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Project Grant Team
John S. Pazdar Patricia L. Hirschy Project Director Principal Investigator Capital Community College Asnuntuck Community College Hartford, Connecticut Enfield, Connecticut
This project was supported, in part, by the Peter A. Wursthorn National Science Foundation Principal Investigator Opinions expressed are those of the authors Capital Community College and not necessarily those of the Foundation Hartford, Connecticut
SPINOFFS
Spinoffs are relatively short learning modules inspired by the LTAs. They can be easily implemented to support student learning in courses ranging from prealgebra through calculus. The Spinoffs typically give students an opportunity to use mathematics in a real world context.
LTA - SPINOFF 11A Newton’s Law of Cooling and Heating:
Vaporization of Liquid Helium
LTA - SPINOFF 11B The Combined Gas Law
Dennis Ebersole - AMATYC Writing Team Member Northampton Community College, Bethlehem, Pennsylvania
George King - AMATYC Writing Team Member North Florida Community College, Madison, Florida
Thomas Elam - NASA Scientist/Engineer Kennedy Space Center, Florida
NASA - AMATYC - NSF
SPINOFF 11A
Newton’s Law of Cooling and Heating: Vaporization of Liquid Helium
Problem
The Kennedy Space Center (KSC) receives helium in liquid form in over-the-road tankers. As needed, this helium is vaporized, compressed, and fed into A high pressure pipeline that distributes gaseous helium to Shuttle Launch Pads A and B the three Orbital Processing Facilities (OPFs), the Vehicle Assembly Building (VAB), and other KSC support Facilities. The helium is stored at – 271.5º C. Assume that the helium tank was removed from its refrigeration unit on a day when the temperature of the surrounding air was 80º F. Also, assume that ten minutes after the helium was removed from the refrigeration unit, you measured its temperature at – 271º C. How long will it take for the temperature of the helium to reach – 269.5º C, the temperature at which helium vaporizes?
In order to solve this problem, we need to know how the temperature of a liquid changes with time. Isaac Newton first described this process in a law which later became known as Newton’s Law of Cooling and Heating. If you are not familiar with Newton’s Law of Cooling and Heating, click here.
NASA - AMATYC - NSF
We rewrite this differential equation as
dy kdt. y
= − Integrating with respect to t , we find that
the solution of the differential equation is ln y = − kt + C. Changing this to exponential form
gives y = e −^ kt^ + C^^ = e − kt^ ⋅ eC. (Recall that x a^ ⋅ x b^ = x a^ + b .) Thus, y = ± e e C^ − kt. Since C represents
any real constant, ± eC represents any non-zero real constant. Thus, y = Ae − kt is a solution of the
differential equation (1) where A is any non-zero constant. In fact, A can even be allowed to equal 0, since the function y = 0 is a solution of the differential equation (1).
At time t = 0, y is the difference between the initial temperature T 0 of the coffee and the room
temperature TS. Thus, at time t = 0, y = T 0 (^) − TS = Ae − k (0)= A.
Since y = T − TS , Newton’s Law of Cooling becomes
( 0 ) or ( 0 ) kt kt T T S T TS e T TS T TS e − = − −^ = + − −
Recall that the coffee was heated to 170º Fahrenheit in a room with an ambient temperature of 70º F. This means that T 0 = 170º F and TS = 70º F. Thus, the temperature vs. time function
becomes:
T = 70 + (170 − 70) e −^ kt^ or T = 70 + 100 e − kt
The function is complete except for the value of k. We will now estimate k by plotting the data from the table and then using trial and error to find the value of k that appears to make the model fit the data best. It will be easiest to use technology to plot the data and to graph cooling functions for trial values of k. The keystrokes to do this are given below.
Enter the data in a TI-83™ graphing calculator using L1 for the time values and L2 for the temperature values. The key strokes to do this are shown below.
STAT 4[ClrList] 2 nd^ L1, L2 ENTER {This clears the two lists.} STAT ENTER {You are now ready to enter the data.} 0 ENTER 2 ENTER {Continue in this way until all times are entered in L1.} → {Now enter the temperature data in L2.} We can now create a scatterplot of the data.
2 nd^ STAT PLOT ENTER ENTER {Turn on the first statistics plot.}
↓ ENTER {Turn on the scatterplot.} ↓ ENTER {Let L1 be the x-values.} → → ENTER {Now change the Window parameters to [0, 50, 5, 0, 200, 10].} GRAPH
NASA - AMATYC - NSF
Now access Y= and type in the function T = 70 + 100 e − kt using a reasonable value of k. Graph the function on the screen with the scatterplot and observe how the curve fits the data. Continue this process using different values of k until you find a value for which the function appears to fit the data best.
The following function appears to fit the data quite well.
T = 70 + 100 e −^ 0.045 t
Now that you have gained some familiarity with Newton’s Law of Cooling and Hearting we turn to another problem.
Warming orange juice: A cup of orange at 34º F is removed from a refrigerator and placed in a room with temperature of 72º F. Twenty minutes later the temperature of the orange juice is 40º F. Find the function that relates temperature T and time t.
Solution
It should be noted that Newton’s Law of Cooling and Heating also applies to a quantity that is
becoming warmer. Thus, the applicable model is T = T S +( T 0 − TS ) e − kt. At time t = 0, the
temperature is 34º F. Thus, T 0 = 34º F and TS = 72º F. Substituting these values into the
equation gives:
T = 72 +(34 − 72) e − kt or T = 72 − 38 e − kt
Since T = 40º F when t = 20 minutes, it follows that 40 = 72 − 38 e − k^20. Solving for k we get:
ln 38 0.
20 20
k
= ^ = =
This implies that T = 72 − 38 e −0.0086 t.
Click here when you are ready to return to Newton’s Law of Cooling and Heating: Vaporization of Liquid Helium.
