# Photoelectric effect

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problems 5.3 – 5.4.

The photoelectric effect occurs when a beam of light (or any electromagnetic radiation) is fired at a metal, causing electrons to be emitted from the metal. Light with a wavelength longer than a critical wavelength fails to dislodge any electrons. Below the critical wavelength, electrons are emitted in proportion to the intensity of the light. However, the energy of the emitted electrons doesn’t increase unless we use light with shorter wavelengths.

As Einstein first proposed, the cause of the photoelectric effect is that light is quantized into photons, and the energy of a photon is related to its frequence ${\nu}$ by Planck’s formula

$\displaystyle E=h\nu=\frac{hc}{\lambda} \ \ \ \ \ (1)$

The product ${hc}$ has the value

 $\displaystyle hc$ $\displaystyle =$ $\displaystyle \left(6.62606957\times10^{-34}\mbox{m}^{2}\mbox{kg s}^{-1}\right)\left(2.99792458\times10^{8}\mbox{m s}^{-1}\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.986446\times10^{-25}\mbox{m}^{3}\mbox{kg s}^{-2}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.986446\times10^{-25}\mbox{J m} \ \ \ \ \ (4)$

As 1 eV is ${1.602\times10^{-19}\mbox{ J}}$ we can write this as

 $\displaystyle hc$ $\displaystyle =$ $\displaystyle 1.239979\times10^{-6}\mbox{ eV m}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle 1240\mbox{ eV nm} \ \ \ \ \ (6)$

The minimum binding energy at which an electron can be dislodged from a metal is known as its work function, denoted ${\phi}$, with some electrons having a larger binding energy. The maximum kinetic energy of an ejected electron is therefore

$\displaystyle K_{max}=\frac{hc}{\lambda}-\phi \ \ \ \ \ (7)$

The dust in interstellar clouds experiences the photoelectric effect when it is irradiated with starlight. Suppose an ultraviolet photon with ${\lambda\approx100\mbox{ nm}}$ strikes a dust grain and the ejected electron has a kinetic energy of around 5 eV. Then the work function of the dust grain is about

$\displaystyle \phi=\frac{1240\mbox{ eV nm}}{100\mbox{ nm}}-5\mbox{ eV}=7.4\mbox{ eV} \ \ \ \ \ (8)$

# Barnard’s star: distance and velocity

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problem 5.1.

By combining measurements of parallax, proper motion (the apparent motion of a star across the sky, usually measured in seconds of arc per year) and the Doppler shifting of spectral lines, we can work out a star’s actual velocity through space. A good example is Barnard’s star, a magnitude 9.5 red dwarf star in the constellation Ophiucus, which has the largest known proper motion of any star.

The parallax is ${p=0.54901\mbox{"}}$ so its distance is

$\displaystyle d=\frac{1}{p}=1.82146\mbox{ pc}=5.9408\mbox{ ly}=5.62032\times10^{16}\mbox{ m} \ \ \ \ \ (1)$

Its proper motion is ${\mu=10.3577\mbox{" yr}^{-1}}$ which works out to

 $\displaystyle \mu$ $\displaystyle =$ $\displaystyle 10.3577\frac{\pi}{3600\times180}=5.021555\times10^{-5}\mbox{ rad yr}^{-1}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5.021555\times10^{-5}}{3600\times24\times365.25}=1.59123\times10^{-12}\mbox{ rad s}^{-1} \ \ \ \ \ (3)$

At the distance of Barnard’s star, this gives a transverse velocity component of

$\displaystyle v_{\theta}=d\mu=8.9432\times10^{4}\mbox{ m s}^{-1} \ \ \ \ \ (4)$

In the spectrum, the hydrogen alpha (${\mbox{H}\alpha}$) line is observed at a wavelength of 656.034 nm, and the rest ${\mbox{H}\alpha}$ wavelength is 656.28 nm. As the star’s wavelength is shorter than the rest wavelength, it is blue-shifted and Barnard’s star is moving towards us. The (relativistic) Doppler shift formula for a speed ${v}$ that is positive for approaching objects is

$\displaystyle \bar{\lambda}=\lambda\sqrt{\frac{1-v/c}{1+v/c}} \ \ \ \ \ (5)$

For non-relativistic speeds, ${v\ll c}$ and we can Taylor-expand this formula to get

$\displaystyle \bar{\lambda}=\lambda-\frac{\lambda}{c}v+\mathcal{O}\left(\frac{v^{2}}{c^{2}}\right) \ \ \ \ \ (6)$

Keeping only terms first order in ${v/c}$ we get

$\displaystyle v=\frac{\lambda-\bar{\lambda}}{\lambda}c \ \ \ \ \ (7)$

For Barnard’s star this gives a radial velocity of

 $\displaystyle v_{r}$ $\displaystyle =$ $\displaystyle \frac{656.28-656.034}{656.28}\left(2.99792\times10^{8}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.12374\times10^{5}\mbox{ m s}^{-1} \ \ \ \ \ (9)$

The actual speed of Barnard’s star through space, relative to Earth, is

$\displaystyle v=\sqrt{v_{r}^{2}+v_{\theta}^{2}}=1.4362\times10^{5}\mbox{ m s}^{-1} \ \ \ \ \ (10)$

# Diffraction gratings in spectroscopy

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 5, Problem 5.2.

Jenkins, Francis A. & White, Harvey E. (1957), Fundamentals of Optics, 3rd Edition; McGraw-Hill – Chapter 17.

One of the main sources of information about celestial objects is their spectrum. In the case of light from a star, the spectrum can tell us a lot about the composition of the star, and the Doppler shift can tell us the star’s radial velocity.

Getting a detailed spectrum of a star is somewhat more involved than just passing its light through a prism, as Newton famously did to show that white light is composed of the colours of the rainbow. In practice, a diffraction grating is used to split starlight into its constituent spectral lines. A complete treatment of the physics of diffraction gratings is beyond the scope of this post, so if you’re interested I’ll refer you to any good textbook on optics. [My own reference, the book by Jenkins and White mentioned at the top, is quite dated now, but I doubt that the physics has changed in the intervening years.] I’ll give a summary of the key concepts here.

A diffraction grating is basically a screen with a large number of closely-spaced parallel slits cut in it (actual gratings are usually made of glass with very fine lines ruled on them, but the principle is the same). Consider a grating with only 2 slits separated by a distance ${d}$. Monochromatic light of wavelength ${\lambda}$ is shone onto the grating and diffracts through each of the slits. [Diffraction is the process by which a light wave spreads out after passing through a narrow gap. The same effect can be seen with water waves as they hit a narrow gap in a barrier; waves spread out in a semi-circular pattern beyond the gap.]

Now look at the light rays that leave the slits at an angle ${\theta}$ to the normal to the plane containing the slits. By drawing a diagram (see Fig 3.3 in Carroll & Ostlie, although in practice the two light rays leaving the slits are parallel and are focussed onto the detecting screen by a convex lens or concave mirror) we can see that the path difference between the two rays is ${d\sin\theta}$. If this path difference is an integral multiple of the wavelength, the two rays will reinforce each other and we’ll see a bright fringe at that angle. On the other hand, if the path difference is an integral multiple of wavelengths plus half a wavelength, the two rays will destructively interfere, cancelling each other out, and we’ll see a dark fringe. That is, the condition for bright fringes is

$\displaystyle d\sin\theta=n\lambda \ \ \ \ \ (1)$

for ${n}$ a non-negative integer.

Now suppose we have a diffraction grating with ${N}$ slits. The same condition applies for light fringes, but because we’re now adding up ${N}$ rays instead of just 2, the intensity of the bright fringes is larger, and is in fact proportional to ${N^{2}}$. The actual formula for the intensity is

$\displaystyle I=a^{2}\frac{\sin^{2}N\gamma}{\sin^{2}\gamma} \ \ \ \ \ (2)$

where ${a}$ is a constant, derived from the intensity passing through a single slit, and

$\displaystyle \gamma=\frac{\pi d\sin\theta}{\lambda} \ \ \ \ \ (3)$

The bright fringes occur when ${\gamma=n\pi}$. This can be seen by finding the limit:

$\displaystyle \lim_{\gamma\rightarrow n\pi}\frac{\sin N\gamma}{\sin\gamma} \ \ \ \ \ (4)$

This limit can be found using l’Hôpital’s rule from calculus, which states that for two functions ${f}$ and ${g}$ where ${f\left(x_{0}\right)=g\left(x_{0}\right)=0}$:

$\displaystyle \lim_{x\rightarrow x_{0}}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x\rightarrow x_{0}}\frac{f'\left(x\right)}{g'\left(x\right)} \ \ \ \ \ (5)$

Therefore

 $\displaystyle \lim_{\gamma\rightarrow n\pi}\frac{\sin N\gamma}{\sin\gamma}$ $\displaystyle =$ $\displaystyle \lim_{\gamma\rightarrow n\pi}\frac{N\cos N\gamma}{\cos\gamma}=\pm N\ \ \ \ \ (6)$ $\displaystyle \lim_{\gamma\rightarrow n\pi}\frac{\sin^{2}N\gamma}{\sin^{2}\gamma}$ $\displaystyle =$ $\displaystyle N^{2} \ \ \ \ \ (7)$

Because of the ${N}$ in the upper sine, however, the numerator has many more zeroes than the denominator, meaning that whenever ${N\gamma=m\pi}$ but ${m}$ is not a multiple of ${N}$, ${I=0}$. Between any two bright fringes there will therefore be ${N-1}$ dark fringes and ${N-2}$ secondary ‘brightish’ fringes between these dark fringes. In practice, these secondary fringes are much fainter than the primary bright fringes and for large ${N}$ they are effectively invisible.

It turns out that the number of slits ${N}$ also determines the angular width of each primary maximum, according to

$\displaystyle \Delta\theta_{width}=\frac{2\lambda}{Nd\cos\theta} \ \ \ \ \ (8)$

That is, the more slits, the sharper the spectral line for a given wavelength. Looked at another way, the smallest wavelength difference ${\Delta\lambda}$ that can be resolved is

$\displaystyle \Delta\lambda=\frac{\lambda}{nN} \ \ \ \ \ (9)$

Not only does increasing the number slits increase the resolving power of the grating, but looking a higher order (larger ${n}$) lines allows greater resolution.

Example The sodium D lines are commonly found in stellar spectra and have wavelengths of ${\lambda_{1}=588.997\mbox{ nm}}$ and ${\lambda_{2}=589.594\mbox{ nm}}$. If light containing these two lines shines on a grating with 300 lines per millimetre, then

$\displaystyle d=\frac{10^{-3}}{300}=3.33\times10^{-6}\mbox{ m} \ \ \ \ \ (10)$

From 1 with ${n=2}$ (second-order spectra) we have

 $\displaystyle \sin\theta_{1}$ $\displaystyle =$ $\displaystyle \frac{2\left(588.997\times10^{-9}\right)}{3.33\times10^{-6}}=0.3533982\ \ \ \ \ (11)$ $\displaystyle \theta_{1}$ $\displaystyle =$ $\displaystyle 20.69531^{\circ}\ \ \ \ \ (12)$ $\displaystyle \sin\theta_{2}$ $\displaystyle =$ $\displaystyle \frac{2\left(589.594\times10^{-9}\right)}{3.33\times10^{-6}}=0.3537564\ \ \ \ \ (13)$ $\displaystyle \theta_{2}$ $\displaystyle =$ $\displaystyle 20.71725^{\circ} \ \ \ \ \ (14)$

The angle between the two lines is therefore

$\displaystyle \Delta\theta_{Na}=\theta_{2}-\theta_{1}=0.02194^{\circ} \ \ \ \ \ (15)$

From 9 we can work out how many lines need to be illuminated in order for these 2 lines to be resolved in second-order. For the wavelength in the formula, we’ll use the average of the two wavelengths of the sodium D lines.

 $\displaystyle N$ $\displaystyle =$ $\displaystyle \frac{\lambda}{n\Delta\lambda}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{589.2955\times10^{-9}}{2\times0.597\times10^{-9}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 494 \ \ \ \ \ (18)$

Thus if only 2 mm of the grating were illuminated, we could resolve the lines, at least at second-order.

# Negative heat capacity in gravitational systems; estimating the Sun’s temperature

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.55.

Suppose we have two identical masses ${m}$ in a circular orbit of radius ${r}$ about their centre of mass. By equating centripetal and gravitational forces, we have

$\displaystyle \frac{mv^{2}}{r}=\frac{Gm^{2}}{4r^{2}} \ \ \ \ \ (1)$

The total kinetic and potential energies of the system are

 $\displaystyle K$ $\displaystyle =$ $\displaystyle 2\times\frac{mv^{2}}{2}=mv^{2}\ \ \ \ \ (2)$ $\displaystyle V$ $\displaystyle =$ $\displaystyle -\frac{Gm^{2}}{2r} \ \ \ \ \ (3)$

Therefore from 1

$\displaystyle K=\frac{Gm^{2}}{4r}=-\frac{V}{2} \ \ \ \ \ (4)$

This is a special case of the virial theorem for gravitational orbits, which gives a relation between the average potential and kinetic energies:

$\displaystyle \left\langle V\right\rangle =-2\left\langle K\right\rangle \ \ \ \ \ (5)$

The total energy of a gravitational system is therefore

$\displaystyle U=\left\langle K\right\rangle +\left\langle V\right\rangle =-\left\langle K\right\rangle \ \ \ \ \ (6)$

The total energy is negative, which indicates that it is gravitationally bound and won’t fly apart with time. However, this has a curious consequence in that, if we increase the energy of the system by an amount such that ${U}$ is still negative, the kinetic energy must actually decrease.

For a system such as a star that is bound by gravitational forces and contains many particles, we can apply the equipartition theorem. Because of the high temperature within a star, the atomic nuclei become dissociated from their electrons, so the only degrees of freedom available to each particle are the translational degrees of freedom, meaning that the average kinetic energy of each particle is ${\frac{3}{2}kT}$. Therefore, for a system with ${N}$ particles

$\displaystyle U=-K=-\frac{3}{2}NkT \ \ \ \ \ (7)$

This means that the heat capacity is actually negative:

$\displaystyle C=\frac{\partial U}{\partial T}=-\frac{3}{2}Nk \ \ \ \ \ (8)$

[Note that for a star, ${C=C_{V}=C_{P}}$ since even if the star’s volume changes as energy is added to it, no expansion work is done since the star expands into a vacuum where the pressure is zero. However, gravitational work is done, so I’m not sure how that will affect the formula.]

At this point, Schroeder asks us to use dimensional analysis to get a formula for the potential energy of a star. I’m not quite sure what he means, but we can get an estimate of the potential energy as follows.

Suppose the star has a uniform density ${\rho}$ and a radius ${R}$. Then the potential energy of a thin shell of the star at radius ${r}$ is determined by the portion of the star inside this radius (the parts of the star outside ${r}$ have no net force on anything inside; the proof of this is similar to that in electrostatics using Gauss’s law since gravity is also an inverse square force). The portion of the star inside the radius ${r}$ acts as a point mass ${M_{r}}$ at the centre of the star. That is

 $\displaystyle dV$ $\displaystyle =$ $\displaystyle -\frac{GM_{r}\;dm}{r}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{G}{r}\left(\frac{4}{3}\pi r^{3}\rho\right)\left(4\pi r^{2}\rho\;dr\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{16\pi^{2}G}{3}\rho^{2}r^{4}dr \ \ \ \ \ (11)$

We can integrate this to get ${V}$:

 $\displaystyle V$ $\displaystyle =$ $\displaystyle -\frac{16\pi^{2}G}{3}\rho^{2}\int_{0}^{R}r^{4}dr\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{16\pi^{2}G}{15}\rho^{2}R^{5}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3G}{5R}\left(\frac{4}{3}\pi R^{3}\rho\right)^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{3GM^{2}}{5R} \ \ \ \ \ (15)$

The average kinetic energy is therefore

$\displaystyle \left\langle K\right\rangle =-\frac{1}{2}V=\frac{3GM^{2}}{10R}=\frac{3}{2}NkT \ \ \ \ \ (16)$

We can use this to get an estimate of the temperature of the Sun, whose mass is ${2\times10^{30}\mbox{ kg}}$ and radius is ${R=7\times10^{8}\mbox{ m}}$. Taking the Sun to be composed of equal numbers of bare protons and electrons (and neglecting the mass of the electron compared that of the proton), we can estimate ${N}$:

$\displaystyle N=\frac{2M}{m_{p}} \ \ \ \ \ (17)$

The temperature estimate is

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{GM^{2}}{5RNk}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{GMm_{p}}{10Rk}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(6.67\times10^{-11}\right)\left(2\times10^{30}\right)\left(1.67\times10^{-27}\right)}{10\left(7\times10^{8}\right)\left(1.38\times10^{-23}\right)}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.31\times10^{6}\mbox{ K} \ \ \ \ \ (21)$

This is within currently accepted values for the interior of the sun. The core of the Sun is estimated to be around ${15\times10^{6}\mbox{ K}}$, but drops to between ${2\times10^{6}\mbox{ K}}$ and ${7\times10^{6}\mbox{ K}}$ as we get beyond half the Sun’s radius from the core.

# Thermodynamics of hiking

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.54.

Suppose a 60 kg hiker wishes to climb a 1500 m tall mountain. If she wants to provide the energy for the climb by eating only corn flakes, how much would she need to eat? We’ll make a lot of oversimplifying assumptions to get an estimate. First, suppose that the only work expended on the hike is the energy required to climb 1500 m. The work is equal to the potential energy gained, so

 $\displaystyle W$ $\displaystyle =$ $\displaystyle mgh\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 60\times9.8\times1500\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8.82\times10^{5}\mbox{ J}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 211\mbox{ kcal} \ \ \ \ \ (4)$

Assuming that she can convert 25% of the food’s energy to work, she will therefore need to consume 844 kcal in corn flakes. The enthalpy change in ‘burning’ corn flakes is 100 kcal per 28 g, so she will need to eat

$\displaystyle \frac{844}{100}\times28=236\mbox{ g} \ \ \ \ \ (5)$

If the remaining 75% of the energy is retained as heat within the hiker’s body, this will cause her temperature to rise. Taking the body to be water (which it is, mostly), 1 kcal of heat will raise 1 kg of water by 1 K, so from retaining the ${844-211=633\mbox{ kcal}}$ of heat, her body temperature will rise by

$\displaystyle \Delta T=\frac{633}{60}=10.55^{\circ}\mbox{ C} \ \ \ \ \ (6)$

That would give her quite a fever, so it’s not surprising that most of the heat is lost through the skin by evaporating sweat. Given that the latent heat of vaporization of water at ${25^{\circ}\mbox{ C}}$ is ${580\mbox{ cal g}^{-1}}$, this excess heat will evaporate an amount of water given by

$\displaystyle \frac{633\times10^{3}}{580}=1091\mbox{ g} \ \ \ \ \ (7)$

To replace this, she’d need to drink just over a litre of water.

# Enthalpy: a few examples

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.51 – 1.53.

Here are a few more examples of enthalpy calculations.

Example 1 The enthalpy for the reaction in which glucose combines with oxygen to produce carbon dioxide and water is given in the appendix to Schroeder’s book as ${\Delta H=-1273\mbox{ kJ mol}^{-1}}$. The reaction, which is the source of most energy in mammals, is

$\displaystyle \mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6}+6\mbox{O}_{2}\rightarrow6\mbox{CO}_{2}+6\mbox{H}_{2}\mbox{O} \ \ \ \ \ (1)$

The net enthalpy change for this reaction is

 $\displaystyle \Delta H$ $\displaystyle =$ $\displaystyle 6\Delta H_{CO_{2}}+6\Delta H_{H_{2}O}-\Delta H_{glucose}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -6\left(393.52+285.83\right)+1273\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2803.1\mbox{ kJ mol}^{-1}=670\mbox{ kcal mol}^{-1} \ \ \ \ \ (4)$

Incidentally, one mole of glucose weighs about ${6\times12+12+6\times16=180\mbox{ g}}$ so this works out to about 372 kcal per 100 g. The food ‘calorie’ is actually a kilocalorie so this means that glucose contains about 372 food calories per 100 g.

Example 2 The combustion of one litre of petrol (or ‘gasoline’ in North America) produces an enthalpy change of -8158 kcal. The current cost of petrol here in Scotland is around £1.15 per litre (yes, that’s a lot more than in North America), so this works out to

$\displaystyle \frac{115\mbox{ pence}}{8158\times10^{3}\mbox{ cal}}=1.4\times10^{-5}\mbox{ pence cal}^{-1} \ \ \ \ \ (5)$

To compare this with human (as opposed to car) food, the ‘combustion’ of corn flakes in the body produces ${\Delta H=-100\mbox{ kcal}}$ per 28 grams. The current cost of a 750 g box of corn flakes here is £1.98 so the cost per calorie is

$\displaystyle \frac{198}{\left(750/28\right)\times10^{5}}=7.4\times10^{-5}\mbox{ pence cal}^{-1} \ \ \ \ \ (6)$

So per calorie, corn flakes are actually about 5 times as expensive as petrol. In North America, where gasoline is a lot cheaper, the difference is probably even greater. Of course, you wouldn’t want to drink petrol.

Example 3 The enthalpy of formation of atomic hydrogen gas from molecular hydrogen ${\mbox{H}_{2}}$ is ${\Delta H=+217.97\mbox{ kJ}}$ per mole of atomic hydrogen. Since this is positive, energy must be added to molecular hydrogen to dissociate it into atomic hydrogen. To get the enthalpy change per molecule, we double this quantity to get the enthalpy change per mole of molecular hydrogen, then divide by Avogradro’s number:

$\displaystyle \Delta H=\frac{2\times217.97\times10^{3}}{6.02\times10^{23}}=7.24\times10^{-19}\mbox{ J} \ \ \ \ \ (7)$

One electron volt is ${1.6\times10^{-19}\mbox{ J}}$ so this is

$\displaystyle \Delta H=\frac{7.24\times10^{-19}}{1.6\times10^{-19}}=4.53\mbox{ eV} \ \ \ \ \ (8)$

# Enthalpy in chemical reactions

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.50.

We can find how much heat is emitted or absorbed by a chemical reaction by working with the enthalpy of the reactants and products. For example, consider the combustion of methane with oxygen at a constant temperature of ${298\mbox{ K}}$:

$\displaystyle \mbox{CH}_{4}+2\mbox{O}_{2}\rightarrow\mbox{CO}_{2}+2\mbox{H}_{2}\mbox{O} \ \ \ \ \ (1)$

Using the table at the back of Schroeder’s book, ${\Delta H}$ for the formation of methane from elemental carbon (solid) and hydrogen (gas) is

$\displaystyle \Delta H_{CH_{4}}=-74.81\mbox{ kJ mol}^{-1} \ \ \ \ \ (2)$

${\Delta H}$ for the dissociation of methane into its elements is therefore the negative of this.

Similarly

 $\displaystyle \Delta H_{CO_{2}}$ $\displaystyle =$ $\displaystyle -393.51\mbox{ kJ mol}^{-1}\ \ \ \ \ (3)$ $\displaystyle \Delta H_{H_{2}O}\left(\mbox{gas}\right)$ $\displaystyle =$ $\displaystyle -241.82\mbox{ kJ mol}^{-1} \ \ \ \ \ (4)$

The total enthalpy change for the creation of the products in reaction 1 from their elements is

$\displaystyle \Delta H_{prod}=-393.51-2\times241.82=-877.15\mbox{ kJ mol}^{-1} \ \ \ \ \ (5)$

${\Delta H}$ for the reaction as a whole is

$\displaystyle \Delta H=\Delta H_{prod}-\Delta H_{react}=-802.34\mbox{ kJ mol}^{-1} \ \ \ \ \ (6)$

We can see this as follows. Suppose the absolute enthalpy of the elemental components (carbon, hydrogen and oxygen) is ${H_{elem}}$ and of the reactants and products ${H_{reac}}$ and ${H_{prod}}$. Then

 $\displaystyle \Delta H_{reac}$ $\displaystyle =$ $\displaystyle H_{reac}-H_{elem}\ \ \ \ \ (7)$ $\displaystyle \Delta H_{prod}$ $\displaystyle =$ $\displaystyle H_{prod}-H_{elem}\ \ \ \ \ (8)$ $\displaystyle \Delta H$ $\displaystyle =$ $\displaystyle H_{prod}-H_{reac}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Delta H_{prod}-\Delta H_{reac} \ \ \ \ \ (10)$

In 1, if all 4 compounds are gases and the temperature is the same on both sides, there is no volume change, as there are 3 moles of gas both before and after the reaction. Therefore the entire enthalpy change is due to change in internal energy ${U}$ and, assuming no ‘other’ work is done, all this energy is emitted as heat. That is, for one mole of methane

$\displaystyle \Delta U=\Delta H=Q=-802.34\mbox{ kJ } \ \ \ \ \ (11)$

If the water is produced as liquid instead of vapour, then (from the table in Schroeder), ${\Delta H_{H_{2}O}=-285.83\mbox{ kJ mol}^{-1}}$ and

$\displaystyle \Delta H=-890.36\mbox{ kJ mol}^{-1} \ \ \ \ \ (12)$

This time, the final volume is ${\frac{1}{3}}$ of the initial volume, since the 2 moles of water has condensed out as liquid with negligible volume compared to the gases. Thus the atmosphere does work

 $\displaystyle -P\Delta V$ $\displaystyle =$ $\displaystyle -RT\Delta n\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(8.31\mbox{ J K}^{-1}\right)\left(298\mbox{ K}\right)\left(-2\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4.953\mbox{ kJ} \ \ \ \ \ (15)$

The change in internal energy is therefore found from

 $\displaystyle \Delta H$ $\displaystyle =$ $\displaystyle \Delta U+P\Delta V\ \ \ \ \ (16)$ $\displaystyle \Delta U$ $\displaystyle =$ $\displaystyle -890.36+4.953\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -885.41\mbox{ kJ} \ \ \ \ \ (18)$

The difference between this value and 6 should be the latent heat of vaporization at 298 K. An approximation for this is

$\displaystyle L\approx2500.8-2.36T_{C} \ \ \ \ \ (19)$

where ${T_{C}}$ is the temperature in centigrade, so ${T_{C}=25}$ here. This gives a value of ${L=2.442\mbox{ kJ g}^{-1}}$ so for 2 moles (around 36 g) of water, we have ${L=88\mbox{ kJ}}$. The difference from the calculations here is only about 83 kJ, but I’m not sure how accurate the various values and formulas are. At least it’s close.

As a final example, suppose the Sun with a mass of around ${2\times10^{33}\mbox{ g}}$ and luminosity of ${3.839\times10^{26}\mbox{ watts}}$ used the combustion of methane and oxygen as its energy source. The molar weights of methane and molecular oxygen around 16 g and 32 g so if the Sun were composed of one part methane to two parts oxygen (by molecular number), then the mass ratio is around methane:oxygen = 1 : 4. The number of moles of methane in the Sun is therefore

$\displaystyle n_{CH_{4}}=\frac{2\times10^{33}}{5\times16}=2.5\times10^{31}\mbox{ moles} \ \ \ \ \ (20)$

Assuming the water is produced as vapour, the Sun could produce a total energy of

$\displaystyle E=2.5\times10^{31}\times802.34=2\times10^{34}\mbox{ kJ} \ \ \ \ \ (21)$

so it would burn out after a time interval of

$\displaystyle t=\frac{2\times10^{34}\mbox{ kJ}}{3.839\times10^{23}\mbox{ kJ s}^{-1}}=5.21\times10^{10}\mbox{ s}=1651\mbox{ years} \ \ \ \ \ (22)$

We can be pretty sure the Sun’s source of power isn’t chemical reactions!

# Enthalpy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.49.

The enthalpy of a system with thermal energy ${U}$, pressure ${P}$ and volume ${V}$ is defined as

$\displaystyle H\equiv U+PV \ \ \ \ \ (1)$

It can be thought of as the energy required to create the system ‘from nothing’, since to do this, we must provide the thermal energy ${U}$ and push aside the atmosphere to create the volume ${V}$ in which to place the new system. Pushing aside the atmosphere (assumed to be at constant pressure ${P}$) requires work ${PV}$, so the total energy required to create the system is its thermal energy plus the energy to push the air out of the volume the system occupies.

Note that when I say the system is created ‘from nothing’, I’m not saying that the actual matter itself is created, since that would require providing the relativistic energy ${mc^{2}}$, which is not part of the thermal energy.

The word ‘enthalpy’ is derived from a Greek word meaning ‘to put heat into’ and the symbol ${H}$ is based on ‘heat’. In practice, it is usually changes in enthalpy that are measured; the absolute enthalpy doesn’t appear in experiments.

Enthalpy is a handy quantity in some calculations since it isolates the heat transfer from the work done. To see this, recall that the energy of a system is

$\displaystyle U=Q-PV+W_{other} \ \ \ \ \ (2)$

That is, the energy is the heat transferred into the system plus the compression or expansion work done on the system (which is ${-PV}$ here, since the system does work ${PV}$ on its surroundings as it is created), plus any other work (from chemical reactions, for example) done on the system. As a result

$\displaystyle H=Q+W_{other} \ \ \ \ \ (3)$

and if no ‘other’ work is done,

$\displaystyle H=Q \ \ \ \ \ (4)$

That is, enthalpy is just the heat added to the system, separated from the compression or expansion work done.

Example The enthalpy change for the reaction where one mole of hydrogen molecules combines with half a mole of oxygen molecues to produce water is ${\Delta H=-2.86\times10^{5}\mbox{ J}}$, assuming that the reactant gases and the resulting water are both at ${25^{\circ}\mbox{ C}}$ and 1 atm pressure. As this is an explosive reaction producing a lot of heat, the water will initially be in the form of vapour, so it will have to give off heat to condense into a liquid and then cool off to room temperature. This results in a decrease in the thermal energy ${U}$ of the system. As well, the atmosphere will fill in the volume originally occupied by the reactant gases, doing work ${PV}$ on the system, which is also given off as heat. The enthalpy change is the total heat emitted by the system as a result of these two mechanisms.

The energy resulting from the ${PV}$ work is (assuming that the volume of the liquid water is negligible compared to the intial volume) is

$\displaystyle PV=nRT \ \ \ \ \ (5)$

We started with 1.5 moles of gas, so

$\displaystyle PV=\frac{3}{2}\left(8.31\mbox{ J K}^{-1}\right)\left(298\mbox{ K}\right)=3.71\times10^{3}\mbox{ J} \ \ \ \ \ (6)$

Therefore the energy released as a result of decreasing ${U}$ is

$\displaystyle -\Delta U=2.86\times10^{5}-3.71\times10^{3}=2.82\times10^{5}\mbox{ J} \ \ \ \ \ (7)$

The ${PV}$ contribution is just over 1% of energy released.

# Latent heat

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 1.47 – 1.48.

In most cases, adding heat to a substance causes its temperature to rise, by an amount governed by its heat capacity. At a phase transition (melting or boiling), however, heat is required just to convert the substance from one phase to another without changing its temperature. [I can still remember the first time I discovered this, in a high school chemistry lab. We were given a test tube filled with wax in which was embedded a thermometer and had to record the temperature as we heated the test tube over a bunsen burner. At first, as you’d expect, the temperature of the (still solid) wax increased, but then it levelled off as the wax started to melt. It was one of those experiments that left a lasting impression on me some 45 years later.]

The amount of heat per unit mass required to effect a phase transition is called latent heat:

$\displaystyle L\equiv\frac{Q}{m} \ \ \ \ \ (1)$

Every substance has its own latent heats at its melting and boiling points, and these latent heats can be quite large compared with the specific heat capacity. For water, the latent heat for melting ice is ${333\mbox{ J g}^{-1}=80\mbox{ cal g}^{-1}}$ and for boiling water it is ${2260\mbox{ J g}^{-1}=540\mbox{ cal g}^{-1}}$. (Recall that the specific heat capacity of liquid water is ${1\mbox{ cal g}^{-1}\mbox{K}^{-1}}$, so heating liquid water from its melting point to its boiling point takes ${100\mbox{ cal g}^{-1}}$.)

Example 1 We have a 200 gram cup of tea at boiling point and wish to cool it down to ${65^{\circ}\mbox{ C}}$ before we drink it, by putting a mass ${m}$ of ice (initially at ${-15^{\circ}\mbox{C}}$) into the tea. Given that the specific heat capacity of ice is ${0.5\mbox{ cal g}^{-1}\mbox{K}^{-1}}$, how much ice do we need?

The tea (assumed to have the same heat capacity as pure water) must decrease by 35 K, so it must give up an amount of heat

$\displaystyle Q=200\times35=7000\mbox{ cal} \ \ \ \ \ (2)$

This heat goes into, first, heating the ice by 15 degrees to its melting point, then melting it, then heating the resulting water to ${65^{\circ}\mbox{ C}}$. The heat required for each of these steps is

$\displaystyle Q=\begin{cases} 15\times0.5m & \mbox{to heat the ice to }0^{\circ}\\ 80m & \mbox{to melt the ice}\\ 65m & \mbox{to heat the water to }65^{\circ}\mbox{ C} \end{cases} \ \ \ \ \ (3)$

The sum of these three heats is equal to the heat lost by the tea, so

 $\displaystyle \left(7.5+80+65\right)m$ $\displaystyle =$ $\displaystyle 7000\ \ \ \ \ (4)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 45.9\mbox{ g} \ \ \ \ \ (5)$

Example 2 Estimating how long it takes the snow pack to melt with the spring thaw. Suppose the snow is composed of 50% ice and 50% air, and that it’s 2 metres deep. The sun provides around ${1000\mbox{ watts m}^{-2}}$ of energy, but around 90% of this is reflected by the snow, so the snow absorbs only about ${100\mbox{ watts m}^{-2}}$. The density of ice at its melting point is ${0.9167\mbox{ g cm}^{3}}$ so in a block of snow with ${1\mbox{ m}^{2}}$ surface area and a depth of 2 m, there is a mass ${m}$ of ice:

$\displaystyle m=1\times2\times0.9167\times\frac{1}{2}\times10^{6}=9.167\times10^{5}\mbox{ g} \ \ \ \ \ (6)$

The amount of heat required to melt the ice is

$\displaystyle Q=80\times9.167\times10^{5}=7.33\times10^{7}\mbox{ cal} \ \ \ \ \ (7)$

The heat is provided by the sun at the rate of

$\displaystyle P=\frac{100}{4.186}=23.89\mbox{ cal s}^{-1} \ \ \ \ \ (8)$

so it takes

$\displaystyle t=\frac{7.33\times10^{7}}{23.89}=3.07\times10^{6}\mbox{ s}\approx5\mbox{ weeks} \ \ \ \ \ (9)$

to melt the snow. Of course, the sun is shining only during the daytime, so we’d expect about double this time would actually be needed. [This is actually not that bad an estimate. I remember when I used to go hiking in the mountains north of Vancouver that even in July there was still a lot of snow. We usually had to wait until August before it was gone.]

# Measuring heat capacity at constant volume

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.46.

Heat capacities of substances can be measured at constant volume or constant pressure. To measure the heat capacity of a gas, it is relatively easy to keep the gas at a constant volume (just enclose it in a sealed container). For solids and liquids, however, it is much easier to measure heat capacity at constant pressure. To see why, we can estimate how much pressure needs to be increased to keep a solid or liquid from expanding as it is heated.

The thermal expansion coefficient is a measure of the relative volume change with temperature at constant pressure:

$\displaystyle \beta\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (1)$

Suppose we heat a solid (or liquid) a bit at constant pressure, so that its volume changes by an amount ${dV}$:

$\displaystyle dV=\beta V\; dT \ \ \ \ \ (2)$

The reciprocal of the bulk modulus is the isothermal compressibility:

$\displaystyle \kappa_{T}\equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T} \ \ \ \ \ (3)$

This measures how much the relative volume changes as we increase the pressure at constant temperature. If we take the substance that we heated a bit and then compress it at the new temperature by just the amount that it expanded, then the volume change is ${-dV}$ so

$\displaystyle dV=\kappa_{T}V\; dP \ \ \ \ \ (4)$

Dividing 4 by 2 we get the deriviatve of pressure with respect to temperature at constant volume, since after the two actions, the volume is unchanged:

$\displaystyle \left(\frac{\partial P}{\partial T}\right)_{V}=\frac{\beta}{\kappa_{T}}=-\frac{\left(\partial V/\partial T\right)_{P}}{\left(\partial V/\partial P\right)_{T}} \ \ \ \ \ (5)$

For an ideal gas, we have

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle \frac{1}{V}\frac{Nk}{P}\ \ \ \ \ (6)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle \frac{NkT}{VP^{2}}=\frac{1}{P}\ \ \ \ \ (7)$ $\displaystyle \left(\frac{\partial P}{\partial T}\right)_{V}$ $\displaystyle =$ $\displaystyle \frac{Nk}{V}=\frac{\beta}{\kappa_{T}} \ \ \ \ \ (8)$

For water, the values given by Schroeder are (at ${25^{\circ}\mbox{ C}}$):

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle 2.57\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (9)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle 4.52\times10^{-10}\mbox{ Pa}^{-1} \ \ \ \ \ (10)$

To keep the volume of some water constant as its temperature is raised from ${20^{\circ}\mbox{ C}}$ to ${30^{\circ}\mbox{ C}}$, the pressure would need to be increased by about

$\displaystyle \Delta P\cong\frac{\beta}{\kappa_{T}}\Delta T=\frac{2.57\times10^{-4}}{4.52\times10^{-10}}\times10=5.69\times10^{6}\mbox{ Pa}\approx56\mbox{ atm} \ \ \ \ \ (11)$

For mercury in the same temperature range

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle 1.81\times10^{-4}\mbox{ K}^{-1}\ \ \ \ \ (12)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle 4.04\times10^{-11}\mbox{ Pa}^{-1} \ \ \ \ \ (13)$

so the pressure increase must be

$\displaystyle \Delta P\cong\frac{\beta}{\kappa_{T}}\Delta T=\frac{1.81\times10^{-4}}{4.04\times10^{-11}}\times10=4.48\times10^{7}\mbox{ Pa}\approx442\mbox{ atm} \ \ \ \ \ (14)$

To measure ${C_{V}}$ thus requires very large pressures for typical liquids (and even more so for solids). Measuring ${C_{P}}$ for liquids and solids is relatively easy as we can just heat them at atmospheric pressure.