A study of basic thermodynamics is necessary to understanding the behavior of heat energy in hot water systems and refrigeration. It also provides an understanding of how houses can be constructed in such a way as to control temperature and ventilation with the home.
There are two common options for providing hot water in the Caribbean. The most common is the electrical water heater. The second option, that is gaining in popularity, is direct solar heating. (The second most common type on the mainland is the natural gas water heater. But natural gas is not available in most of the Caribbean. On the other hand, propane gas is available. Although it is not economical to heat water with a propane heater, many homes use propane for cooking. We'll examine that in a later section.)
The proper choice of building materials, surface preparations and architectural design can greatly improve the ability to control the environment within your home.
Temperature and Heat
Although they are closely related, heat (denoted DQ) and temperature (denoted T) are NOT the same. Temperature is a measure of the kinetic energy of vibration of the molecules that make up a substance. Because the molecules are connected to one another in vastly different ways from one substance to the next, the simplest and most useful way to think of temperature is based upon a comparison system. The idea is the following. If two substances are at the same temperature, then nothing will happen when they are placed in thermal contact with one another. That is, neither will heat up of cool down. If they are at different temperatures, then the hotter one will cool down and the cooler one will heat up ... until they reach the same temperature! We say the two objects are in thermodynamic equilibrium. (Doesn't that sound more impressive than the same temperature?) This simple idea is encoded in the Zeroeth Law of Thermodynamics:
You can think of object B as a thermometer. The two most commonly used temperature scales are the Fahrenheit scale and the Celsius scale. On the Fahrenheit scale, water freezes at 32 oF and boils at 212 oF. On the Celsius scale (also referred to as the centigrade scale) water freezes at 0 oC and boils at 100 oC. To convert from one scale to the other, use
For a quick estimate of the Celsius temperature, subtract 30 from the Fahrenheit temperature and divide by two. For the reverse conversion estimate, multiply the Celsius temperature by 2 and add 30. There is a third scale used in science that is important. It is the Kelvin or Absolute scale. The Kelvin temeprature is found from the Celsius temperature by simply adding 273.
On the Kelvin scale, water freezes at the scorching temperature of 273 oK. The coldest temperature possible is 0 oK, often referred to as absolute zero. (Absolute zero is the temperature at which all atomic and molecular motion ceases, in the "classical theory". The more accurate "quantum theory" paints a more complicated picture.)
Heat
Heat is one of the many forms of energy. We do not usually refer to the absolute amount of heat energy in an object. That can be complicated and is generally not a useful quantity. Instead, we refer to the amount of heat that "flows into" or "flows out of" an object. Let's consider an experiment where we place a hot block of aluminum (Al) in contact with a cold block of iron (Fe). For now, assume the blocks have exactly the same mass. (So why are they different sizes in the picture?) Heat flows out of the hot aluminum (and it's temperature decreases) and flows into the cold iron (and it's temperature increases). Of course, heat could also flow into and out of the surrounding environment. But we eliminate this by carefully insulating the system. This process continues until the block reach the same temperature. Since the masses are the same, is the final temperature exactly halfway between the original temperatures? NO! How much the temperature increases or decreases depends upon the amount (specifically the mass) of the substance AND upon the substance itself. It takes more energy to heat up or cool off the block of Al than it does for Fe. We say the Al has a greater heat capacity. For a substance of mass m, the amount of heat energy DQ required to raise it's temperature by an amount DT, is given by:
The symbol c is the specific heat (capacity) of the substance. It is different for every substance and is determined by experiment. The units for heat energy are the same as for any energy. In the SI system, energy is given in joules and the specific heat has units of J / kg oC. Although we will use these units whenever possible, there are other units that are common. Before it was recognized that heat was indeed energy, the calorie was defined as the amount of heat required to raise 1 gram (that's 1 cubic centimeter in volume) of water by exactly 1 degree Celsius. The specific heat of water is cwater = 1.00 cal / g oC. As it turns out, the specific heat of water is very high compared to most other substances and hence the seas have a major enfluence on the world's weather. To convert to the SI system,
Here's an important bit of trivia. The food Calorie, written with a capital "C", is 1000 calories as defined above. The average human body consumes about 2000 Calories/day or about 8.4 million Joules/day!
The specific heat of several substances in both sets of units are given below.
| Substance | c (J / kg oC) | c (cal / g oC) |
| Water | 4,190 | 1.00 |
| Aluminum | 910 | .22 |
| Iron | 470 | .11 |
| Concrete | ~ 1000 | ~ .2 |
| Pine | ~ 1800 | ~ .4 |
| Marble | 860 | .21 |
Note that it takes nearly 5 times more heat energy for the same temperture increase of 1 kg of water compared to 1 kg of aluminum. You might have already had experience with the low specific heat of aluminum. When you grab a piece of aluminum foil inside a hot oven, it will feel very hot. But once out of the oven, it cools very quickly. Although hot, it does have much heat energy. This is also why hot bread is less likely to burn your mouth than a piece of cheese at the same temperature. Which one one has the higher specific heat?
Problem: How much energy is required to heat 1 liter of water in a kettle from room temperature (about 20 oC) to boiling (100 oC)? How long will this take if the burner is rated at 500 W?
Solution: One liter of water has a mass of 1 kg. The specific heat of water is 4190 J / kg oC. So we have DQ = m c DT = 1.0 x 4190 x (100 - 20) = 3.3 x 105 J. How long will it take to do this? We will assume that all 500 W of power is directed into the water. This 500 J/s = energy / time = 3.3 x 105 / t. Solving for t we get t = 670 seconds or about 11 minutes. Of course, the heating element or flame is also heating the surrounding air and the kettle itself radiates away some of the heat. Do you recall from the Appliances Lab the efficiency you calculated for the hotplate? For most groups it was in the 30% range. That is, only 30% of the electrical energy was transformed into heat energy. That would increase the time we calculated above by a factor of about 3.
Heats of Fusion and Vaporization
It is appropriate that we also mention heat exchange is involved with a tranformation of state, even without a change in temperature. When ice at 0 oC melts into water at 0 oC, heat must flow into the ice. This is called the latent heat of fusion. It is about 80 calories for every gram of ice, or 3.35 x 105 J/kg. Similarly, about 540 cal/g = 2.26 x 106 J/kg must flow into water at 100 oC to convert it into steam (water vapor) at 100 oC. This is the latent heat of vaporization. This is a very large number and is why steam can burn so badly. For every gram of steam that condenses on your cold arm, 540 calories (about 2000 J!) of energy flows out of the steam and into your arm. As you learned in SCI 100, it is the immense energy in the latent heat of vaporization that fuels hurricanes.
Problem: In the previous problem we calculated that it takes 3.3 x 105 J of heat energy to raise the temperature of 1 liter of water from 20 oC to boiling at 100 oC. How much would it take to boil all of this water away into steam? How long would that take?
Solution: The solution is easy. One liter is 1000 grams and every grams needs 540 Joules to evaporate. So DQ = 1.0 kg x 2.26 x 106 J/kg = 2.26 x 106 J. That's 2.26 x 106 / 3.3 x 105 = 6.8 times as much energy! That means it will take 6.8 times longer to boil the water away than to heat it from 20 oC to 100 oC.
Conduction
How does heat "flow" from one place to another? It does so through some combination of three basic processes. One method you studied in SCI 100 and is called convection. In this process, the heat is transferred via a flowing of a fluid. The earth's flowing magma, blowing winds, and the movement of radiator fluid in your car's engine are all examples of this process. But the fluid is heated in one area and cooled in another through direct physical contact in a process called conduction. The third process is radiation and is discussed below.
Conduction is the process in which heat "flows" from a hot region to a cooler one through direct physical contact. Consider a rectangular block of thickness Dx that has one face at temperature T1 and the opposite face at a lower temperature T2. Heat flows from the hot face to the cooler one. But what exactly is flowing? Early in the study of heat, it was thought that a substance called caloric flowed. An the terminology has stuck. What is actually happening is that the vibrational energy (heat energy) of the molecules are being transferred from the hotter regions (faster vibrations) to the cooler regions (slower vibrations). This completely natural process is what causes heat to flow from hot to cold. (Although energy is conserved, this is more than just conservation of energy. If you are interested in the First and Second Laws of Thermodynamics and how they are related to the fascinating topic of Entropy, check out Dave Slaven's Page of Entropy.)
The rate at which heat energy flows from a hot region to a colder one is a function of the temperature difference, the distance over which the heat flows, and the substance through which it flows. This last element should be no suprise. If you are at a campfire and you are given the choice of roasting marshmallows on the end of a wooden stick or a metal coat hanger, which do you choose? The metal has a much higher thermal conductivity than the wood. It will heat up at the end you are holding much more quickly. Thermal conductivity is typically denote as "k". (It is not the same as electrical conductivity, although there is generally a strong correlation between the two.) Consider the rectangular block again. The rate at which heat flows (in units of power, joules/s = watts) through from the hot surface, through the block and to the cold surface is given by
where DT = T1 - T2. The thermal conductivity is high for good conuctors such as metals and low for insulators such as dry wood, concrete and fiberglass. Below is a table with the values for a few substances.
| Substance | k (Watts / m oC) |
| Aluminum | 200 |
| Steel | 40 |
| Concrete | .84 |
| Wood | .1 |
| Fiberglass | .048 |
| Dry Air | .023 |
[A very extensive table of values of many of the material properties we have studied so far can be found at Conversion Factors, Material Properties and Constants provided by Stanford University and MIT.]
Thermal conductivity is an important consideration in several areas. Both the water heater and the refrigerator depend upon good thermal insulation for efficient operation. The walls of most water heaters and refrigerators are filled with fiberglass to reduce heat conduction from or to the outside. Fiberglass is a common insulation for walls in houses in temperate climates, necesary to keep heat from flowing out during the winter months. In the tropics, such considerations are not necessary for insulating from the cold outside. However, if you wish to air condition your home, you should definitely use insulation to reduce the heat conduction into your home. But you may also wish to consider heat conduction through the walls even without air conditioning. Let's look a typical problem.
Problem: At what rate does heat energy pass through a 10 m by 10 m roof that is constructed of 1) 3 mm thick galvanized steel, 2) 2" thick plywood, and finally 3) 4" concrete? Assume the temperature on top of the roof is 90 oF = 32 oC and inside the house is 80oF = 27 oC
Solution: The area is 100 m2 and DT = 5 oC in each case. The solutions can be found by using the table and plugging into our heat conduction equation, but we have to be careful with units. Note that 3 mm = .003 m, 2" = .051 m and 4" = .10 m. From this we get
1) for galvanized steel, DQ / Dt = 6.7 x 106 Watts
2) for plywood, DQ / Dt = 980 Watts
3) for concrete, DQ / Dt = 4,200 Watts
At first glance, the wood roof clearly seems superior for insulation from the heat. But there are several complications that we have not considered. First, we assumed the same 5 degree temperature difference for all three. What actually happens is that the sun begins to heat the top of the roof in the morning and the temperature difference between the top and bottom slowly grows. But it will not be the same for each roof. The galvanize, with a low mass and low specific heat, will heat up much faster, but will not necessarily maintain a 5 degree temperature difference, even with a steady steam of cooler air constantly blowing over the bottom surface. While the wood roof offers better insulation than concrete, a concrete roof typically has a much greater mass than than a wood roof, and consequently takes much longer to heat up. Regardless, a simple galvanized roof is undoubtely the worst option. And there's one more factor to fold into the mix ... absorption and radiation.
Absorption and Radiation
All objects above a temperature of absolute zero radiate electromagnetic energy. At room temperature, almost all of this energy is in the infra-red (IR) region and is perceived as heat. The thermal vibrations of the electrons in the substance create the electromagnetic radiation. As objects become hotter, the frequency at which most of the radiation is emitted increases. The heating elements in an electric stove begin to glow dull red and eventually become orange. If they became even hotter, they would glow yellow and eventually white.
Objects also absorb radiation energy. When radiation strikes an object, one can think of the alternating electric field in the electromagnetic wave as shaking the electrons back and forth, creating thermal or heat energy. It is just the reverse of the radiation process. Objects radiate energy as well as they absorb it. The rate at which objects radiate depends upon several factors. It depends upon the surface area (A), the temperature of the object (Tobj), and upon the emissivity, e, of the material. The emissivity is a number between 0 and 1. Very reflective materials (typically white) have e close to zero, while very absorptive surfaces (typically black) have e close to one. The rate of absorption depends upon these same factors, but it is the temperature of the surrounding environment (Tenv) that conributes to absorption. For an object at a Kelvin temperature Tobj, the rate of radiation (measured in watts) is
where A is the surface area and s = 5.67 x 10-8 W / m2 oK4 is Stefan's constant. The exact same equation applies to absorption when the object is surrounded by an environment of temperature Tenv. The net heat flowing into the object is the difference between radiation and absorption. So it is
Note that when the object is at the same temperature as its surroundings, there is no net heat flow and the object stays at the same temperature ... good common sense.
We are particularly interested in the absorption of heat from the sun. As we pointed out in the previous module, the sun provides an intensity of about 1400 W/m2 just outside the earth's atmosphere. How much of that reaches the earth is a function of such variables as cloud cover, humidity and where the sun is in the sky. (These are subjects to be considered in the Direct Solar Heating special project.) But the most you can expect to reach the earth's surface at noon on a clear day here in the Caribbean is about 1000 W/m2. The amount that a surface directly facing the sun can absorb is given by the expression:
The 1000 W/m2 value represents the maximum. Areas north of latitude 23oN or south of latitude 23oS will always have less than this. If the surface is not perpendicular to the sun's rays, the area A should be modified accordingly. Despite all these possible variations, this equation will allow us to make some good estimates.
Problem: At what rate is energy absorbed by a 10 m by 10 m roof that is painted white (emissivity = .1)? What if it was painted dark red (emissivity = .8)? Estimate the rate at which the roof would radiate energy away.
Solution: For a white roof at noon on a clear day, the energy abosrbed from the sun would be DQ / Dt = (1000 W/m2) .1 (10 x 10) = 10,000 W. This is a lot of energy and would heat the house up very quickly. But as the roof begins to heat up, it also begins to radiate energy. We could make a conservative estimate that the roof was at a very modest 100 oF = 38 oC = 311 oK, while the surrounding air is at 70oF = 21oC = 294 oK from which we get
DQ / Dt = es A (Tobj4 - Tenv4) = .1 x 5.67 x 10-8 x (10 x 10) x (3114 - 2944) = 1000 W
The total energy flow into the roof would be 10,000 - 1000 = 9000 W. Note that the absorption is much higher than the radiation. The roof would heat up until it radiated the same amount of energy as it absorbed. But it is important to remember that 10,000 W would be at noon only and that 100oF is only an estimate. As the sun rises and sets, the effective incident radiation goes from zero to the maximum amount and back to zero again. The average is closer to half this value. Note also that the rate of conduction through the roof is important. A galvanize or thin wood roof will result in a much hotter home interior than a concrete or insulated roof. And finally, we did not include heat radiating from the inside surface.
If we used dark red, the absorption increases by a factor of eight. The roof is going to heat up quickly and will have to get very hot before it can radiate this much energy away.
And now, for some questions and problems.
| RETURN |