Specific Heat Capacity: Definition, Units, Formula & Examples

Suppose you poured a fixed amount of water into two different beakers. One beaker is tall and narrow, and the other beaker is tall and wide. If the amount of water poured into each beaker is the same, you would expect the water level to be higher in the narrow beaker.

The width of these buckets is analogous to the concept of specific heat capacity. In this analogy, the water being poured into the buckets can be thought of as the heat energy being added to two different materials. The rise in level on the buckets is analogous to the resulting rise in temperature.

The specific heat capacity of a material is the amount of heat energy required to raise a unit mass of that material by 1 Kelvin (or degree Celsius). The SI units of specific heat capacity are J/kgK (joules per kilogram × Kelvin).

The specific heat varies depending on physical properties of a material. As such, it is a value you typically look up in a table. The heat ​Q​ added to a material of mass ​m​ with specific heat capacity ​c​ results in a temperature change ​ΔT​ determined by the following relationship:

\(Q=mc\Delta T\)

The specific heat capacity of granite is 790 J/kgK, of lead is 128 J/kgK, of glass is 840 J/kgK, of copper is 386 J/kgK and of water is 4,186 J/kgK. Note how much larger water's specific heat capacity is compared to the other substances in the list. It turns out that water has one of the highest specific heat capacities of any substance.

Substances with larger specific heat capacities can have much more stable temperatures. That is, their temperatures will not fluctuate as much when you add or remove heat energy. (Think back to the beaker analogy at the beginning of this article. If you add and subtract the same amount of liquid to the wide and the narrow beaker, the level changes a lot less in the wide beaker.)

It is because of this that coastal towns have much more temperate climates than inland cities. Being close to such a large body of water stabilizes their temperatures.

Water's large specific heat capacity is also why, when you take a pizza out of the oven, the sauce will still burn you even after the crust has cooled. The water-containing sauce has to give off a lot more heat energy before it can drop in temperature compared to the crust.

Suppose 10,000 J of heat energy is added to 1 kg of sand (​c​_s = 840 J/kgK) initially at 20 degrees Celsius, while the same amount of heat energy is added to a mixture of 0.5 kg sand and 0.5 kg of water, also initially at 20 C. How does the final temperature of the sand compare to the final temperature of the sand/water mixture?

​Solution:​ First, solve the heat formula for ​ΔT​ to obtain:

\(\Delta T=\frac{Q}{mc}\)

For the sand, then, you get the following change in temperature:

\(\Delta T=\frac{10,000}{1\times 840}=11.9 \text{ degrees}\)

Which gives a final temperature of 31.9 C.

For the mixture of sand and water, it's a little more complicated. You can't just divide the heat energy equally among the water and sand. They are mixed together, so they must undergo the same temperature change.

While you know the total heat energy, you don't know how much each one gets at first. Let ​_Q_s​ be the amount of energy from heat that the sand gets and ​Q_w​ be the amount of energy the water gets. Now use the fact that ​Q =​ ​Q_s + Q_w_​ to get the following:

\(Q=Q_s+Q_w=m_sc_s\Delta T+m_wc_w\Delta T=(m_sc_s+m_wc_w)\Delta T\)

Now it is straightforward to solve for ​ΔT:​

\(\Delta T = \frac{Q}{m_sc_s+m_wc_w}\)

Plugging in numbers then gives:

\(\Delta T = \frac{10,000}{0.5\times 840+0.5\times 4,186} = 4 \text{ degrees}\)

The mixture only rises by 4 C, for a final temperature of 24 C, significantly lower than the pure sand!