Syllabus coverage: 2.2.2 Specific heat capacity

Learning outcomes

By the end of this lesson, students should be able to:

  • describe internal energy as particle kinetic plus potential energy
  • explain temperature rise as an increase in average particle kinetic energy
  • define specific heat capacity correctly
  • use c = ΔE/(mΔθ) and ΔE = mcΔθ
  • compare materials using specific heat capacity and heating curves

5.1 Internal energy

The internal energy of an object is the total microscopic energy of its particles. It includes their kinetic energy due to random motion and their potential energy associated with forces and separation. It does not include the kinetic energy of the object moving as a whole or gravitational potential energy due to its height above the ground.

Heating a substance without a change of state usually raises particle average kinetic energy, so temperature rises. During a change of state, energy can increase particle potential energy while temperature remains constant. This distinction explains the flat sections of heating curves.

5.2 Definition of specific heat capacity

Specific heat capacity c is the energy required per unit mass per unit temperature increase. In words, it tells us how difficult it is to raise the temperature of a material. The equation is c = ΔE/(mΔθ), or equivalently ΔE = mcΔθ.

The SI unit is J/(kg °C) or J/(kg K). Because a temperature interval has the same size in degrees Celsius and kelvin, either may be used for Δθ. Use kilograms if c is given in J/(kg °C). A frequent error is to insert mass in grams without converting.

Original KG2UNI thermal physics diagram

Figure 11. Original KG2UNI diagram.

5.3 Interpreting high and low values

A high specific heat capacity means a large amount of energy is needed for a given mass and temperature rise. Water has a high specific heat capacity, so it warms and cools relatively slowly. This helps moderate coastal climates and makes water useful as a coolant or heat store.

A low specific heat capacity means a material changes temperature more readily for the same energy input. Metals often heat quickly, though thermal conductivity is a separate property. Specific heat capacity concerns the amount of energy required; conductivity concerns the rate at which energy passes through the material.

5.4 Energy accounting

In an ideal calculation, all supplied energy raises the internal energy of the chosen object. In a real experiment, some energy heats the container, heater and thermometer, and some is transferred to the surroundings. Therefore, using electrical energy supplied as though all of it enters the sample often gives an experimental specific heat capacity larger than the accepted value.

Always identify the system. If a metal block and embedded heater are treated together, some energy stored in the heater is not part of the metal sample. If a liquid is heated in a calorimeter, the calorimeter heat capacity may need to be considered in advanced work, although O Level questions often simplify the situation.

Original KG2UNI thermal physics diagram

Figure 12. Original KG2UNI diagram.

5.5 Rearranging and checking answers

Write the equation before substituting. Use Δθ = final temperature – initial temperature. A temperature change cannot be found by adding the two readings. Check that the answer is physically sensible: heating 1 kg of material by only a few degrees normally requires thousands of joules, not a fraction of a joule.

If the question asks for final temperature, first calculate Δθ = ΔE/(mc), then add the rise to the initial temperature. If energy is removed, temperature falls; use the magnitude of the energy and describe the direction clearly.

Worked examples

Energy to heat water

Calculate the energy needed to heat 0.50 kg of water by 12 °C. Using c = 4200 J/(kg °C), ΔE = 0.50 × 4200 × 12 = 25 200 J.

Finding specific heat capacity

A 0.80 kg block receives 7200 J and warms by 15 °C. c = 7200/(0.80 × 15) = 600 J/(kg °C).

Final temperature

A 2.0 kg material of c = 900 J/(kg °C) receives 18 000 J. Δθ = 18 000/(2.0 × 900) = 10 °C. If it began at 23 °C, the final temperature is 33 °C.

Practical focus

Investigation

Use equal-mass blocks of different materials, supply equal energy with identical heaters and compare temperature rises. Insulate each block similarly and start at the same temperature. A smaller temperature rise indicates a larger specific heat capacity, provided the energy losses are comparable.

Examination guidance

  • Separate specific heat capacity from thermal conductivity.
  • Convert mass to kilograms when the unit of c contains kg.
  • Use temperature change, not final temperature, in ΔE = mcΔθ.

Check your understanding

  1. Define specific heat capacity.
  2. What are the two microscopic parts of internal energy?
  3. A 0.25 kg block of c = 800 J/(kg °C) warms by 20 °C. Find the energy supplied.
  4. Why does water warm more slowly than many metals for equal mass and energy?
  5. Why may an experiment give a value of c that is too high?

Answers

  1. Energy required per unit mass per unit temperature increase.
  2. Total particle kinetic energy and total particle potential energy.
  3. 4000 J.
  4. Water has a higher specific heat capacity.
  5. Some supplied energy is lost to the surroundings or heats the apparatus, yet all is assumed to heat the sample.