144, 147–148ThermodynamicsCore

Internal Energy and Specific Heat Capacity

Heat a pan of water and its temperature rises steadily — until it boils, when the temperature stops rising even though you keep supplying energy. Explaining both stages needs one idea: internal energy\textbf{internal energy}, the total kinetic and potential energy of the particles in a substance.

35 min Topic 9 · Thermodynamics

What you'll be able to do

  • Define internal energy as the sum of random kinetic and potential energies of particles
  • Use ΔE=mcΔθ\Delta E = mc\Delta\theta and describe how to measure specific heat capacity
  • Use ΔE=LΔm\Delta E = L\Delta m for changes of state
  • Explain absolute zero and convert between Celsius and kelvin
1

Internal energy and temperature

The internal energy\textbf{internal energy} of a substance is the sum of the randomly distributed kinetic and potential energies of all its particles.

The kinetic part comes from particles moving and vibrating. The potential part comes from the forces between them, and depends on how far apart they are.

Temperature\textbf{Temperature} is a measure of the average kinetic energy of the particles, not of the total internal energy. A bath of warm water has more internal energy than a cup of boiling water, though it is at a lower temperature.

Energy flows by heating from a region of higher temperature to one of lower temperature, until they reach thermal equilibrium\textbf{thermal equilibrium}.

Tip — Say "randomly distributed" in the definition of internal energy. The ordered kinetic energy of a moving object does not count.

2

Absolute zero and the kelvin scale

Absolute zero\textbf{Absolute zero} is the temperature at which particles have the minimum possible internal energy. It is 0 K0\ \mathrm{K}, or −273.15∘C-273.15^{\circ}\mathrm{C}.

The kelvin scale starts at absolute zero, and one kelvin is the same size as one degree Celsius. So a temperature change is numerically the same on both scales.

Any equation involving gases or particle energy needs temperatures in kelvin. Heating equations only use Δθ\Delta\theta, so Celsius is fine there.

T/K=θ/∘C+273T/\mathrm{K} = \theta/^{\circ}\mathrm{C} + 273
Converting to kelvin.
3

Specific heat capacity

The specific heat capacity\textbf{specific heat capacity} cc of a substance is the energy needed to raise the temperature of 1 kg by 1 K, with no change of state. Its unit is J kg−1 K−1\mathrm{J\,kg^{-1}\,K^{-1}}.

Water has a very high value, about 4200 J kg−1 K−14200\ \mathrm{J\,kg^{-1}\,K^{-1}}, which is why it is used in heating systems and why coastal climates are mild.

To measure it, heat a known mass with an electrical heater in an insulated block, measuring the energy supplied with a joulemeter or from E=VItE = VIt, and plot temperature against time. The gradient Δθ/Δt\Delta\theta/\Delta t gives c=Pm (Δθ/Δt)c = \dfrac{P}{m\,(\Delta\theta/\Delta t)}. Using the gradient avoids errors from the initial lag while the heater warms up.

ΔE=mcΔθ\Delta E = mc\Delta\theta
Energy to change temperature.
Example 1A 2.0 kW kettle heats 1.5 kg of water from 15∘15^{\circ}C to 100∘100^{\circ}C. Ignoring losses, how long does it take? Use c=4200 J kg−1 K−1c = 4200\ \mathrm{J\,kg^{-1}\,K^{-1}}.
1ΔE=1.5×4200×85=5.355×105\Delta E = 1.5 \times 4200 \times 85 = 5.355\times10^{5} J.
2t=5.355×1052000t = \dfrac{5.355\times10^{5}}{2000}.
Answer268 s268\ \mathrm{s}, about 4.5 minutes.
4

Latent heat and changes of state

During melting or boiling, energy is supplied but the temperature stays constant. The energy increases the potential\textbf{potential} energy of the particles as bonds are broken or weakened; their average kinetic energy does not change.

The specific latent heat\textbf{specific latent heat} LL is the energy needed to change the state of 1 kg without changing temperature. Fusion refers to melting; vaporisation to boiling.

Latent heat of vaporisation is much larger than of fusion. Melting only loosens the structure, but boiling separates particles completely and does work pushing back the atmosphere as the gas expands.

ΔE=LΔm\Delta E = L\Delta m
Energy to change state.
Example 1How much energy is needed to turn 0.40 kg of ice at 0∘0^{\circ}C into water at 20∘20^{\circ}C? Lf=3.34×105 J kg−1L_{f} = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}.
1Melting: 0.40×3.34×105=1.336×1050.40 \times 3.34\times10^{5} = 1.336\times10^{5} J.
2Warming: 0.40×4200×20=3.36×1040.40 \times 4200 \times 20 = 3.36\times10^{4} J.
3Total =1.336×105+0.336×105= 1.336\times10^{5} + 0.336\times10^{5}.
Answer1.67×105 J1.67\times10^{5}\ \mathrm{J}

Tip — On a heating curve, sloping sections use mcΔθmc\Delta\theta and flat sections use LΔmL\Delta m. Split multi-stage problems at every change of state.

Equation recap

ΔE=mcΔθ\Delta E = mc\Delta\theta
Temperature change.
ΔE=LΔm\Delta E = L\Delta m
Change of state.
T=θ+273T = \theta + 273
Kelvin.

Common mistakes to avoid

Saying temperature measures internal energy.
Temperature measures average kinetic energy per particle.
Saying particles speed up while ice melts.
Kinetic energy is constant; potential energy increases.
Using mcΔθmc\Delta\theta across a change of state.
Add a separate LΔmL\Delta m term for each phase change.
Forgetting energy losses make cc too large.
Unmeasured losses mean more energy supplied for the same rise.

Key takeaways

  • Internal energy is the sum of random kinetic and potential energies of particles.
  • Temperature measures average particle kinetic energy; 0 K is absolute zero.
  • ΔE=mcΔθ\Delta E = mc\Delta\theta for a change in temperature.
  • ΔE=LΔm\Delta E = L\Delta m for a change of state at constant temperature.
  • During a change of state, potential energy changes but kinetic energy does not.

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