2.2.2 Specific heat capacity
- Syllabus
- 0625–2026–2027
- Topic
- 2.2.2
- Level
- —
Learning objectives
2.2.2.1A rise in the temperature of an object• Know: a rise in the temperature of an object increases its internal energy2.2.2.2An increase in temperature of an• Describe an increase in temperature of an object in terms of an increase in the average kinetic energies of all of the particles in the object2.2.2.3Specific heat capacity as energy• Define specific heat capacity as energy required per unit mass per unit temperature increase; recall/use: c = ΔE/(mΔT)2.2.2.4Experiments to measure the specific• Describe experiments to measure the specific heat capacity of a solid and a liquid
A temperature rise increases internal energy
When an object's temperature rises, its internal energy increases. Internal energy is energy stored by all the particles in the object; temperature is a measurement, not an energy store.
The statement describes a change in one object: compare it before and after heating. Energy transferred to the object raises its internal energy, and a temperature rise is evidence of that increase.
Temperature and internal energy are not the same quantity. A large amount of cooler material can have more internal energy than a small hot object, so temperature alone does not give an object's total internal energy.
Temperature tracks average particle kinetic energy
An increase in temperature means an increase in the average kinetic energy of all the particles in the object.
| State | What the greater average kinetic energy looks like |
|---|---|
| solid | particles vibrate more energetically about their fixed positions |
| liquid | particles move around one another faster on average |
| gas | particles travel faster on average between collisions |
‘Average’ matters: particles do not all have identical kinetic energy or move at one identical speed. The distribution changes so that the mean kinetic energy of the whole collection is greater.
Do not say that heating makes particles themselves larger. The temperature change is explained by their motion and kinetic energy, not by a change in particle size.
Define and calculate specific heat capacity
Specific heat capacity, c, is the energy required per unit mass of a substance for a unit increase in temperature. Its unit is J/(kg °C), equivalently J kg⁻¹ °C⁻¹.
c=\frac{\Delta E}{m\Delta T}\qquad\text{and therefore}\qquad\Delta E=mc\Delta T
Here ΔE is the increase in internal energy in joules, m is mass in kilograms, and ΔT is the temperature rise in °C. A temperature interval has the same numerical size in °C and K.
Example: a 0.80 kg aluminium block receives 18 000 J and warms by 25 °C.
c=0.80×2518000=900 J/(kg °C)
The substitution uses the temperature change, not the final temperature.
A larger c means more energy is needed to produce the same temperature rise in the same mass. It does not mean the substance automatically heats faster; with the same energy input, a larger c gives a smaller temperature rise.
Measure the specific heat capacity of a solid or liquid
Measure the sample's mass m, the electrical energy supplied ΔE, and its temperature rise ΔT, then calculate c=ΔE/(mΔT). If heater power P is known, ΔE=Pt; otherwise use ΔE=VIt.
| Feature | Solid block | Liquid |
|---|---|---|
| sample mass | measure the block on a balance | measure filled container minus empty container |
| heater and thermometer | place them in snug holes with good thermal contact | immerse both in the liquid; keep them clear of the container wall |
| limiting heat transfer | wrap the block in insulation | use an insulated container and lid |
| uniform temperature | allow the block to conduct energy through it | stir the liquid gently while heating |
- Measure m and the initial temperature.
- Switch on the heater and start the timer together. Record P and t, or record V, I and t.
- Record the final temperature and calculate ΔT=Tfinal−Tinitial.
- Calculate the supplied energy and then c=ΔE/(mΔT).
- Repeat and compare values; use a moderate temperature rise so the change is clear without making heat loss excessive.
Insulate the sample, use a lid for a liquid, ensure good heater contact in a solid, and stir a liquid. Some electrical energy warms the heater, thermometer and container or escapes to the surroundings. Treating all supplied energy as energy gained by the sample makes the calculated c too large.