝ T), temperature increases.
3Isothermal compression: temperature constant → for ideal gas, ΔU = 0. By first law: 0 = Q + W → Q = −W = −3000 J. Heat flows OUT of the gas to the surroundings.
Adiabatic: ΔU = +3000 J, temperature rises. Isothermal: ΔU = 0, 3000 J of heat flows out.
In a calorimetry experiment, 0.10 kg of copper (c = 385 J kg⁻¹ K⁻¹) at 100°C is dropped into 0.15 kg of water at 20°C in a 0.05 kg aluminium calorimeter (c = 897 J kg⁻¹ K⁻¹) also initially at 20°C. Find the final temperature, assuming no heat losses.
1Heat lost by copper = heat gained by water + calorimeter
2Let final temperature = T °C. Heat lost: Q_Cu = 0.10 × 385 × (100 − T)
Final temperature T ≈ 24.3°C. The large heat capacity of water means the temperature rise is small.
Q1. The internal energy of an ideal gas is:
Q2. 500 J of heat is added to a system, and 200 J of work is done on the system. What is the change in internal energy?
Q3. Why does the temperature of a substance remain constant during melting even though heat is being supplied?
Q4. Which has the greater specific latent heat — vaporisation or fusion? Why?
Q5. A 2.0 kg block of iron (c = 450 J kg⁻¹ K⁻¹) cools from 300°C to 50°C. How much energy does it release?
Challenge 1. A steam power station generates electricity by converting heat from burning fuel into mechanical work. In one cycle, 50 MJ of heat is supplied from the boiler and 35 MJ is rejected to the cooling towers. (a) Calculate the work output per cycle. (b) Calculate the thermal efficiency. (c) Identify which law of thermodynamics sets an upper limit on efficiency, and state why 100% efficiency is impossible.
✓ (a) By conservation of energy (1st law): W = Q_in − Q_out = 50 − 35 = 15 MJ. (b) Efficiency η = W/Q_in = 15/50 = 0.30 = 30%. (c) The Second Law of Thermodynamics (not First) sets the upper limit. The Carnot efficiency η_Carnot = 1 − T_cold/T_hot gives the theoretical maximum. 100% efficiency is impossible because some heat must always be rejected to the cold reservoir — converting all heat into work would require T_cold = 0 K (absolute zero), which is unattainable. In this case, η_Carnot ≥ 30%, meaning some irreversibility reduces efficiency below the Carnot limit.
Challenge 2. A student measures the specific heat capacity of aluminium using an electrical immersion heater. They use a 50 W heater for 300 s and measure a temperature rise of 36 K in a 1.1 kg block. (a) Calculate the value of c obtained. (b) The accepted value is 897 J kg⁻¹ K⁻¹. Find the percentage difference. (c) State two ways to reduce the systematic error.
✓ (a) Energy supplied: E = Pt = 50 × 300 = 15 000 J. Using E = mcΔθ: c = E/(mΔθ) = 15 000/(1.1 × 36) = 15 000/39.6 = 378.8 J kg⁻¹ K⁻¹. (b) % difference = |378.8 − 897|/897 × 100% = 518.2/897 × 100% = 57.8%. Very large error! (c) Improvements: (i) Insulate the block thoroughly — the main source of error is heat loss to surroundings, which means less energy goes into the aluminium than assumed; (ii) Embed the heater and thermometer fully in the block to ensure good thermal contact; (iii) Start timing when steady heating begins and read temperature precisely; (iv) Use a longer heating time to reduce the fractional uncertainty in temperature rise. The large error here is dominated by heat loss — real c measurements require good calorimetry technique.
Challenge 3 (Synoptic). An adiabatic process involves a gas being compressed from volume 4.0 × 10⁻³ m³ to 1.0 × 10⁻³ m³. The initial pressure is 1.0 × 10⁵ Pa and the final pressure is 6.0 × 10⁵ Pa. Using the work done approximation W ≈ −½(p₁+p₂)ΔV, find (a) the work done on the gas, (b) the change in internal energy, (c) the expected change in temperature given n = 0.16 mol of a monatomic ideal gas.
✓ (a) ΔV = 1.0×10⁻³ − 4.0×10⁻³ = −3.0×10⁻³ m³ (volume decreases). Average pressure = ½(1.0+6.0)×10⁵ = 3.5×10⁵ Pa. Work done ON gas: W = −p_avg × ΔV = −3.5×10⁵ × (−3.0×10⁻³) = +1050 J. (b) Adiabatic: Q = 0. ΔU = Q + W = 0 + 1050 = +1050 J. (c) For monatomic ideal gas, U = (3/2)nRT → ΔU = (3/2)nRΔT → ΔT = 2ΔU/(3nR) = 2×1050/(3×0.16×8.314) = 2100/3.991 = 526 K. Temperature rises by ~526 K due to compression (no heat loss in adiabatic process).
Challenge 4. Explain why sweating is an effective cooling mechanism for the human body, using the concept of latent heat. Why is sweating less effective in humid conditions? Calculate the mass of sweat that must evaporate to remove 1.0 MJ of heat from the body (L_v,sweat ≈ 2.4 × 10⁶ J kg⁻¹).
✓ Sweating works because evaporation requires a large latent heat of vaporisation (L_v ≈ 2.4×10⁶ J kg⁻¹ for sweat ≈ water at body temperature). When sweat evaporates from skin, it takes this energy from the body's surface, cooling the skin and blood flowing near it. The latent heat is the energy needed to break hydrogen bonds between water molecules, transferring thermal energy from the body to the water vapour. In humid conditions: air near skin already contains high concentrations of water vapour, so the rate of evaporation slows dramatically. Evaporation rate ∝ (concentration gradient of water vapour between skin surface and air). High humidity reduces this gradient → less evaporation → less cooling. This is why 35°C in humid conditions feels far hotter than 35°C in dry conditions (wet-bulb temperature effect). Mass: m = Q/L_v = 1.0×10⁶ / 2.4×10⁶ = 0.417 kg ≈ 0.42 kg ≈ 420 ml of sweat. This is achievable during intense exercise — hence the importance of hydration.