Why is the Earth Still Hot Inside?

Students will learn about heat and temperature.

Students will learn about rates of heat transfer as a function of the size of the object: doubling the radius of a sphere increases its surface area by a factor of 4 but increases its volume by a factor of 8. Therefore it transfers heat at a rate ½ the rate of the smaller sphere. The rate of heat transfer is directly proportional to surface area but inversely proportional to volume.

Students will gain proficiency in using lab equipment to make repeatable measurements.

This activity is best utilized with students in 11th grade chemistry or 12th grade physics. Students would benefit from prior exposure to thermochemistry and/or basic concepts of heat capacity and measurement of heat using calorimetry. This is an inquiry lab experience that may be completed in as little as two hours or may take four hours or longer. The length of time is up to the instructor's discretion and should depend on whether students are making productive discoveries. Students should work in groups of 2 to 4. The complexity of the activity can be modified for more advanced students to include calculations of heat content at different temperatures so that a plot of heat loss rate vs. area and vs. volume may be plotted to directly establish the nature of the proportions.

Background

The oldest structures in the solar system are small spheres called calcium-aluminum inclusions (CAIs). These have been found in certain classes of meteorites that represent material that has remained unchanged since the formation of the solar system. CAIs have been dated to 4.567 billion years ago and the Earth is thought to have completed its formation between 30 and 90 million years later. This puts the formation of the Earth at about 4.53 to 4.47 billion years ago. The process of forming rocky planets is a violent one in which the planet grows larger through collisions with other bodies. Some of these collisions are extremely energetic and some are so large that the entire surface of the planet becomes a magma ocean. The Earth's Moon is thought to have formed as the result of such a collision.

The heat within the Earth can only be lost to space by radiation (mostly in the infrared part of the electromagnetic spectrum). The process of planet formation creates an enormous amount of heat through two mechanisms. First, the collisions that lead to the growth of a planet are very energetic and the largest of these can release more energy than any of the nuclear bombs in military arsenals. This is a powerful source of heat. Second, as the planet forms the denser materials sink toward the center of the planet. This releases gravitational potential energy as heat. In addition to these ancient heat sources, heat continues to be produced within the Earth by the radioactive decay of uranium, thorium, and potassium-40, though this source has diminished in size since the formation of the Earth. So why has this heat not all been lost to space in the past 4.5 billion years?

The Earth is still so hot inside in large part because its surface is too small, relative to its volume, for this heat to escape. The amount of heat energy is proportional to the mass (or the volume, since mass is proportional to volume). The rate of heat loss is proportional to the surface area. Small objects can radiate heat away quickly because the surface area is large relative to the volume compared to larger objects. For example, smaller planets such as Mercury and Mars are thought not to contain a liquid outer core, as does the Earth. Small moons of Jupiter and the other gas giant planets are normally expected to have completely solidified. Recent findings indicate that Europa (an icy moon of Jupiter) and Enceladus (a smaller icy moon of Saturn) both show signs of current melting and the possible extistence of sub-surface liquid water. These bodies demand a different explanation of the interior heating evidenced by recent observations by the Galileo and Cassini spacecraft. They are much smaller than the Earth and so should not have much internal heat. Scientists are currently unsure what leads to the internal heating of these small bodies (perhaps tidal interactions play a role) but for the Earth it is certain that the small surface-to-volume ratio plays an important part in preventing heat loss from the Earth.

When a sphere is scaled up by doubling its radius the surface area is scaled up by a factor of 4 and the volume is scaled up by a factor of 8. This can be understood perhaps most simply by examining the formulas for calculating surface area and volume. Surface area is proportional to the square of the radius (SA = 4πr²) and volume is proportional to the cube of the radius (V = 4/3πr³). Generally, if a body is scaled up by a length factor of k then the area scales up by k² and the volume scales up by k³. The area/volume ratio scales down by a factor of 1/k (k²/k³) as an object is scaled up.

It is for this reason that polar bears are larger than black bears. Polar bears (8 – 10 ft) stand twice as tall as black bears (4 – 6 ft). As a result they have about 2³ = 8 times as much volume. Both animals must maintain a constant internal temperature throughout their bodily volume. All body heat that escapes must escape through the skin and polar bears do not have 8 times as much skin; they have only 4 times as much. This means that heat escapes from a polar bear's body only ½ as efficiently (4/8) as from a black bear's body. Polar bears have a clear advantage in their cold arctic habitat because of this reduced efficiency of heat loss. Of course, bears are not planets but this analogy may help in understanding the relevance of scaling effects on the rate of heat loss.

In this activity students will be given an opportunity to heat an equal mass of either small glass spheres or larger spheres (the number of spheres will not be the same: clearly there will be a larger number of small spheres than large ones). Each group may choose the size it will test. Since the mass is equal, the amount of heat absorbed by either type of sphere will be the same, as long as they are made of the same material. These heated spheres will be placed in a calorimeter containing a set amount of water. The amount of temperature gain by the water will be the same for both the small spheres and the large spheres but the larger spheres will require a longer time to heat the water fully. Different groups will each have only one size of sphere at first. After they have collected several trials' data a class discussion will be held in which the data are plotted together from different groups. Ideas about total change in temperature, rate of heat loss, and size will be discussed. Heat loss rate can be calculated at change in temperature over change in time:

∆T/∆t 1.

If one or more groups have very different plots then the issue should undergo troubleshooting. It is a great opportunity to talk about how important it is to run down sources of error. At times they reflect a blunder of one kind or another but at other times an outlier value may hint at some as-yet unexplored phenomenon.

Note for advanced users: First have students measure the heat capacity of the glass spheres and ensure that it is the same for the small and large spheres. Then have them control their experiments based on the amount of heat, not the mass of glass. Heat can be calculated using the following equation:

Q = mC∆T 2.

Where Q is the heat in joules (J), m is the mass in grams, C is the heat capacity in J/g-ºC, and ∆T is the change in temperature measured in ºC. Since the heat capacity of water is 4.18 J/g-ºC the amount of heat transferred to water can be calculated using this equation. Then, with the known value of heat, mass of glass, and ∆T the heat capacity of the glass can be calculated. This allows students to calculate heat loss rate as ∆Q/∆t rather than as ∆T/∆t (where the small t represents time in seconds). Ultimately, this may lead to more understandable results. than as ∆T/∆t (where the small t represents time in seconds). Ultimately, this may lead to more understandable results.

In-class Activities

The teacher directs the class in the following steps with or without a written procedure, as per the preference of the instructor.

1. Students prepare their lab notebooks to collect temperature data as a function of time. Time intervals should be no longer than 20 sec and no shorter than 5 sec.
2. Students are given a choice of pre-measured samples of large or small spheres.
3. Students make a 100ºC hot water bath by boiling water over a burner or electric heater. They add the spheres to this bath carefully, to avoid splashing.
4. They allow the spheres to come up to the temperature of the water bath.
5. While they wait they prepare a styrofoam cup (which can be nested for further thermal isolation) with 50 mL of water and a thermometer. The thermometer can be clamped to a stand for convenience.
   Once sufficient time has passed to ensure that the spheres have been heated through they are transferred to the styrofoam cup (this may require some creativity to enable this transfer with a minimum of time out of the water―perhaps one at a time with tongs or tweezers).
6. As soon as the glass is in the styrofoam cup students record the temperature as a function of time at intervals between 5 sec and 20 sec.
7. They plot this data as a graph in their lab notebooks with time on the x-axis and temperature on the y-axis.
8. Students should repeat this procedure a few times in each group to see whether they can get relatively consistent results. They should be encouraged to troubleshoot any trials which have big deviations.
9. Finally, they should calculate the change in temperature over the change in time (or change in heat, if the class is using that option).
10. The teacher can call the students together for a group discussion once all groups have collected a few trials' of data. First, a class-wide plot of data for each size of sphere should be made. Then the teacher can ask questions such as the following:
• Are there any problems with these data? What are they and how can we explain them or fix them?
• What is happening? (Heat is transferred from glass to water at different rates).
• What stays the same for both the small and large spheres?
• What is different?
• Let's try to explain both of these.
• Are there any experiments we can try to test our explanations? Let's go try them.
11. Go back to the lab and try the new experiments suggested by the data and discussion.
12. Conduct further discussion as necessary to work toward the conclusion that the rate of heat loss by the larger sphere is lower by a factor of 1/k where k is the number of times larger the radius of the larger sphere is than the radius of the smaller sphere.In-Class Activities

At Home Assignments

Students will write a formal lab write-up as described at http://kaffee.50webs.com/Science/labs/Lab-Report.Writing.html

The report will describe the experiments done within the group and within the class. It will display the shared data. Each student must write their own interpretation of the data and discuss its relevance to the formation of planets and to comparing planets of different sizes.

Materials

250-mL beakers
burners or heaters
large glass spheres (sufficient for several groups)
small glass spheres (sufficient for several groups)
styrofoam cups
thermometers
lab notebooks
safety goggles
tongs and/or forceps (small tweezers)

Standards

This is not particularly relevant in my state.

I calculated that 25 g of glass with a heat capacity of 0.75 J/g-ºC at 100ºC would be able to raise the temperature of 25 mL of water about 13.5º. If 50 mL of water is used then the ∆T would be about 7º. This seems do-able. Also, if the density of the glass is near 2.2 g/mL then it would have a volume of 11.4 mL and should be able to be submerged in 25 – 50 mL of water.

I have not yet determined whether I can get enough marbles (glass spheres) of sizes that would be useful in this activity. Ideally, the larger spheres should have a radius at least 3 times the radius of the smaller spheres.

It may also be problematic to arrange samples of large and small marbles that have the same mass. In that case it may be necessary to measure the heat capacity of the glass and use the heat content to calculate ∆Q/∆t rather than ∆T/∆t.

Assessment will be based on the formal lab write-up produced by individual students.