Specific Heats — Definition & Overview

As we know, every object has a heat capacity. This gives a measure of how much the temperature changes when we put a certain amount of heat energy into the object. Heat capacity on its own, however, is not incredibly useful. It needs to be measured individually for every object. Instead, a little trick is done to make it way more useful. We divide the heat capacity by the mass, and our result is the specific heat capacity.

Imagine we have a hunk of aluminum. If we know the specific heat capacity (which can be easily found in databases), all we need to know is that mass, then we can find the heat capacity by multiplying the specific heat capacity by the mass. Alternatively, we could measure the heat capacity and mass of our aluminum and find the specific heat capacity that is valid for all samples of aluminum

If the temperature rises 1 kelvin every 10 joules, then the heat capacity is $10 \text{ } \frac{\text{J}}{\text{k}}$. Heat capacity is the product of mass and specific heat capacity, thus, $$4.184 \text{ }\frac{\text{J}}{\text{g}\cdot\text{k}} \cdot m\text{ g}=10 \text{ } \frac{\text{J}}{\text{k}} \\ m=\frac{10}{4.184}=2.39$$ Thus, we need 2.39 grams of water.

(Video) specific heat capacity explained

by PhysicsHigh

In this video, PhysicsHigh explains the specific heat capacity. He starts by defining the specific heat and demonstrating that it varies for different substances. Then, he explains the units and equations. Finally, he gives a few real-world illustrations. This video is a great explanation that will help you understand specific heat!

In our previous lesson, we investigated heat capacity. It was the energy required to raise an object's temperature by one kelvin. So, it has units of joules per kelvin ($\frac{J}{K}$). But there's a problem. How do we define an object? It's quite vague. We'd have to take individual measurements for everything in the universe. That's not reasonable! So instead, we developed the specific heat capacity. That's the energy it takes to raise the temperature of one gram of a substance by one kelvin. So, its units are joules per gram-kelvin ($\frac{J}{gK}$).

The Specific Heat Capacity of Water

In general chemistry, we commonly apply specific heat capacity to heating water. Liquid water has a specific heat of $4.18\frac{J}{gK}$. You'll see this number so often that it's helpful to memorize it. Let's explore how specific heat influences the properties of water.

First, what causes water to have such a large heat capacity? Water's hydrogen bonding. Remember, hydrogen bonds are the most powerful Van der Waal's force. They occur when a very electronegative element forms a covalent bond with hydrogen. The molecule also needs the correct molecular geometry. For water, we have oxygen-hydrogen bonds, and a bent geometry. But what does this have to do with the heat capacity? Remember, temperature measure the system's average kinetic energy. Water's hydrogen bonds will resist molecular movements. So, we have to put in more energy to increase the temperature.

So, why does it matter that water has a high specific heat? It's paramount for life on earth! For example, water has precise freezing and vaporization temperatures. Why does this matter? It helps to create our seasons. In winter, water gets hit less by solar radiation. As the energy diminishes, its temperature drops. We see this very gradually during the fall. Soon, water begins to freeze. But remember, liquid water is more dense than solid. So, ice floats! The ice provides a protective layer over lakes and rivers. It also allows vegetation to undergo a period of dormancy. The dormancy gets triggered by the freezing of water, lack of sunlight, and lower temperatures. As we approach Spring, we quickly reach the melting temperature of ice. Rather quickly, rivers, lakes, and plants become active again. All of this gets influenced by water's specific heat.

Specific Heat Calculations

Thankfully, specific heat capacity calculations are easy! They basically amount to dimensional analysis problems. The standard formula for these problems is $Q=m\times c_{p}\times\Delta T$.

• $Q$ is the heat in joules.

• $m$ is mass in grams.

• $c_{p}$ is the constant pressure heat capacity in $\frac{J}{gK}$.

• As a side note, the heat capacity is different for constant volume ($c_{v}$). It's an application of path functions. The way the process occurs matters!

• $\Delta T$ is the change in temperature in kelvin.

• Remember, the change in kelvin and degrees celsius is equal.

Let's look at an example. How much energy is required to raise the temperature of 100.0 grams of water from $0.0^{o}C$ to $100.0^{o}C$?

For this problem, we need to remember that the specific heat of water is $4.18\frac{J}{gK}$. Once we know it, we can plug everything into the equation.

$Q=m\times c_{p}\times\Delta T \rightarrow Q=100.0g\times 4.18\frac{J}{gK}\times 100.0K=41,800.0 J$

Do you recognize this calculation? It's part of our calculations for heating and cooling curves.