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.

#### Rotation

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.

### 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
J/g·K
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

Compound Spec. Heat
J/g·K
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

A simple example

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

X

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

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