Thermal Conductivity Printer Friendly Version
Background

Heat can be transferred from one point to another by three common methods: conduction, convection and radiation. Each method can be analyzed and each yields its own specific mathematical relationship. We will use Pasco’s Thermal Conductivity Apparatus to investigate the rate of thermal conduction through three common materials used in building construction. The equation giving the amount of heat conducted through a material is:

Solving for the conductivity constant, k, we have the equation

As you see from our equation, to calculate k we need to know the mass of ice melted in grams (m), the thickness of the insulting plate (), and the area of the ice cylinder that was in contact with the insulating plate (A). The latent heat of fusion is a given, Lf = 334 J/g, and the temperature differential between the ice temperature (0ºC) the steam’s temperature (100ºC) is known.

We will conduct the experiment six times. Three times for ambient heat to calibrate the apparatus and then three times with steam.

Directions for Part I

1. Measure and record , the thickness of the insulating plate.

2. Mount the insulting plate onto the steam chamber taking care that it is flush against the water channel (to insure that no water will leak) and then tighten the thumbscrews. The manufacturer recommends the use of a little Vaseline between the channel and the sample to help create a good seal.

3. Initially let the ice sit for several minutes so it begins to melt and comes in full contact with the sample. Don't begin taking data before the ice begins to melt, because it may be at a lower temperature than 0°C. Obtain data for determining the ambient melting rate of the ice, as follows:

4. Determine the mass of the beaker used for collecting the melted ice and record it below.

5. Collect the melting ice in the beater for a measured time (approximately 5 minutes).

6. Determine the mass of the beaker plus the melted ice-water and record it. Then subtract your first measured mass from your second to determine the mass of the melted ice.

Repeat this process for all three insulating plates. Record your results in Table I below.

Data from Part I (ambient temperature)

 mass of empty beaker in grams

 temperature of room in Celsius

Table I - Ambient Temperature

 insulating platedescription total waterand beaker (g) ambient water(g)
 glass
 wood
 gypsum

Directions for Part II

7. Next run steam into the steam chamber. Let the steam run for several minutes until temperatures stabilize so that the heat flow is steady. Don't forget to place a container under the drain spout of the steam chamber to collect the water that escapes as the steam cools.

8. Measure the diameter of the ice block. Record this value as d in Table II given below. Place the ice on top of the insulating plate. Leaving the ice in its mold, just place the open end of the mold against the insulting plate, and let the ice slide out as the experiment proceeds.

9. Calculate the surface area between the ice and the insulting plate.

10. Repeat step 5, but this time with the steam running into the steam chamber. As before, measure and record the mass of the melted ice in grams and the time during which the ice melted (5 minutes).

Repeat this process for all three insulating plates. Record your results in Table II below.

Data from Part II (steam)

Table II - Steam-Filled Chamber

 insulating platedescription thickness(m) diameter(cm) area(m2) total waterand beaker (g) melted ice fromsteam (g) k(watt/mK)
 glass
 wood
 gypsum

Conclusions

 Rank your insulating plates in order of best insulator to worst insulator.

 Based on your results, calculate the conductivity of a wall made from our gypsum dry wall covered with our wood paneling.

 Based on your results, calculate the heat transferred over an 8 hour day through a glass window that measures 3 feet by 3 feet if the interior temperature is 72ºF and the exterior temperature is 106ºF. NOTE: There are 3.28 feet in a meter and the conversion between Fahrenheit and Celsius is ºC = 5/9 (ºF - 32).

 If your local power company charges \$0.085 per kwh, how much money is lost to cooling the room in which this window is located?

Related Documents