#### xaktly | Thermodynamics

Heat capacity The sections on thermodynamics ought to be read and worked through in roughly this order:

• Heat The atomic and molecular nature of heat
• Temperature About measuring temperatures, and temperature scales
• Heat transfer How heat moves from one place to another
• Heat capacity (this page) Substances differ in their abilities to absorb heat and reflect it as a temperature rise.
• Thermodynamics The laws of thermodynamics
• Enthalpy Heat content of a system
• Entropy A measure of the disorder of the universe, or a part of it
• Gibbs energy Energy at constant temperature & pressure

### 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. ### 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, each container represents a degree of freedom. The situations for a 3-atom and a 10-atom

molecule are shown. If the same total amount of heat energy is added to each molecule, the 3-atom molecule ends up with more energy in its translational degrees of freedom. Because the 10-atom molecule has more vibrational modes in which to store kinetic energy, less is available to go into the translational modes, and it is mostly the translational energy that we measure as temperature. ### 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 ΔT is

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

where C is the specific heat and Tf and Ti 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 J/mol·K (J·mol-1K-1) or J/g·K (J·g-1·K-1). Remember that it's OK to swap ˚C for K because the size of the Celsius degree and the Kelvin are the same.

### 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 ΔT 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 (ΔHf), 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, ΔHv, 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.

X

### 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, ΔHv, or the enthalpy of fusion, ΔHf 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 ΔHf and ΔHv 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
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  