Physics 14.5-14.10 - Heating and changes of state

Physics 14.5-14.10 - Heating and changes of state

Heating can raise temperature or change state. Learn how to choose and use the two thermal-energy equations, explain conservation of mass, and reduce unwanted transfer with insulation.

Physical changes and energy

Choose a system before describing an energy change. The system might be the water in a sealed container, or the water together with its container. Energy can cross the system boundary, and matter can cross it if the boundary is open.

Changes of state conserve mass

A substance can change state by:

  • melting: solid to liquid;
  • freezing: liquid to solid;
  • evaporating or boiling: liquid to gas;
  • condensing: gas to liquid;
  • sublimating: solid directly to gas.

Evaporation occurs at the liquid surface and can happen below the boiling point. Boiling occurs throughout a liquid at its boiling point, forming bubbles of vapour.

These are physical changes. The particles remain particles of the same substance, so mass is conserved. If the change is reversed, the substance recovers its original properties. This differs from some chemical changes, in which new substances form and reversing the conditions does not simply recover the starting material.

An open beaker can appear to lose mass as water evaporates because water vapour leaves the chosen system. If the system includes both the beaker and all the vapour, total mass is still conserved.

What heating changes

Heating transfers energy into a system. That transfer increases the energy stored in the system, but it can have two different observable outcomes:

  1. The temperature rises. The particles' average kinetic energy increases, so their random motion or vibration becomes more vigorous.
  2. The state changes. The energy changes the particles' arrangement and the energy associated with their attractions. During the change of state, the average kinetic energy does not increase, so the temperature stays approximately constant.

Do not say that energy is "used up" or disappears during melting. The system stores more energy even though a thermometer may show no temperature rise.

Two thermal properties

QuantityMeaningWhat changes?Unit
Specific heat capacityEnergy needed to raise the temperature of 1 kg1\,\mathrm{kg} of a substance by 1 ∘C1\,{}^{\circ}\mathrm{C}, without changing its stateTemperatureJ kg−1 ∘C−1\mathrm{J\,kg^{-1}\,{}^{\circ}C^{-1}}
Specific latent heatEnergy needed to change the state of 1 kg1\,\mathrm{kg} of a substance without changing its temperatureStateJ kg−1\mathrm{J\,kg^{-1}}

A high specific heat capacity means that a large energy transfer produces only a small temperature rise for a given mass. A high specific latent heat means that a large energy transfer is needed to change the state of a given mass.

Thermal energy calculations

The thermal-energy equations are required for both Foundation and Higher tiers. Choose the equation from what happens to the substance.

Temperature change without a change of state

Change in thermal energy

ΔQ=mcΔθ\Delta Q=mc\Delta\theta

  • ΔQ\Delta Q = change in thermal energy, in joules (J\mathrm{J})
  • mm = mass, in kilograms (kg\mathrm{kg})
  • cc = specific heat capacity, in joules per kilogram degree Celsius (J kg−1 ∘C−1\mathrm{J\,kg^{-1}\,{}^{\circ}C^{-1}})
  • Δθ\Delta\theta = temperature change, in degrees Celsius (∘C{}^{\circ}\mathrm{C}), calculated as final temperature minus initial temperature

Use this equation when the substance remains in one state over the temperature interval. A temperature change is final minus initial temperature. If cooling, the change is negative; the magnitude gives the energy released.

Rearrange to find a missing quantity: c=ΔQmΔθc=\frac{\Delta Q}{m\Delta\theta}, m=ΔQcΔθm=\frac{\Delta Q}{c\Delta\theta} or Δθ=ΔQmc\Delta\theta=\frac{\Delta Q}{mc}.

Worked example 1

A 0.60 kg0.60\,\mathrm{kg} metal block has a specific heat capacity of 900 J kg−1 ∘C−1900\,\mathrm{J\,kg^{-1}\,{}^{\circ}C^{-1}}. Its temperature rises from 18 ∘C18\,{}^{\circ}\mathrm{C} to 43 ∘C43\,{}^{\circ}\mathrm{C}. Calculate the change in thermal energy.

Solution

  1. Temperature change: Δθ=43−18=25 ∘C\Delta\theta=43-18=25\,{}^{\circ}\mathrm{C}.
  2. Equation: ΔQ=mcΔθ\Delta Q=mc\Delta\theta.
  3. Substitution: ΔQ=0.60×900×25\Delta Q=0.60\times900\times25.
  4. Calculation: ΔQ=13 500 J\Delta Q=13\,500\,\mathrm{J}.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account