Temperature

All three subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

14.1 Thermal equilibrium

• Understand that (thermal) energy is transferred from a region of higher temperature to a region of lower temperature.
• Understand that regions of equal temperature are in thermal equilibrium.

14.2 Temperature scales

• Understand that a physical property that varies with temperature may be used for the measurement of temperature and state examples of such properties, including the density of a liquid, volume of a gas at constant pressure, resistance of a metal, e.m.f. of a thermocouple.
• Understand that the scale of thermodynamic temperature does not depend on the property of any particular substance.
• Convert temperatures between kelvin and degrees Celsius and recall that T/K = θ/°C + 273.15.
• Understand that the lowest possible temperature is zero kelvin on the thermodynamic temperature scale and that this is known as absolute zero.

14.3 Specific heat capacity and specific latent heat

• Define and use specific heat capacity.
• Define and use specific latent heat and distinguish between specific latent heat of fusion and specific latent heat of vaporisation.

Temperature and thermal equilibrium

Thermal energy is always transferred from a region of higher temperature to a region of lower temperature, and never the other way of its own accord. Two regions at the same temperature are in thermal equilibrium: there is no net transfer of energy between them, though energy still passes in both directions in equal amounts.

That is what temperature actually tells you. It is not a measure of how much energy an object contains; it tells you the direction in which energy will flow when the object is placed in contact with another. A bath at 40 °C holds far more internal energy than a spark at 1000 °C, but energy still flows from the spark to the bath.

At the microscopic level, temperature is a measure of the mean kinetic energy of the random motion of the molecules. That is why two objects at the same temperature exchange no net energy: their molecules, on average, carry the same kinetic energy, so collisions at the boundary transfer as much one way as the other.

Note the words mean, kinetic and random. Not total, because a large object at a low temperature can hold more energy than a small one at a high temperature. Not internal, because internal energy also includes potential energy due to intermolecular forces. Not the bulk motion of the object, because a moving brick is not hot.

Measuring temperature

Any physical property that varies with temperature can be used to measure it. Such a property is called a thermometric property, and the syllabus names four:

Each has its uses. A thermocouple has a very small junction and low heat capacity, so it responds quickly and can measure the temperature of a small object without cooling it, and it works over a very wide range. A resistance thermometer is precise but slow. A liquid-in-glass thermometer is cheap and direct but limited by the freezing and boiling points of its liquid.

Why an empirical scale is not enough

The problem with defining temperature from any one of those properties is that they do not agree with each other. Calibrate a mercury thermometer and a resistance thermometer so that both read 0 at the ice point and 100 at the steam point, put them in the same bath in between, and they give slightly different readings, because the resistance of platinum and the volume of mercury do not vary with temperature in the same way.

So a scale defined by one substance's property is arbitrary: it makes that substance correct by definition and everything else wrong.

The thermodynamic scale avoids this. It does not depend on the property of any particular substance, and it is why the kelvin is the SI unit of temperature.

Absolute zero and the kelvin

The lowest possible temperature on the thermodynamic scale is zero kelvin, called absolute zero. It is the temperature at which molecules have their minimum possible kinetic energy and no more energy can be removed from a substance. It cannot be reached, only approached.

Conversion:

T / K = θ / °C + 273.15

Worked examples.

25 + 273.15 = 298.15

so 25 °C is 298 K to three significant figures.

Going the other way, for a temperature of 350 K:

350 - 273.15 = 76.85

so 350 K is 76.9 °C.

Two things follow from the fact that the two scales have the same size of degree:

• A change in temperature is the same number in kelvin as in degrees Celsius. A rise of 20 °C is a rise of 20 K, and no conversion is needed. This matters constantly in specific heat capacity calculations, where only the change appears.
• An absolute temperature must always be converted. Every gas law and every equation involving mean kinetic energy needs kelvin, and putting Celsius into pV = nRT is one of the most reliable ways to get a wrong answer in topic 15.

Specific heat capacity

Specific heat capacity c is the energy required to raise the temperature of unit mass of a substance by one kelvin without a change of state.

E = mcΔθ

with c in J kg⁻¹ K⁻¹. Water has an unusually large value, 4180 J kg⁻¹ K⁻¹, which is why it is used as a coolant and why coastal climates are milder than inland ones.

Be careful with the word specific: it means per unit mass. Heat capacity without "specific" is for the whole object, in J K⁻¹, and equals mc.

Worked example. How much energy is needed to raise the temperature of 0.50 kg of water from 20 °C to 80 °C?

Δθ = 80 - 20 = 60

E = 0.50 × 4180 × 60 = 125400

so 1.25 × 10⁵ J, or 125 kJ.

Worked example, an electrical method. A 60 W immersion heater is placed in 0.40 kg of a liquid and run for 5.0 minutes. The temperature rises by 18 K. Find c, assuming no energy is lost.

t = 5.0 × 60 = 300

E = 60 × 300 = 18000

c = 18000 / (0.40 × 18) = 2500

so c is 2500 J kg⁻¹ K⁻¹.

Every real version of this experiment loses energy to the surroundings, so less energy reaches the liquid than the heater supplies, and the measured temperature rise is too small. That makes the calculated value of c too large. Reasoning about the direction of an error, rather than just naming the error, is what separates a full answer from a partial one. Lagging the container, starting below room temperature and finishing the same amount above it, and extrapolating a cooling curve all reduce the effect.

Specific latent heat

Specific latent heat L is the energy required to change the state of unit mass of a substance without any change of temperature.

E = mL

with L in J kg⁻¹. Two kinds:

• Specific latent heat of fusion: solid to liquid, or liquid to solid. For water it is 3.34 × 10⁵ J kg⁻¹.
• Specific latent heat of vaporisation: liquid to gas, or gas to liquid. For water it is 2.26 × 10⁶ J kg⁻¹.

Why the temperature does not change

This is the conceptual heart of the topic. During a change of state the energy supplied does not increase the mean kinetic energy of the molecules, so the temperature stays constant. Instead it does work against the intermolecular forces, increasing the potential energy of the molecules by separating them.

Why vaporisation is so much larger than fusion

For water, the latent heat of vaporisation is about seven times the latent heat of fusion, and the reason follows from what happens to the molecules:

• In melting, the molecules are only partially separated. The lattice breaks down and they become free to move past one another, but they stay roughly the same distance apart, so the change in volume is very small and little work is done against intermolecular forces.
• In boiling, the molecules are completely separated, so the intermolecular forces must be entirely overcome. The volume increases by a factor of over a thousand, and work is also done pushing back the atmosphere as the vapour expands.

Worked example. How much energy is needed to turn 0.20 kg of ice at 0 °C into water at 100 °C?

Melt it first:

E = 0.20 × 3.34 × 10⁵ = 66800

Then heat the water:

E = 0.20 × 4180 × 100 = 83600

Total:

66800 + 83600 = 150400

so 1.5 × 10⁵ J. Notice that melting the ice takes nearly as much energy as heating the resulting water through the whole 100 K, which is why a drink stays cold as long as any ice is left in it.

Worked example, finding L. A 50 W heater is run for 200 s in a funnel packed with ice at 0 °C, and 30 g of water is collected, of which 6.0 g was collected in a control experiment with the heater switched off.

The control measures the melting caused by the room, so subtract it:

mass melted by the heater = 30 - 6.0 = 24

E = 50 × 200 = 10000

Working in kilograms, 24 g is 0.024 kg:

L = 10000 / 0.024 = 416667

so L is about 4.2 × 10⁵ J kg⁻¹. The control experiment is the whole point of the method: without it, the melting caused by the warm room would be credited to the heater, and L would come out far too small.

Common mistakes

• Saying temperature measures the total energy or the internal energy of an object, rather than the mean kinetic energy of its molecules.
• Leaving out the word "mean", or the word "random", in a definition of temperature.
• Converting a temperature change from Celsius to kelvin by adding 273. A change is the same number on both scales.
• Failing to convert an absolute temperature to kelvin in a gas law.
• Using 273 instead of 273.15 when the question asks for the relationship precisely, though 273 is fine for most calculations.
• Saying absolute zero is where molecules have no energy at all, rather than the minimum possible energy.
• Using E = mcΔθ across a change of state, where the temperature does not change and c does not apply.
• Saying the energy supplied during boiling increases the kinetic energy of the molecules. It increases their potential energy.
• Explaining why latent heat of vaporisation exceeds that of fusion without mentioning that the molecules are fully separated and that work is done against the atmosphere.
• Assuming a specific heat capacity experiment loses no energy, and so not explaining why the measured value comes out too high.
• Omitting the control in a latent heat of fusion experiment, then blaming the discrepancy on random error.