The only remaining heat transport mechanism that we will study in this section is heat conduction. We will give the students an opportunity to see head conduction in action and to make measurements of this heat transport property.
By watching cold water become warm and hot water become cool via heat conduction, the student will see that heat can be transported by a thermal conductor. The students will hopefully see that the rate of heat transport can be altered by the difference in temperature of the thermal sources, the cross-sectional area of the thermal conductor and the length of the thermal conductor.
When the students arrive, the instructor should have a beaker of boiling water on a ring stand and a beaker of ice water on a ring stand with several heavy copper wires between them.
Invite the students to touch the copper wire. Start and the end that is submerged in the ice water and carefully move down the wire. BE CAREFUL. The rod can be very hot. Ask them to notice what is happening to the ice. It should be melting.
When learning new concepts, we are often held back by old concepts. In this case, the students may struggle with the idea that we are not looking at an equilibrium case but a dynamic system. We are not interested in the end result but the rate at which the systems change.
Review the steps in the scientific method. Ask the students to devise a hypothesis or hypotheses about the way the heat moves through a thermal conductor. They may have trouble getting to the point where they realize that they have to measure the temperature of the water as a function of time. Encourage them to devise and perform experiments to test their hypothesis. Some may choose to use a Bunsen burner to keep one end of the copper hot. Other may choose to start with cold water on one and hot (but unheated) water at the other. Don't let them waste too much time trying to find an equilibrium situation. Help them to understand this is a dynamic system and we are interested in the rate of change of temperature.
Here is a quick math review for the instructor. The students should not see this because it involves calculus notation and they are to devise equations similar to these.
The rate of heat transfer should be proportional to the difference in temperature between the two ends of the copper wire.
The rate of heat transfer should be proportional to the cross sectional area of the wire.
The rate of heat transfer should be inversely proportional to the length of the wire.
In class today, we discovered that the transfer of heat was faster when the temperature between the ends of the copper wire was larger. We also discovered that when we used more wires or a thicker wire, the transfer of heat was faster. We might have discovered that when the copper wires were longer the transfer was slower. Let's think about the implications of these relationships in our houses. (If you did not devise the mathematical expressions for these ideas, make some reasonable guesses as to what they should be. Write them down and use them to answer the questions below.)
Suppose it costs $1/hr to keep our house at 20°C when the temperature outside is 0°C. What is the hourly cost to keep our house at 20°C when the temperature outside is -20°C? Assume we lose all our house heat through thermal conduction through the walls, ceiling and windows.
Suppose it costs $1/hr to keep our house at 20°C when the temperature outside is 0°C. We decide that we can reduce our expenses by improving the insulation. We make the walls twice as thick and use the same insulation type as the house originally had. We also double the thickness of insulation in the ceiling/attic. Now, what is the hourly cost to keep our house at 20°C when the temperature outside is 0°C?
Suppose it costs $1/hr to keep our house at 20°C when the temperature outside is 0°C. We decide that we can reduce our expenses by moving into a smaller house with identical construction (same materials and wall thickness) and identical insulation. Our original house was 10m by 20m in size. Our new house is 8m by 15m in size. What is the hourly cost to keep our house at 20°C when the outside temperature is 0°C? Be sure to include the roof in your calculation. Omit the floor.
Suppose it costs $1/hr to keep our house at 20°C when the temperature outside is 0°C. We decide that we can reduce our expenses by lowering the temperature inside the house. We decide we can withstand a temperature of 15°C. What is the resulting hourly cost for heating our house?
You can imagine that heat is transferred more rapidly through some materials. For example, you might think that wood transfers heat poorly when compared to copper. Glass might conduct better than wood but not as well as copper. Let's assume that glass windows conduct heat 10 times "better" than the walls of our house. Make a few approximations about your own house and estimate the ratio of heat that is transferred though the windows as compared to the walls.
Using the ideas we developed today, and making some reasonable assumptions, show the ratio of heat transfer between single pane glass windows, double pane glass windows, and triple pane glass windows. Double pane windows use two glass panes separated by a thin layer of air. Triple pane windows use three panes of glass.
Note to the instructor: This is a fairly difficult problem. It is probably more appropriate to use this problem as a classroom example problem.
Concepts: In the double pane situation, there are three layers, glass, air, and glass. For the problem, treat each layer as thermally conductive, although one could argue that the air layer is convective. The thermal conductivity of each layer can be described with an equation of the following form:
The rate at which heat flows through each layer must be equal since all the heat has to flow through each of the three layers. This fact forces the ΔQ/Δt term in each layer to be the same. The factor of k is a different constant for air and glass. It describes the thermal conductivity, the area, and the thickness of the material. When we take all of this into consideration, we can write two equations.
The Ti are the temperatures of the glass surfaces from inside the room, T0, to outside, T3. The ka represents the conductivity of air and kg represents the conductivity of glass. We know the inside and outside temperatures so we have two equations and two unknowns. Solve for the unknown temperatures.
The last part of the problem asks the students to compare the double (or triple) pane heat transfer rate to the single pane heat transfer rate. They will find that the double pane problem produces an equation that is exactly like the single pane equation except that the temperatures are different. So the ratio of the rates of heat transport is the ratio of the temperatures across a single pane of glass. That is,
The denominator of the equation above represents the single pane case where the difference in temperature is calculated from the inside and outside temperatures used in the double pane case. The numerator represents the heat flow through the "outer" layer of the double pane system. Any layer could be used in the numerator because the heat flow is the same through all the layers. I like to use the glass layers because the value of kg divides out leaving only the temperatures.