xaktly | Thermodynamics

Heat capacity

Heat capacity & specific heat

Heat capacity is the ability of a material to absorb heat without directly reflecting all of it as a rise in temperature. You should read the sections on heat and temperature as background, and the water section would help, too.

As heat is added uniformly to like quantities of different substances, their temperatures can rise at different rates. For example, metals,

good conductors of heat, show fast temperature rises when heated. It is relatively easy to heat a metal until it glows red. On the other hand, water can absorb a lot of heat with a relatively small rise in temperature. Insulating materials (insulators) are very poor conductors of heat, and are used to isolate materials that need to be kept at different temperatures — like the inside of your house from the outside.

Substances absorbing the same amount of heat can have different temperature rises.

This graph shows the rise in temperature as heat is added at the same rate to equal masses of aluminium (Al) and water (H2O). The temperature of water rises much more slowly than that of Al.

In the metal, Al atoms only have translational kinetic energy (although that motion is coupled strongly to neighbor atoms). Water, on the other hand, can rotate and vibrate as well. These degrees of freedom of motion can absorb kinetic energy without reflecting it as a rise in temperature of the substance.

Translation — an atomic gas

How can the atoms of an atomic gas (a gas composed of single atoms rather than molecules) manifest their kinetic energy? The answer is translation. The can move right, left, up or down, or any combination of those.

Here's a diagram of an argon (Ar) atom, which is really just a sphere of electrons surrounding a positive nucleus buried deep inside — deep enough that an argon atom really looks, to other argon atoms, like a ball of negative charge.

The only way an argon atom can have kinetic energy is to move along one of these three directions (x, y or z; see the coordinate axes), or any combination of them. We say that the three translational motions of argon form a basis for any kind of translational motion.

On the coordinate axes, you can see how the motion along the blue arrow (vector) can be represented by a little bit of motion along x, along y and along z.

You might think that argon atoms could also rotate, but we can look at that in two ways. First, if an argon did rotate, how would we know; it's a sphere (more or less). But the better way of thinking about it is that rotations of atoms really correspond to the energies of the electrons in the atoms, and those don't really contribute to the current subject.

Atoms can only translate along the x, y or z axes or any combination of those motions.

Kinetic energy of molecules

As a proxy for all molecules, we'll use water (H2O). Why not? It's arguably the most important solvent on Earth, many of our future discussions will focus on water, and water has some very interesting properties, especially for a molecule of its size.

Water — and any molecule — can undergo the same three translations that an atom can. Without rotating or changing its shape, a water molecule can move along the x, y or z axes, or any combination of those motions. But water (molecules) can express their kinetic energy in more ways, too.


Molecules can also rotate. Check out the three rotations of a water molecule below. Water — like any object — rotates around its center of mass, which lies within the heavy oxygen (O) atom, but nearer to the hydrogens (H).

Water molecules can rotate around the x axis (drawn here coming out of your screen), the y-axis or the z-axis.

Any crazy rotation of a water molecule can be represented by a little bit of each of these rotational motions. We say that water has "three degrees of rotational freedom,$ giving water three more (so far) degrees of freedom than argon atoms.


Molecules have one more way of storing kinetic energy, and that's vibration. Any molecule, which is a collection of atoms connected by chemical bonds, which I've drawn in these molecules as springs. Bonds, like springs, can stretch and bend. All of the bonds of all molecules are constantly stretching or bending to some extent — there's no such thing as an atom or molecule standing still.

In molecules, just as we did for translation and rotation, we can always find some basis vibrations from which any crazy vibration of the molecule can be represented.

In water we have three vibrations (larger molecules, having more bonds, have more). Each of these represents another degree of freedom of motion. Water can bend symmetrically, as shown on the left. Its O—H bonds can also stretch symmetrically (center picture) or asymmetrically (right). These three vibrational motions form the three degrees of vibrational freedom of water.

So our final result in comparing the modes of kinetic energy of atoms and water is that

Water has nine degrees of freedom of motion while argon (or any atomic gas) only has three.

This difference will explain the difference in the heating graphs we started with. It takes a lot more heat input to increase the temperature of a sample of water than it does to heat an atomic substance (like a gas or many metals) to the same temperature. We just need one more concept. Keep on ...

Equipartition of Energy

Most substances obey the law of equipartition of energy over a broad range of temperatures. The law says that energy tends to be distributed evenly among all of the degrees of freedom of a molecule — translation, rotation and vibration. This has consequences for substances with more or fewer atoms. In the diagram below, the container on the left represents a fixed amount of energy that we're going to put into a sample of argon gas (just atoms) and into a sample of the same number of water molecules.

In the gas, the energy has only three places to go, represented by the three containers. These are the three translational degrees of freedom. Notice that the containers are equally full. When we put the same amount of energy into water molecules, there are nine containers, and the energy gets equally partitioned among them, thus they're all less full than the argon containers. The 3-atom molecule ends up with less energy in its translational degrees of freedom. It is the translational energy that we measure as temperature, thus we expect the water to rise to a lower temperature than the gas with the same input of energy.

Aside: How microwave ovens work

The figure below shows roughly how microwave ovens work — it's a study in equipartition of energy. A microwave oven uses a relatively cheap device called a magnetron to generate microwave radiation (same as visible light, just lower frequency/longer wavelength) of a frequency of about 2.45 GHz. That's just the right frequency to cause water molecules — ubiquitous in food — to rotate faster. Here we're just putting energy into the rotational degrees of freedom (upper row of energy containers). But in a short time, those rapidly-rotating water molecules will bump into other waters, causing them to translate and vibrate faster, too, until it all evens out, and the energy in the translational degrees of freedom (what we measure as temperature) is greater. Soup warmed!

Specific heat

There's one more refinement left to make to heat capacity. Obviously, the amount of heat required to raise the temperature of a large quantity of a substance is greater than the amount required for a small amount of the same substance.

To control for the amount, we generally measure and report heat capacities as specific heat, the heat capacity per unit mass.

Specific heats of a great many substances have been measured under a variety of conditions. They are tabulated in books an on-line.

We generally choose units of J/gram or KJ/Kg. The specific heat of liquid water is 4.184 J/g, which is also 4.184 KJ/Kg. The calorie is a unit of heat defined as the amount of heat required to raise the temperature of 1 cm3 of water by 1˚C.

Specific heat

Specific heat is the heat capacity per unit mass.

The specific heat of water is 1 cal/g˚C = 4.184J/g˚C

Calculating Heat and Temperature Changes

The heat, $q$, required to raise the temperature of a mass, $m$, of a substance by an amount $\Delta T$ is

$$q = mC \Delta T = mC (T_f - T_i)$$

where C is the specific heat and $T_f$ and $T_i$ are the final and initial temperatures.

The slope of a graph of temperature vs. heat added to a unit mass is just 1/C.

Using this formula, it's relatively easy to calculate heat added, final or initial temperature or the specific heat itself (that's how it's measured) if the other variables are known.

Heat added to achieve a temperature change

The heat q added or evolved for a temperature change of a mass m of a substance with specific heat C is

$$q = mC \Delta T = mC (T_f - T_i)$$

The units of specific heat are usually   $\frac{J}{\text{mol}\cdot K}$   or   $\frac{J}{g \cdot K}$. Remember that it's OK to swap ˚C for K because the size of the Celsius degree and the Kelvin are the same. Note: You might not be familiar with the "mol" part — that's from chemistry, the mole, an indispensible unit in that field.

Example 1

Calculate the amount of heat (in Joules) required to change the temperature of 1 Liter of water (1 L = 1 Kg) from 20˚C to 37˚C.

The specific heat capacity (C) of water is 4.184 J/g˚C (or J/g·K — as long we work with Celsius degrees or Kelvins, the $\Delta T$ will be the same because the size of the two are the same. It's Fahrenheit that's a smaller-sized degree). The equation we need is:

$$q = m C \Delta T$$

Plugging in 1000 g for the mass of 1 L of water (the gram is defined as the mass of 1 mL of water), and the temperature change (37˚C - 20˚C), we get:

$$= (1000 \, g) \left( 4.185 \frac{J}{g\cdot ˚C} \right) (37 - 20)˚C$$

The result is

$$= 71,128 \; J = \bf 71 \; KJ$$

When the number of Joules of energy runs over 1,000, we generally express the amount in KiloJoules (KJ) in order to simplify the number.

Practice problems

(Use the table below to look up missing specific heats.)

1. How much heat (in Joules) does it take to raise the temperature of 100 g of H2O from 22˚C to 98˚C?
2. If it takes 640 J of heat energy to increase the temperature of 100 g of a substance by 25˚C (without changing its phase), calculate the specific heat of the substance.
3. If 80 J of heat are added to 100 ml of ethanol [density (ρ) = 789 Kg·m-3] initially at 10˚C, calculate the final temperature of the sample.

Heat (enthalpy) of phase change

... or, what if we heat or cool through a phase-change temperature

Phase changes are a big source or sink of heat. Here, for example, is the heating curve of water.

It shows the rise in temperature as heat is added at a constant rate to water. Here's what's going on in regions A-E:

A. Heat is added to solid water (ice) below 0˚C, and its temperature rises at a constant rate.

B. Solid ice is melted to liquid water. During the addition of the latent heat of fusion $(\Delta H_f)$, no temperature rise is observed, but hydrogen bonds holding the ice together break.

C. Heat is added to liquid water above 0˚C, and its temperature rises at a constant rate until the boiling point at 100˚C.

D. Water at 100˚C absorbs a great deal of heat energy at 100˚C as it undergoes a phase transition from liquid to gas. This is the latent heat of vaporization, $(\Delta H_v)$, the energy it takes for water to have no more cohesive force.

E. Finally, gaseous water above 100˚C absorbs heat, increasing its temperature at a constant rate. Water has no more phase transitions after this.

The relatively large attractive intermolecular forces between water molecules gives water very high heats of fusion and vaporization. Compared to most other substances, it takes a large amount of heat to melt water ice and to boil or evaporate water.

Enthalpies of fusion and vaporization are tabulated and can be looked up. The Wikipedia page of a compound is usually a good place to find them. Below we'll do an example of a heat calculation as the temperature of a substance rises through a phase change.


Cohesive forces

Cohesive forces are forces that hold a substance together. When water hits a waxy or hydrophobic surface, it forms small sphere-like drops – "beads." These beads of water minimize the contact with the surface and with the air, and maximize the contact of water with itself. Liquid water is very cohesive. It forms intermittent, but relatively strong bonds with itself.

Other substances like CO2 lack such strong intermolecular attractions, and don't form liquids or solids unless very cold or at very high pressure.

Heat of phase transition

The heat absorbed or released upon a phase transition is calculated by multiplying the enthalpy of vaporization, $(\Delta H_v)$, or the enthalpy of fusion, $(\Delta H_f)$ by the number of moles of substance:

$$ \begin{align} q &= m \, \Delta H_f \\[5pt] q &= m \, \Delta H_v \end{align}$$

The enthalpy of fusion is often called the "latent heat of fusion" and the enthalpy of vaporization is often called the "latent heat of vaporization."

The units of $(\Delta H_f)$ and $(\Delta H_v)$ are Joules/mole (J·mol-1) or J/g (J·g-1).

Example 2

Calculate the amount of heat needed to heat 18 g ice from -20˚C to water at 25˚C.

Solution: There is a phase transition of water in this temperature range, so this problem will comprise three steps:

  1. Raise the temperature of ice from -20˚C to the melting point, 0˚C, using the specific heat of ice, C = 2.010 J·g-1K-1.

  2. Convert ice to water at 0˚C, using the molar enthalpy of fusion, ΔHf = 333.5 J·g-1.

  3. Raise the temperature of liquid water from 0˚C to 25˚C, using the specific heat of water, C = 4.184 J·g-1K-1
  4. .

Here are the calculations for each of our steps:

Step 1:The amount of heat required to raise the temperature of ice (before it melts) by 20˚C is:

$$ \begin{align} q &= m C \Delta T \\ &= (18 \, g) \left( 2.01 \frac{J}{g\cdot K} \right)(273 \, K - 263 \, K) \\ &= \bf 723.6 \; J \end{align}$$

Note that we've converted the Celsius temperatures to Kelvin.

Step 2: The amount of heat required to melt 18 g of ice is:

$$ \begin{align} q &= m \Delta H_f \\ &= (18 \, g) \left( 2.01 \frac{J}{g} \right) \\ &= 36.18 \; J \end{align}$$

Step 3: The amount of heat required to raise the temperature of liquid water by 25˚C is:

$$ \begin{align} q &= m C \Delta T \\ &= (18 \, g) \left( 4.184 \frac{J}{g\cdot K} \right)(298 \, K - 273 \, K) \\ &= \bf 1882.8 \; J \end{align}$$

Adding all of these energies up, we get the total, q = 2642 J,

Now let's compare this to a similar calculation, but this time we'll heat liquid water through its boiling point to a gas.

Example 3

Calculate the heat required to raise the temperature of 18 g of liquid water at Ti = 80˚C to steam at 125˚C.

Solution: This is also a three-step problem, but this time we're vaproizing water. Here are the steps:

  1. Raise the temperature of liquid water from 80˚C to the boiling point, 100˚C, using the specific heat of water, C = 4.184 J·g-1K-1.

  2. Convert water to steam (gaseous water) at 100˚C, using the molar enthalpy of vaporization, ΔHf = 2258 J·g-1.

  3. Raise the temperature of steam from 100˚C to 125˚C, using the specific heat of steam, C = 2.010 J·g-1K-1
  4. .

Here are the calculations for each of our steps:

Step 1:The amount of heat required to raise the temperature of water (before it vaporizes) from 80˚C to 100˚C is:

$$ \begin{align} q &= m C \Delta T \\ &= (18 \, g)\left(4.184 \frac{J}{mol \, ˚C}\right)(100 - 80)˚C \\ &= 1505 \; J \end{align}$$

Step 2: Convert the liquid water to steam at 100˚C. Here we use the heat of vaporization of water:

$$ \begin{align} q &= m \Delta H_v = (18 \, g)\left( 2258 \, \frac{J}{g} \right) \\ &= 180,640 \; J \end{align}$$

Step 3: Finally, we calculate the amount of heat required to change the temperature of the 80 g of steam from 100˚C to 125˚C:

$$ \begin{align} q &= m C \Delta T \\ &= (18 \, g)\left( 2.010 \, \frac{J}{mol ˚C} \right)(125 - 100)˚C \\ &= 904 \; J \end{align}$$

Almost there. The last step is to add all of these energies together:

$$ \begin{align} q_{total} &= 1505 \, J + 180,640 \, J + 904 \, J \\ &= 183,050 \; J = 183 \; KJ \end{align}$$

Notice that the largest contribution to this energy, by far, is in evaporating the water — changing it from liquid to gas. This process takes a tremendous amount of energy, and that energy accounts for the large amount of energy it takes to boil water to make steam in electrical generating plants of all kinds (including nuclear), and for the efficient means humans have of cooling our bodies: perspiration.

Practice problems

(Use the table below to look up missing specific heats; heats of fusion or vaporization are given in the problems.)

1. How much heat (in Joules) does it take to change 120g of ice at -10˚C to water at 37˚C? (ΔHf = 334 KJ·Kg-1)? Note that this is a three-step problem: First heat the ice to 0˚C, then convert all 120g to liquid, then raise the temperature of the water to 37˚C (human body temp.).
2. How much heat is released when 1 Kg of steam at 300˚C is cooled to liquid at 40˚C? (ΔHv = 2260 KJ·Kg-1)
3. Is there enough heat in 100 ml of water at 25˚C to completely melt 50g of ice at 0˚C? (ΔHf = 334 KJ·Kg-1)

Specific heats of selected compounds

Compound Spec. Heat
Water ice (H2O) 2.11
Water liquid 4.184
Water gas 2.08
Aluminum (s) (Al) 0.897
Copper (s) (Cu) 0.385
Iron (s) (Fe) 0.450
Lead (s) (Pb) 0.129

Compound Spec. Heat
Methanol (l) (CH3OH) 2.14
Ethanol (l) (C2H5OH) 2.44
Ethylene glycol (l) (C2O2H6) 2.2
Hydrogen (g) (H2) 14.267
Benzene (l) (C6H6) 1.750
wood (typical) 1.674
glass (typical) 0.867

Refinement: Cp and Cv

Until now, in order to keep things simple, I've been referring to specific heat as C. But heat capacity (specific heat if it's per mole or per gram) can change depending on whether the thermodynamic variables of pressure or temperature are held constant during heating or cooling.

Generally in the laboratory we work at constant pressure — atmospheric pressure, so Cp is the most commonly used specific heat.

In systems held at constant volume, such as in a gas-cylinder, where pressure can change but volume can't, we use the heat capacity at constant volume, Cv.

Relative heat capacities — Cp of water is high

One of the most interesting and important things about water is its high specific heat compared to other molecules of its size. It is anomalously high, and that feature is thought to be coupled in a few ways to the origin of life on Earth. The uniqueness of water in this sense is partly why we look for water on other planets and moons when searching for signs of extraterrestrial life.

A conservation of energy example

10 g of ice at -10˚C is added to 100 g of water at 50˚C. Calculate the final temperature of this mixture once the ice has melted and the temperature has equilibrated. Assume no exchange of heat with the surroundings (adiabatic conditions).

This problem is solved using the principle of conservation of energy. If no heat is lost to the surroundings, and none is allowed to enter the system, then all of the heat used to melt the ice and heat the resulting liquid must come from the 50˚C water.

Heat lost by water = heat gained by ice

One common problem with working out problems like this is bookkeeping — keeping track of the signs for heat loss and heat gain. You'll have better results if you work it like this:

$$q_{lost \; by \; water} + q_{gained \; by \; ice} = 0$$

The heat lost by the water is

$$ \begin{align} q &= m C \Delta T \\ &= (100 \, g) \left( 4.184 \frac{J}{g ˚C} \right)(T_f - 50)˚C \\ &= 418.4 \, T_f - 20,920 \end{align}$$

where Tf is the unknown final temperature.

The heat gained by the ice includes the heat to bring the temperature of the solid to 0˚C, the heat required to melt it, and the heat to bring it to the final temperature, Tf.

Those are:

$$ \begin{align} q &= m C_{ice} \Delta T + m\Delta H_v + m C_{liq} \Delta T \\ \\ &= (10 \, g)(2.11 \, J/g˚C)(0 - (-10))˚C \\ \\ &\; \; \; + (10 \, g)(334 J/g) \\ \\ &\; \; \; + (10 \, g)(4.184 \, J/g˚C)(T_f - 0)˚C \\ \\ &= 211 + 3,340 + 41.84 \, T_f \end{align}$$

Now if we add the heat lost by the water to the heat gained by the ice and set the sum equal to zero, we get:

$$418.4 T_f - 20920 + 211 + 3340 + 41.84 T_f = 0$$

Then we gather like terms and solve for the final temperature:

$$ \begin{align} 460.24 \, T_f &= 20920 - 211 - 3340 \\ \bf T_f &= 37.7 ˚C \end{align}$$

This temperature makes sense. The water was cooled as it melted the ice, but not too much because there's 10 times as much water by mass.

Video examples

1. Heat required to raise the temperature of water

Minutes of your life: 2:08
A simple example

2. Heat required to raise the temperature of water, convert it to steam, then heat the steam

Minutes of your life: 2:32

3. Heat required to convert ice at -20˚C to steam at 150˚C, a five-step process

Minutes of your life: 3:09



An adiabatic system is one in which no heat enters or leaves.

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