K323 / 2027
Thermal properties of matter overview

Topic 2 of 6

Heat capacity and specific heat capacity

The energy needed for a temperature rise depends on what is being heated and how much material there is.

Heating transfers energy into a sample's internal store. This page considers a temperature change without a change of state, with specific heat capacity treated as constant over the interval.

A property of the body and a property per unit mass

Heat capacity, C
The energy needed to raise the temperature of the specified body by one degree Celsius or one kelvin. Its unit is J/°C or J/K. A larger amount of the same material has a larger heat capacity under the same conditions.
Specific heat capacity, c
The energy needed to raise the temperature of unit mass of a substance by one degree Celsius or one kelvin. A common unit is J/(kg °C), equivalent to J/(kg K). The word specific means that the energy is given per unit mass.
E = CΔT = mcΔTE is the energy transferred to the sample in J; m is its mass; ΔT is its temperature change. For a uniform sample under these conditions, C = mc.

Here ΔT denotes a temperature difference: final temperature minus initial temperature on one scale. It can be expressed in °C or K, matching the heat-capacity unit. A change of 1°C is the same size as a change of 1 K. This does not mean that a temperature of 1°C equals 1 K.

Use the temperature rise of the sample

A 0.20 kg metal sample warms from 20 to 50 °C without changing state. Its supplied specific heat capacity is 450 J/(kg °C).

A temperature rise from 20 to 50 degrees Celsius is 30 degrees CelsiusA single temperature scale runs from zero to 60 degrees Celsius, with equally spaced ten-degree divisions. The initial reading is 20 and the final reading is 50. A bracket spans those two readings and marks the temperature rise as 30 degrees Celsius. The final temperature of 50 is not the temperature rise. The scale measures temperature, not energy.Temperature / °C0102030405060Final: 50 °CTemperature rise30 °CInitial: 20 °C50 - 20 = 30 °C

ΔT = 50 - 20 = 30 °C.

E = mcΔT = 0.20 × 450 × 30 = 2700 J.

Heat capacity of this sample: C = mc = 90 J/°C.

A rise of 30 °C is also a rise of 30 K. This is a temperature interval, not a conversion of 50 °C to kelvin.

The sample warms from 20°C to 50°C. The marked interval is 30°C, or 30 K. Use this change in the energy calculation, not the final reading of 50°C.

Worked example

Warm a metal sample

A 0.20 kg sample has c = 450 J/(kg °C). It warms from 20°C to 50°C without changing state.

  1. Find the interval: ΔT = 50 - 20 = 30°C.
  2. Find the sample's heat capacity: C = mc = 0.20 x 450 = 90 J/°C.
  3. Find the energy: E = mcΔT = 0.20 x 450 x 30 = 2700 J.

Each degree of temperature rise requires 90 J for this sample. The same material's specific heat capacity is 450 J per kilogram per degree; these are different quantities with different units.

Match the mass unit to the supplied value

A value of 4200 J/(kg °C) means that 1 kg requires 4200 J for a 1°C rise. The equivalent value is 4.2 J/(g °C), because 1 kg contains 1000 g. This is also 4.2 J/(g K), since a temperature interval of 1 K equals an interval of 1°C.

For 100 g of that material warming by 10°C, either use 0.100 x 4200 x 10 or use 100 x 4.2 x 10. Both give 4200 J. Using 100 with the kilogram value would mix incompatible units.

c = E / (mΔT)
ΔT = E / (mc)Rearrange according to the unknown. For equal energy input and equal material, a larger mass has a smaller temperature rise.

Separate energy reaching the sample from heater input

Power is energy transferred per unit time. If a 120 W heater transfers all its output to the metal sample above at a constant rate, the ideal time is t = E/P = 2700/120 = 22.5 s.

If some heater energy warms the apparatus or transfers to the surroundings, less reaches the sample in that time. Reaching the same 30°C rise then requires more input energy or a longer time. A heater's output is not automatically the sample's energy gain.

Determine a specific heat capacity

For a suitable metal sample, use an embedded heater of known constant power and a temperature sensor in good thermal contact with the sample. Insulation reduces transfer to the surroundings. Keep the sample below any change of state.

  1. Measure the mass: use a balance for the sample itself, excluding separate supports and instruments.
  2. Measure the temperature rise: record the initial and final sample temperatures over the chosen heating interval. Place the sensor in the sample's intended sensing position, rather than measuring the exposed heater. Choose suitable range and resolution, and allow for response lag.
  3. Find the supplied energy: with constant heater power P and measured time t, Einput = Pt.
  4. Use the energy account: the input warms the sample and apparatus and may escape to the surroundings. If these other transfers are negligible or have been estimated and subtracted, c = Esample / (mΔT).

Supplied practical data

Correct the energy account

In a separate run, a 20 W heater operates for 180 s while a 0.20 kg sample warms by 30°C. Independent calibration supplies two estimates: 300 J warms the apparatus and 600 J transfers to the surroundings.

  1. Heater input: Einput = 20 x 180 = 3600 J.
  2. Energy reaching the sample: Esample = 3600 - 300 - 600 = 2700 J.
  3. Specific heat capacity: c = 2700 / (0.20 x 30) = 450 J/(kg °C).

Ignoring both other transfers gives 3600 / (0.20 x 30) = 600 J/(kg °C), which is too high. The correction values are supplied for this example. An unpowered setup that stays near room temperature would not automatically measure the heat loss from a sample following a different, hotter temperature history.

If all input is wrongly treated as energy reaching the sample, the numerator is too large and the calculated c tends to be too high. Better insulation and accounting for apparatus heating address that cause. Repeating the same imperfect setup may show variation but does not remove the systematic error.

For cooling, say which energy you are calculating

If the same 0.20 kg sample cools from 50°C to 20°C without changing state, the positive amount of energy transferred out is mc(50 - 20) = 2700 J. Under the stated model, the sample's internal-energy change is -2700 J. The minus sign describes a loss; it does not mean that a negative amount was delivered to the surroundings.

Choose the temperature interval and the system. E = mcΔT does not cover melting or boiling, and Pt is the heater input unless the question says all of it reaches the sample.

Optional check A 0.20 kg metal sample has a specific heat capacity of 450 J/(kg degree C). It warms from 20 to 50 degrees C without changing state. How much energy reaches the sample?
A 0.20 kg metal sample has a specific heat capacity of 450 J/(kg degree C). It warms from 20 to 50 degrees C without changing state. How much energy reaches the sample?