91 Cotton Kills

Conduction

Cotton is a great thermal insulator – as long as it’s dry. Once wet, cotton becomes a poor insulator and does a poor job of preventing -hence the old adage, “cotton kills”.  To understand why cotton is sometimes  good insulator and sometimes a poor one, we need to understand , which is the transfer of between objects by direct contact. The average of the molecules in a hot body is higher than in a colder body. If two molecules collide, an energy transfer from the molecule with greater kinetic energy to the molecule with less kinetic energy occurs. The cumulative effect from many collisions occurring at a surface of contact between the two objects results in transfer of heat from the hot object to the colder object, (or colder environment) in agreement with the .

The figure shows a vertical line labeled “surface” that divides the figure in two. Just below the line is a horizontal rightward wavy arrow labeled Q, heat conduction. The area left of the surface line is labeled higher temperature and the area right of the surface line is labeled lower temperature. One spherical object, labeled “high energy before collision” is on the left bottom side, with an arrow from it pointing to the right and up toward the vertical midpoint of the surface line. There is another spherical object at the top left side close to the surface line with an arrow from it pointing to the left and up. A third spherical object labeled “low energy before collision” appears on the right top side with an arrow pointing from it to the left and down toward the vertical midpoint of the surface line. There is a final spherical object at the lower right side close to the surface line with an arrow pointing from it to the right and down. There are dotted lines coming from all the four particles, merging at the midpoint on the surface line.
The molecules in two objects at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions, in agreement with the second Law of Thermodynamics. In this illustration, a molecule in the lower temperature region (right side) has low energy before collision, but its energy increases after colliding with the contact surface. In contrast, a molecule in the higher temperature region (left side) has high energy before collision, but its energy decreases after colliding with the contact surface.

Contact Area

The number of molecular collisions increases with increasing contact area (A_c), so the rate of depends on the contact area.   More collisions, and thus more time, will be needed to transfer the same amount of heat across a thicker material, so heat transfer rate also depends on thickness, or length across which the heat is transferred (d). This model explains why thick clothing is warmer than thin clothing in winters, and why thin people are typically more susceptible to hypothermia.

Temperature Difference

If the object temperatures are the same, the net transfer rate falls to zero, and is achieved. As the difference in temperature increases, the average kinetic energy transferred from fast to slow molecules during each collision also increases. Therefore the temperature difference also affects the rate of heat transfer by .

Thermal Conductivity

Lastly, some materials conduct thermal energy faster than others. In general, good conductors of electricity (metals like copper, aluminum, gold, and silver) are also good heat conductors, whereas insulators of electricity (wood, plastic, and rubber) are poor heat conductors. The effect of a material’s properties on conductive heat transfer rate is described by the coefficient of (k), which is sometimes shortened to just thermal conductivity.  The following table shows values of k for some common materials in units of , per meter, per (bold{W}/(bold{m cdot K})).

Thermal Conductivities of Common Substances
Substance Thermal conductivity (bold{frac{W}{(m cdot K)}})
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel iron 80
Steel (stainless) 14
Ice 2.2
Glass (average) 0.84
Concrete brick 0.84
Water 0.6
Fatty tissue (without blood) 0.2
Asbestos 0.16
Plasterboard 0.16
Wood 0.08–0.16
Snow (dry) 0.10
Cork 0.042
Glass wool 0.042
Wool 0.04
Down feathers 0.025
Air 0.023
Styrofoam 0.010

Conduction Equation

We can summarize all of the factors effecting the conductive heat transfer rate, which is an amount of energy transferred per time, in a single diagram:

Two rectangular blocks are shown with the right one labeled T one and the left one labeled T two. The blocks are placed on a surface at a distance d from each other, so that their largest face faces the opposite block. The block T one is cold and the block T two is hot. The blocks are connected to each other with a conducting rectangular block of thermal conductivity k and cross-sectional area A. A wavy line labeled Q is inside the conducting block and points from the hot block to the cold block.
Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. The temperature of the material is T_2 on the left and T_1 on the right, where T_2 is greater than T_2. The rate of heat transfer by conduction is directly proportional to the surface area A, the temperature difference, and the substance’s conductivity k. The rate of heat transfer is inversely proportional to the thickness d.

The rate of transfer across a material with temperature T_1 on one side and temperature T_2 on the other can be modeled by the equation:

(1)   \begin{equation*} \frac{Q}{t} = \frac{kA(T_2-T_1)}{d} \end{equation*}

The (Q/t)refers to an amount of energy transferred per unit time so it has the same units as (P). The term power is usually applied to a rate of transformation of energy from one form to another, but is a transfer of thermal energy from one place another, not a change in type of energy, thus we stick with frac{Q}{t} instead of  P as the notation for this rate. This notation also reminds us that if we want to know how much energy was transferred during a certain time, we need to multiply the by the time.

Insulation Technology

We can see from the equation that combining thickness (loft) with low provides the greatest insulating effect. Goose down is the gold-standard insulation clothing for extreme-cold environments because it recovers loft after being compressed and it has low conductivity. As with most insulating materials, the effectiveness of down at reducing conduction is not due to the low conductivity of the feathers themselves, but rather their ability to trap air which has a very low conductivity. Double-pane windows,  Styrofoam, animal hair, and the fiberglass insulation used in buildings all rely on this same strategy of trapping air to provide .
The figure shows two thick rectangular pieces of fiberglass batt lying one upon the other.
The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the inside of the building and the outside environment.

Everyday Example: Down Insulation

Let’s compare the insulating properties of a cotton sweatshirt and a down jacket, such as the one worn in the photograph below.

The author approaches the summit of Mt. Washington in the Central Oregon Cascades, in snowstorm, but wearing a down jacket. February 2017.

We will start with the conduction equation:

    \begin{equation*} \frac{Q}{t} = \frac{kA(T_2-T_1)}{d} \end{equation*}

The outside was about -10\,\bold{^{\circ}C} and skin temperature is roughly 37 \,\bold{^{\circ}C}, so the temperature difference was about 47\,\bold{^{\circ}C}. When dry, the down jacket pictured is roughly 4 \,\bold{cm} thick, which 0.04 \,\bold{m}. The  of dry down is roughly 0.025\, \bold{\frac{W}{m\cdot K}}. Using the methods in Chapter 17 we estimate the surface area of the upper body to be 0.5\,\bold{m}. Entering these values into the equation we find:

    \begin{equation*} \frac{Q}{t} = \frac{(0.025\, \bold{W/(m \cdot K)})(0.5\, \bold{m^2})(47\,\bold{C^{\circ}})}{0.04 \,\bold{m}} \approx 15 \,\bold{W} \end{equation*}

So through the upper body  in this situation is 30 W. That means 15 J of thermal energy is released to the environment as heat through the upper body each second. The typical human is 100 W so it appears that there is plenty of thermal energy available to maintain temperature. However, the 15 W value we found does not account for heat loss from the lower body, which we can see from the picture were not covered with down. In fact the single thin layer worn on the legs easily allowed for the other 85 W to escape so that body temperature was not elevated. In fact, doing the extra of climbing the mountain likely raised the the closer to 300 W so actually something closer to 285 W was being exhausted through the lower body. As a result, body temperature dropped quickly when climbing stopped and the fell back to about 100 W.

If the down were to get wet, things would change drastically. The down would lose its loft and end up with roughly  0.5 cm thickness. Even worse, the water would fill in the air spaces between the down fibers so the would essentially be the same as for water, which is 0.6\, \bold{\frac{W}{m\cdot K}}. Entering these values into the conduction equation we find:

    \begin{equation*} \frac{Q}{t} = \frac{(0.6\, \bold{W/(m \cdot K)})(1\, \bold{m^2})(47\,\bold{C^{\circ}})}{0.01 \,\bold{m}} \approx 5600 \,\bold{W} \end{equation*}

Now that’s a problem. The wet heat loss rate is nearly 190x faster than for the dry down in this situation, so to keep up we would need to eat 188 candy bars every six hours, or 31 candy bars per hour! Our bodies would not be able to digest and convert to fast enough to keep warm in that situation and eventually would occur. Later in this unit we will be able to estimate how long it would take for the body temperature to drop to dangerous levels in this situation. Down is clearly a poor insulator when wet, but even after drying the fibers do not naturally recover their original loft. Down is a poor choice of insulation in wet environments, though it does well in snow storms as long as the air temperature is cool enough that the snow doesn’t melt when it lands on the jacket.

Everyday Examples: Cotton Kills

Just as with down, water can permeate the spaces between other fabrics like wool, synthetics, and cotton. The majority of  water can be wrung out of wool and synthetics, partially restoring their insulating properties and helping them to dry out quickly and recover  full insulation value. In the other hand, water fills space between cotton fibers and also saturates the fibers themselves. As a result  cotton does not wring out well and dries slowly, so its remains much closer to that of water than wool or synthetic.  Cotton is a poor choice of insulation in wet environments.

Reinforcement Exercise:  Single Pane Windows

Calculate the through a window with dimensions 2 m x 3 m with thickness of 5 mm when the inside is 22 °C and the outdoor temperature is 12 °C. Use the of glass listed in the table found in this chapter.

Find the heat lost through the window in one day, in units of kW·hr, by first converting your answer to kilowatts (kW) and then multiply by 24 hours .

If electricity is $0.12 per kW·hr, how much money is spent each day to replace the heat loss from the window using and electric heater?

Significant content in this chapter, including the table of conductivities, was adapted from College Physics, by BC Open Textbooks [1]


License

Share This Book