7.4: Transport by Random Walks

When a hot-oven temperature front diffuses into a slab of fish or the no-slip, zero-momentum front diffuses into a fluid, it comes with a heat or momentum flow. With our random-walk model, we can estimate the magnitude of these flows, and thereby understand the drag forces on fog droplets and bacteria (Problem 7.26) and why we feel cold on a winter day without thick clothing (Section 7.4.4).

7.4.1 Diffusion speed

The essential property of a random walk is that the distance traveled is not proportional to the time, as it would be in a regular walk, but rather to the square root of the time. This change in scaling exponent means that the speed at which heat, momentum, or particles diffuse depends on the diffusion distance. When the distance is L, the diffusion time is \(t \sim L^{2}/D\), where D is the appropriate diffusion constant. Thus, the transport speed is comparable to D/L:

\[v \sim \frac{L}{t} \sim \frac{L}{L^{2}/D} = \frac{D}{L}.\]

This speed depends inversely on the distance. This scaling is consistent with our calculations showing that diffusion is a terribly slow means of transport overlong distances (for example, for perfume molecules diffusing across a room) but fast over short distances (for example, for neurotransmitter molecules diffusing across a synaptic cleft).

When the diffusing quantity is momentum, the appropriate diffusion constant is \(\nu\), and the diffusion speed is \(\nu/L\). Thus, the Reynolds number, \(v_{flow}L/\nu\), is the ratio \(v_{flow}/v_{diffusion}\)—for the same reason that it is the ratio of times \(t_{diffusion}/t_{flow}\) (as you will find in Problem 7.32). Using the diffusion speed, we can estimate fluxes and flows.

7.4.2 Flux

Transport is measured by the flux:

\[\textrm{flux of stuff} = \frac{\textrm{amount of stuff}}{\textrm{area} \times \textrm{time}}.\]

As we found in Section 3.4.2, the flux is also given by

\[\textrm{flux of stuff} = \frac{\textrm{stuff}}{\textrm{volume}} \times \textrm{transport speed}.\]

When the stuff travels by diffusion, the transport speed is the diffusion speed D/L (from Section 7.4.1). The resulting flux is

\[\textrm{flux of stuff} \sim \frac{\textrm{stuff}}{\textrm{volume}} \times \frac{\textrm{diffusion constant}}{\textrm{distance}}.\]

In symbols,

\[F \sim n \frac{D}{L},\]

where n is the concentration (stuff per volume), D is the appropriate diffusion constant (depending on what is diffusing), and L is the distance

An important application is diffusion across a gap. The gap could be a synaptic cleft, with a gap width L ∼ 20 nanometers and with different neurotransmitter concentrations on its two sides. Or it could be a shirt (L ∼ 2 millimeters) with different temperatures—concentrations of energy—on the inside and outside. On one side, the density of stuff is n₁; on the other side it is n₂. Then there are two fluxes in the gap, left to right and right to left:

\[F_{1 \rightarrow 2} \sim n_{1} \frac{D}{L} // F_{2 \rightarrow 1} \sim n_{2} \frac{D}{L}.\]

The net flux F is their difference:

\[F \sim (n_{2}-n_{1}) \frac{D}{L} = D \frac{n_{2}-n_{1}}{L}.\]

The concentration difference n₂ - n₁ divided by the gap size L is an important abstraction: the concentration gradient. It measures how rapidly the concentration varies with distance. Using it, the flux becomes

\[\textrm{flux} = \textrm{diffusion constant} \times \textrm{concentration gradient}.\]

This result, called Fick’s law, is exact (hence the equals sign instead of ~). In calculus form, the concentration gradient is \(\Delta n/ \Delta x\), so

\[F = D \frac{\Delta n}{\Delta x}.\]

If the diffusing stuff is particles, then the appropriate diffusion constant is just D, and the result can be used as is. If the diffusing stuff is momentum, then the diffusion constant is the kinematic viscosity \(\nu\), and the flux is closely connected to a drag force (Problem 7.25).

Exercise \(\PageIndex{1}\): Momentum flux produces drag

If the diffusing stuff is momentum, then the diffusion constant is the kinematic viscosity \(\nu\), and the concentration gradient is the gradient of momentum density. Thus, explain why Fick’s law becomes

\[\textrm{viscous stress} = \rho \nu \times \textrm{velocity gradient},\]

where viscous stress is viscous force per area (\(\rho \nu\) is the dynamic viscosity \(\eta\)).

Exercise \(\PageIndex{2}\): Stokes drag

In this problem, you use momentum flux (Problem 7.25) to estimate the drag force on a sphere of radius r in a flow at low Reynolds number (Re ≪ 1). If Re ≪ 1, the boundary layer (Section 7.3.4)—the region over which the fluid velocity changes from zero to the free-stream velocity \(v\)—is comparable in thickness to r. Using that information, estimate the viscous drag force on the sphere.

If the diffusing stuff is heat (energy), the diffusion constant is the thermal diffusivity \(\kappa\), and the concentration gradient is the gradient of energy density. Thus, heat flux is

\[\textrm{heat flux} = \textrm{thermal diffusivity} \times \textrm{energy-density gradient}.\]

To figure out the meaning of energy-density gradient, let’s start with energy density itself, which is energy per volume. We usually measure it using temperature. Therefore, to make the eventual formula applicable, let’s rewrite energy density in terms of temperature:

\[\frac{\textrm{energy}}{\textrm{volume}} = \frac{\textrm{energy}}{\textrm{volume} \times \textrm{temperature}} \times \textrm{temperature}.\]

The complicated quotient on the right splits into two simpler factors:

\[\frac{\textrm{energy}}{\textrm{volume} \times \textrm{temperature}} = \frac{\textrm{mass}}{\textrm{volume}} \times \frac{\textrm{energy}}{\textrm{mass} \times \textrm{temperature}}.\]

The first factor, mass per volume, is just the substance’s density \(\rho\). The second factor is called the specific heat c_p. The most familiar specific heat is water’s: 1 calorie per gram per ^∘C. That is, raising 1 gram of water by 1 ^∘C requires 1 calorie (≈ 4 joules)

Using these abstractions,

\[\frac{\textrm{energy}}{\textrm{volume} \times \textrm{temperature}} = \rho c_{p},\]

To get energy density, or energy per volume, we multiply by temperature

\[\frac{\textrm{energy}}{\textrm{volume}} = \rho c_{p} T.\]

Now that we have energy density, we can find the energy-density gradient. Because \(\rho\) and c_p are constants (at least for small temperature and position changes), any gradient of energy density is due to the temperature gradient \(\Delta T/ \Delta x\):

\[\textrm{energy-density gradient } = \rho c_{p} \times \textrm{temperature gradient}.\]

Fick’s law, when energy is the diffusing stuff, tells us that

\[\textrm{energy flux} = \textrm{thermal diffusivity } \times \textrm{energy-density gradient},\]

so

\[\textrm{head (energy) flux} = \kappa \times \underbrace{\rho c_{p} \times \textrm{temperature gradient.}}_{\textrm{energy-density gradient}}\]

The shaded combination \(\rho c_{p} \kappa\) occurs in any heat flow driven by a temperature gradient. It is a powerful abstraction called the thermal conductivity K:

\[\underbrace{\textrm{thermal conductivity}}_{K} = \underbrace{\textrm{density}}_{\rho} \times \underbrace{\textrm{(specific heat)}}_{c_{p}} \times \underbrace{\textrm{(thermal diffusivity)}}_{\kappa}. \]

Thermal conductivity has dimensions of power per length per temperature, and is often quoted in units of watts per meter kelvin (Wm⁻¹ K⁻¹). Using K, Fick’s law for energy flux in terms of temperature gradient \(\Delta T/ \Delta x\) becomes

\[F = K \frac{\Delta T}{\Delta x},\]

where \(\Delta T\) is the temperature difference, and \(\Delta x\) is the gap size.

Now let’s estimate a few important thermal conductivities in order to understand heat flow around us.

7.4.3 Thermal conductivity of air

To estimate our heat loss standing outside on a cold winter’s day, we need to estimate the thermal conductivity of air

Why do we estimate the thermal conductivity of air rather than of clothing?

The purpose of clothing is to trap air so that heat flows via conduction—that is, by diffusion—rather than via the faster process of convection. (If the perfume molecules of Section 7.3.1 could be similarly limited to diffusion, the perfume aromas would travel very slowly.)

Because \(K \equiv \rho c_{p} \kappa\), estimating K splits into three subproblems, one for each factor. The density of air \(\rho_{air}\) is just 1.2 kilograms per cubic meter (slightly more accurate than 1 kilogram per cubic meter). The thermal diffusivity \(\kappa_{air}\) is 1.5×10⁻⁵ square meters per second.

The specific heat c_p is not as familiar, but we can estimate it. As for water, it measures the thermal energy per mass per temperature:

\[c_{p} = \frac{\textrm{thermal energy}}{\textrm{mass} \times \textrm{temperature}}.\]

The thermal energy per particle is comparable to k_BT, where k_B is Boltzmann’s constant, so the energy per temperature is comparable to k_B. Thus,

\[c_{p} \sim \frac{k_{B}}{\textrm{mass}}.\]

Our thermal energy, which is comparable to k_BT, is for one particle. The corresponding mass in the denominator is the mass of one particle, which is an air molecule.

To convert k_B and the mass to human-sized values, we multiply each by Avogadro’s number N_A. Then we replace k_BN_A with the universal gas constant R, and mass x N_A with the molar mass m_molar. The result is

\[c_{p} \sim \frac{k_{B}N_{A}}{\textrm{molecular mass} \times N_{A}} = \frac{R}{m_{molar}},\]

where m_molar is the molar mass of air.

This expression is correct except for a dimensionless prefactor. Airis mostly nitrogen, for which the prefactor is 7/2. This magic number can be made slightly less magical as follows:

\[\frac{7}{2} = \frac{\textrm{spatial dimensions}}{2} + \frac{\textrm{rotational directions}}{2} + 1.\]

Each spatial dimension contributes one 1/2 through a term in the translational kinetic energy of a nitrogen molecule; thus, the translational piece contributes 3/2 to the prefactor. Each rotational direction contributes a 1/2 through a term in the rotational energy of a nitrogen molecule. Because nitrogen is a linear molecule, there are only two rotational directions, so the rotational directions contribute 2/2 to the prefactor. If we stop here, at 5/2, we would have the prefactor to find c_v, the specific heat while holding the volume constant. The final term, +1, accounts for the energy required to expand a gas as it is heated and held at constant pressure. Therefore, c_p gets a prefactor of 7/2. (For a more detailed discussion of the reasoning behind this calculation, see the classic Gases, Liquids and Solids and Other States of Matter [43, pp. 106].)

Air is mostly diatomic nitrogen, so m_molar is roughly 30 grams per mole. Then c_p is roughly 10³ joules per kilogram kelvin:

\[c_{p} = \frac{7}{2} \times \frac{R}{m_{molar}} \approx \frac{7}{2} \times \underbrace{\frac{8 \: J}{\textrm{mol K}}}_{R} \times \underbrace{\frac{1 \textrm{mol}}{3 \times 10^{-2} \textrm{ kg}}}_{m_{molar}^{-1}} \approx \frac{10^{3} \: J}{\textrm{kg K}}.\]

Putting together the three pieces,

\[K_{air} \approx \underbrace{1.2 kg \: m^{-3}}_{\rho_{air}} \times \underbrace{10^{3} J \: kg^{-1}}_{c_{p}} \times \underbrace{1.5 \times 10^{-5} m^{2} s^{-1}}_{\kappa_{air}} \approx 0.02 \frac{W}{m \: K}.\]

Before using the thermal conductivity, let’s try out the specific heat of air on an old method of air conditioning. One summer I lived in a tiny Manhattan apartment (30 square meters). Summers are hot in New York City, and the beautiful people flee for the cooler beach areas—cooler thanks partly to the high specific heat of water (Problem 7.27). Because of global warming and the old electrical wiring in the apartment building, too old to handle an air-conditioning unit, the apartment reached 30^∘C at night. A friend who grew up before air conditioning suggested taking a wet sheet and using a fan to blow air past it.

How much does this system cool the apartment?

As heat from the room air evaporates the water from the sheet, the room air cools, as if the room were sweating. The energy required to evaporate the water is

\[E = m_{water}L_{vap},\]

where m_water is the mass of water held by the wet sheet and L_vap is the heat of vaporization of water. This energy is the demand.

To lower the room’s temperature by \(\Delta T\) requires using up a thermal energy

\[E \sim \rho_{air} V C_{p} \Delta T,\]

where V is the volume of the room and c_p is the specific heat of air. This energy is the supply

Equating supply and demand gives an equation for \(\Delta T\):

\[\rho_{air} V c_{p} \Delta T \sim m_{water} L_{vap}.\]

Its solution is

\[\Delta T \sim \frac{m_{water}L_{vap}}{\rho_{air}V c_{p}}.\]

Now we estimate and plug in the needed values. A typical room is roughly 3 meters high, so the apartment’s volume was roughly 100 cubic meters:

\[V \sim \underbrace{30 m^{2}}_{area} \times \underbrace{3m}_{height} \approx 100 m^{3}.\]

To estimate m_water, I imagined that the sheet was about as wet as it would be after coming out of the washing machine with its fast spin cycle. That mass feels like slightly more than 1 kilogram, so m_water ∼ 1 kilogram. Finally, the heat of vaporization of water is about 2×10⁶ joules per kilogram (as we estimated in Section 1.7.3 using a home experiment)

Putting in all the numbers, \(\Delta T\) is about 20^∘C (or 20 K):

\[\Delta T \times \frac{\overbrace{1kg}^{m_{water}} \times \overbrace{2 \times 10^{6} J kg^{-1}}^{L_{vap}}}{\underbrace{1 \: kg m^{-3}}_{\rho_{air}} \times \underbrace{100 m^{3}}_{V} \times \underbrace{10^{3} \: J \: kg^{-1} K^{-1}}_{c_{p}}} = 20K.\]

This change would have turned the hot 30^∘C room into a cold 10^∘C room, if the cooling had been 100-percent efficient. Because some heat comes from the walls (and from the fan motor), ΔT will be less than 20^∘C—perhaps 10^∘C, leaving the room at a pleasant and sleepable temperature of 20^∘C. This calculation shows not only that evaporative cooling is a reasonable method of air conditioning, but also that our estimate for the specific heat of air is reasonable.

Exercise \(\PageIndex{3}\): Dimensionless specific heat of water

The specific heat per air molecule is 3.5k_B, so the dimensionless specific heat of air, measured relative to k_B, is 3.5. What is the dimensionless specific heat of water?

7.4.4 Keeping warm on a cold day

Now we have assembled the pieces to understand why we dress warmly on a cold day. Our starting point is the heat flux:

\[F = K \frac{\Delta T}{\Delta x},\]

where \(\Delta T = T_{2} -T_{1}\) is the temperature difference across a gap, \(\Delta x\) is the gap size, and K is the thermal conductivity of the gap material. Here, the gap material is air—the clothing serves to trap the air

Let’s say that the air outside is at T₁ = 0^∘C and that skin is at T₂ = 30^∘C (slightly lower than the internal body temperature of 37^∘C). Then \(\Delta T = 30 K\). Against the advice of your elders, you dress in a thin T-shirt—for decency, a very long one. A thin T-shirt has thickness \(\Delta x\) of roughly 2 millimeters. With these parameters, the heat flux through the shirt becomes 300 watts per square meter:

\[F \approx \underbrace{0.02 \frac{\textrm{W}}{\textrm{m K}}}_{K_{air}} \times \frac{\overbrace{30 \textrm{ K}}^{\Delta T}}{\underbrace{2 \times 10^{-3} \textrm{ m}}_{\Delta x}} = 300 \textrm{ W m}^{-2}.\]

Flux is power per area, so the energy flow—the power—is the flux times a person’s surface area. A person is roughly 2 meters tall and 0.5 meters wide, with a front and a back, so the surface area is about 2 square meters. Thus, the power (the energy outflow) is 600 watts.

Is this heat loss worrisome?

Even though 600 might seem like a large number, we cannot therefore conclude that 600 watts is a large heat loss: As we learned in Chapter 5, a quantity with dimensions, such as heat flux, cannot be large or small on its own. It needs to be compared to a relevant quantity with the same dimensions. A relevant quantity is a normal human power output. When sitting around, a person produces 100 watts of heat; that’s our basal metabolic rate. If 600 watts is escaping through your clothing, you are losing heat much faster than the basal metabolism is producing it. No wonder you feel so cold on a winter day wearing only a thin shirt and pants. Eventually, your core body temperature falls. Then essential chemical reactions in your body slow down, because the enzymes lose their shape optimized for body temperature and thereby become less efficient catalysts. Eventually you get hypothermia and, if it goes on too long, die.

One solution is to generate heat to make up the difference: by shivering or exercising. Cycling hard, which generates, say, 200 watts of mechanical power and another 600 watts of heat (thanks to the one-fourth metabolic efficiency), should be vigorous-enough exercise to keep you warm, even on a winter day in thin clothing.

Another simple solution is to dress warmly by putting on thick layers. Let’s recalculate the power loss if you put on a jacket and thick pants, each 2 centimeters thick. We could redo the power calculation from scratch, but that approach is brute force. It is simpler to notice that the gap thickness Δx has increased by a factor of 10, yet nothing else changed. Because flux is inversely proportional to the gap size, the flux and the power drop by the same factor of 10. Therefore, wearing thick clothing reduces the energy outflow to a manageable 60 watts—comparable to the basal metabolism. As a result, your body heat can keep you warm. Indeed, when wearing thick clothing, only areas exposed directly to cold air, such as your hands and face, feel cold. Those regions are protected by only a thin layer of still air (the boundary layer analyzed in Section 7.3.4).

A thick gap means a small heat flux: When it is cold, bundle up!

Exercise \(\PageIndex{4}\): Thermal conductivity of helium gas

Estimate the thermal conductivity of helium at standard temperature and pressure. The following fact will help you estimate the mean free path: The density of liquid helium is 125 grams per liter.

Exercise \(\PageIndex{5}\): Comfortable outdoor temperature

You wear only that long thin T-shirt in which the winter temperature of 0 ∘C felt too cold. Estimate the outside temperature that would feel most comfortable.

7.4.5 Getting your clothes wet: More thermal conductivities

If your thick warm coat gets wet, you feel very cold. Let’s use our knowledge of heat flow to explain why the coat becomes so useless. As we know from Section 7.4.4, you feel cold when the dimensionless ratio

\[\frac{\textrm{energy flow through your coat}}{\textrm{rate at which your body generates heat}}\]

is significantly larger than 1. The dry coat kept the energy flow (a power) comparable to the rate at which your body generated heat. Wetting the coat must increase the energy flow significantly. To see how, let’s look again at the terms in the energy flow and apply proportional reasoning.

\[\textrm{energy flow} = \textrm{area} \times \underbrace{K \frac{\Delta T}{\Delta x}}_{flux}.\]

In comparing the wet to the dry coat, the area is unchanged. The temperature difference \(\Delta T\) is also unchanged: It is still the 30^∘C between skin temperature and winter-air temperature. The gap \(\Delta x\), which is the thickness of the coat, is also unchanged.

The remaining possibility is that the thermal conductivity K of the gap material has increased significantly. If so, the thermal conductivity of water must be much higher than the thermal conductivity of air. Rather than studying water directly, let’s first estimate the thermal conductivity of nonmetallic solids. (Metals have an even higher thermal conductivity, as we will discuss after studying water.) Using that estimate, we will estimate the thermal conductivity of water.

The problem is again comparative (proportional reasoning):

\[\frac{\textrm{thermal conductivity of nonmetallic solids}}{\textrm{thermal conductivity of air}}.\]

The ratio breaks into three ratios, corresponding to the three factors in the thermal conductivity (divide-and-conquer reasoning):

\[K = \underbrace{\textrm{density}}_{\rho} \times \underbrace{\textrm{specific heat}}_{c_{p}} \times \underbrace{\textrm{thermal diffusivity}}_{\kappa}.\]

Rather than using these factors directly, let’s remix the first two ( \(\rho c_{p}\)) into a more insightful combination. The specific heat itself is

\[c_{p} = \frac{\textrm{energy}}{\textrm{mass} \times \Delta T},\]

where \(\Delta T\) is the temperature change.

Multiplying by the density \(\rho\) points us toward an interpretation of \(\rho c_{p}\):

\[\rho c_{p} = \underbrace{\frac{\textrm{mass}}{\textrm{volume}}}_{\rho} \times \underbrace{\frac{\textrm{energy}}{\textrm{mass} \times \Delta T}}_{c_{p}} = \frac{\textrm{energy}}{\textrm{volume} \times \Delta T}.\]

On the right side, the ratio of energy to volume can be subdivided as

\[\frac{\textrm{energy}}{\textrm{volume}} = \frac{\textrm{energy}}{\textrm{mole}} \times (\frac{\textrm{volume}}{\textrm{mole}})^{-1}.\]

After all these reinterpretations, our remix of \(\rho c_{p}\) becomes

\[\rho c_{p} = \frac{\textrm{energy}}{\textrm{mole} \times \Delta T} \times (\frac{\textrm{volume}}{\textrm{mole}})^{-1}.\]

The first factor is the molar specific heat; it is usually denoted C_p (with a capital "C" to distinguish it from the usual, per-mass specific heat c_p). The volume per mole is also called the molar volume; it is usually denoted V_m. Therefore, \(\rho c_{p} = C_{p}/V_{m}\). Then the thermal conductivity becomes

\[K = \rho c_{p} \kappa = \frac{C_{p}}{V_{m}} \kappa.\]

In words, the thermal conductivity K is

\[\frac{\textrm{molar specific heat} C_{p}}{\textrm{molar volume} V_{m}} \times \textrm{thermal diffusivity} \kappa.\]

This remix is more insightful than simply \(\rho c_{p} \kappa\) because C_p varies less between substances than c_p does, and V_m varies less between substances than \(\rho\) does. The remixed form produces less quantity whiplash, in which the product swings wildly up and down as we include each factor. Another way to express the same advantage is that C_p andV_m are less correlated than are \(\rho\) and c_p; therefore, the remixed abstractions C_p and V_m offer more insight into the thermal properties of materials than do c_p and \(\rho\).

Using the remixed form, let’s apply divide-and-conquer reasoning and estimate the ratio corresponding to each factor.

1. Ratio of molar specific heats C_p. For air, C_p was 3.5R. For most solids, whether metallic or nonmetallic, it is similar: 3R (where the 3 reflects the three spatial dimensions). Thus, this ratio is close to 1.

2. Ratio of molar volumes V_m. For any substance, 1 mole has a mass of A grams, where A is the dimensionless atomic mass (roughly, the number of protons and neutrons in the nucleus). Because a typical solid density is A grams per 18 cubic centimeters (Section 6.4.1), the molar volume of the solid is 18 cubic centimeters per mole. In contrast, for air or any ideal gas (at standard temperature and pressure), 1 mole occupies 22 liters or 22 000 cubic centimeters. Thus, the ratio of molar volumes (solid to air) is 18/22 000 or roughly 10⁻³.

3. Ratio of thermal diffusivities \(\kappa\). From Section 7.3.3, this ratio is roughly 0.1:

\[\frac{\textrm{thermal diffusivity of nonmetallic solids}}{\textrm{thermal diffusivity of air}} \approx \frac{10^{-6}m^{2}s^{-1}}{10^{-5}m^{2}s^{-1}} = 0.1.\]

The ratio of thermal conductivities is the product of the three ratios. As long as we remember that the molar volume appears in the denominator (the thermal conductivity is inversely proportional to the molar volume), we get a ratio of 100:

\[\frac{K_{\textrm{nonmetallic solid}}}{K_{air}} \sim \underbrace{1}_{C_{p} \textrm{ ratio}} \times \underbrace{10^{3}}_{(V_{m} \textrm{ ratio})^{-1}} \times \underbrace{10^{-1}}_{\kappa \textrm{ ratio}} = 10^{2}.\]

Thus, in contrast to the thermal conductivity of 0.02 watts per meter kelvin for air, the typical (nonmetallic) solid has a thermal conductivity of about 2 watts per meter kelvin.

Now let’s use proportional reasoning to compare this thermal conductivity to the thermal conductivity of water, which is the gap material in your wet coat. We’ll estimate \(K_{water}/K_{\textrm{nonmetallic solid}}\). The comparison has the same three ratios: molar specific heat, molar volume, and thermal diffusivity.

1. Ratio of molar specific heats C_p. We can find the specific heat of water from the definition of a calorie, as the energy required to raise 1 gram of water by 1 degree (celsius or kelvin). Thus, the regular specific heat is simple to state:

\[c_{p}^{\textrm{water}} = \frac{1 \: cal}{g \: K}.\]

The resulting molar specific heat is 72 joules per mole kelvin:

\[C^{\textrm{water}}_{p} \approx \underbrace{\frac{1 cal}{g \: K}}_{c_{p}} \times \underbrace{\frac{4J}{1 \: cal}}_{1} \times \underbrace{\frac{18g}{mol}}_{m_{molar}} = \frac{72J}{mol \: K} .\]

The dimensionless specific heat \(C_{P}^{water}/R\) is therefore approximately 9 (as you found in Problem 7.27):

\[\frac{C_{p}^{water}}{R} \approx \frac{72 J \: mol^{-1} K^{-1}}{8 J \: mol^{-1} K^{-1}} = 9.\]

For a nonmetallic solid, the dimensionless specific heat was only 3; it is a factor of 3 smaller than for water. Water stores heat very efficiently, which is why it is used as coolant fluid and why coastal weather is milder than inland weather. The ratio of molar specific heats is 3.

2. Ratio of molar volumes V_m. The molar volume of water, 18 cubic centimeters per mole, is identical to the canonical molar volume of a solid. Thus, the molar-volume ratio is 1

3. Ratio of thermal diffusivities \(\kappa\). As we found in Section 7.3.3 by comparing phonon mean free paths and propagation speeds, the thermal diffusivity \(\kappa_{water}\) is roughly a factor of 10 smaller than the thermal diffusivity of a typical solid (10⁻⁷ compared to 10⁻⁶ square meters per second). Thus, the thermal-diffusivity ratio contributes a factor of 0.1.

These three ratios result in the following comparison:

\[\frac{K_{\textrm{water}}}{K_{\textrm{nonmetallic solid}}} \approx \underbrace{3}_{C_{p} \textrm{ ratio}} \times \underbrace{1}{(V_{m} \textrm{ratio})^{-1}} \times \underbrace{0.1}_{\kappa \textrm{ ratio}} \approx 0.3.\]

In absolute terms,

\[K_{water} \approx 0.3 \times \underbrace{\frac{2 \textrm{ W}}{\textrm{m K}}}_{K_{\textrm{nonmetallic solid}}} = \frac{0.6 \textrm{ W}}{\textrm{m K}}.\]

This conductivity is a factor of 30 larger than K_air. As a result, wearing wet clothes on a cold day is so unpleasant and can even be dangerous. Thick clothing (a coat) allowed a comfortable 60 watts of heat flow—a factor of 10 lower than the T-shirt allowed. Wetting the thick coat increases the thermal conductivity by a factor of 30. The heat loss therefore increases by a factor of 30—making it higher even than the heat loss through the dry T-shirt. When you hike in the hills and mountains, bring waterproof clothing!

The table gives thermal conductivities for everyday substances (at room temperature). Our predictions for nonmetals match the data quite well. Having examined gases (in particular, air) and nonmetallic solids and liquids (water), let’s turn to the remaining category of material. As the table shows, metals have an even higher thermal conductivity than a typical nonconducting solid. (For the unusually high thermal conductivity of diamond, try Problem 7.33.)

Metals, similarly, have a higher thermal diffusivity than most other substances. The reason, as we discussed in Section 7.3.3, is that, in a metal, heat is conducted not only by phonons but also by electrons. The electrons move much faster than the speed of sound, and the electron mean free path is much greater than the phonon mean free path.

Countering these increases, which increase the diffusivity, only a small fraction of the free electrons participate in heat conduction. However, this effect is not enough to overcome the greater speed and mean free path. Thus, metals will have a higher thermal conductivity than nonmetallic liquids or solids. As a rule of thumb, the typical K_metal is 200 watts per meter kelvin: a factor of 100 higher than that of nonmetallic solid.

For this reason, a hot piece of metal, such as a seat-belt clip in a car outside on a hot day, feels much hotter than the plastic button on the same seat-belt clip, even though the plastic and metal are at the same temperature. The large heat flow from the metal into your finger pulls the surface temperature of your finger close to the temperature of the hot metal. Ouch!

Exercise \(\PageIndex{6}\): Stone versus wood floors

Why, on a winter morning, do wood floors feel more comfortable than stone floors?

Exercise \(\PageIndex{7}\): Mercury is special

Why does mercury (Hg) have such a low thermal conductivity for a metal?