Among the more important and practical aspects of Basic Thermodynamics, one finds heat conductivity. This is especially useful in the design and construction of buildings to ensure the optimum materials are used, say to make possible staying warm in harsh winters, or staying cool in incendiary summers (which we'll soon see with global warming).
A very basic laboratory experiment for the investigation of heat conductivity is shown in Fig. 1. Also included is the corresponding diagrammatical layout showing the components parts, including: different thermometers (which will be at different temperatures t1, t2, t3 and t4), steam inlet and outlet pipes (left side), steam jacket and water jacket.
In the experiment we pass steam through the steam jacket and adjust the flow of water through the water jacket to a small stream. After a while, a steady-state flow (indicated by a constant difference (t2 - t1) will be achieved, whereupon the flow of water is adjusted to give a difference betwen thermometers t3 and t4 of about 10 F. One continues observations of the readings of all 4 thermometers until a steady state condition is reached.
Once this is established, we read and record: t1, t2, t3 and t4, and catch all the warm water flowing out of the water jacket for a time interval T ~ 10 mins. (The longer the duration of a given trial, the more accurate the results. Needless to say, the thermometers ought to be scrutinized carefully throughout and if any marked fluctuations occur, a new trial should be started, because otherwise the experimental errors will be too large. Finally, one determines the mass of water collected, records the time interval T, and the readings of the four thermometers. The distance L between the thermometers t1 and t2 will also be measured, as well as the diameter d of the test rod.
During each test trial, the value of heat Q transferred to the water is determined, which will be estimated by:
Q = k A T(t2 - t1)/L
where k denotes the 'thermal conductivity' of the material (which will be provided), A is the cross sectional area, L the length, and (t2 - t1) the temperature difference. If a known mass of water (m) passes through the jacket then the total heat received from the end of the test rod will be:
Q = mc(t4 - t3)
Of course, the experiment can also be performed with the ojective of determining k, the thermal conductivity. If this is the case one will make use of the relationship:
k A T(t2 - t1)/L = mc(t4 - t3)
so that, on solving for k:
k = mc(t4 - t3)L/ A T(t2 - t1)
In Fig. 2, a simple diagram is shown which describes the basic principle of heat conductivity. The temperature gradient is defined: (T2 - T1)/ x and the heat passing thorugh per second: Q/t = k(T2 - T1)/x. That is, the product of the thermal conductivity by the temperature gradient.
Example Problem:
Find the quantity of heat Q, transferred through 2 square meters of a brick wall 12 cm thick in 1 hour, if the temperature on one side is 8 C, and the temperature is 28 C on the other. (k = 0.13 W/mK). Then:
Q = kAt[T2 - T1]/ x
Q = (0.13 W/mC)(2 m^2)(3600 s) [20 C/ 0.12 m]
= 156, 000 Joules
Problems:
1) A student performs the heat conductivity experiment as shown in Fig. 1, and determines the thermal conductivity of copper to be 390 W/mC. If he then measures the thermometer differences (t4 - t3) = 5 C and (t2 - t1) = 2C, using 0.5kg of warmed water, and his copper test rod for the experiment is 0.5 m long, what would its cross-sectional area A have to be? (Take the specific heat capacity of water = 4200 J/kg K). Also, obtain the % of error in the student's result by looking up the actual thermal conductivity of copper.
2)A plate of copper 0.4 cm thick has a temperature difference of 60 C between its faces. Find: a) the temperature gradient, and b) the quantity of heat that flows through each square centimeter of one face each minute?
3)How many calories per minute will be conducted through a window glass 80 cm x 100 cm by 2mm thick if the difference between the two sides is 20C?
4) A group of 4 astronauts lands on Mars with solar radiation collection material of total area 2000 m^2. If the efficiency of the material is 30%, and the ambient night time temperature on Mars for their base location (Isidis Planitia) is -40 C (10C day time), will they have adequate collecting material if the solar constant on Mars is 620 W/m^2? (Assume insulating material with a thermal conductivity of 0.08 W/mC, and a need to keep the inside area of their domecile at least at 10 C, requiring solar radiant energy collected of at least 1,200 W per minute for an area of 10 m x 10 m.) Estimate the thickness of insulating material they're likely to need in order to make it work. Comment on whether this expedition is even feasible given the limits of their materials, and that no more than 100 m^3 of insulating material can be taken.
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