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FACULTY NOTES
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FACULTY NOTES
SPINOFF 11A
Newton’s law of Cooling and Heating: Vaporization of Liquid Helium
Solution
We start with the following general function that relates the temperature, T , of a liquid to the time, t, that it has been cooling down or heating up.
kt T T S T TS e = + − − (1)
We are given that the ambient temperature is 80º F. This is equivalent to 26.66º C as shown in the conversion below.
5 (80 F 32 F) 26.6 C 9
o (^) − o (^) = o
The ambient temperature, TS = 26.66º C and the temperature at time 0, T 0 = –271.5º C.
Substituting into equation 1, gives:
T = 26.66 + −( 271.5 − 26.66) e − kt (2)
Since the temperature is –271º C when t = 10 minutes, we can use equation 2 to write the following:
− 271 = 26.66 + −( 271.5 − 26.66) e −^ k ⋅^10 (3)
We now solve equation 3 for k to obtain:
10 ln 0. 298.16 10
k k
−^ −
−^ −
Thus, the equation that describes the relationship between the temperature of the helium and the time is:
T = 26.66 −298.16 e −^ 0.000168 t
We are now able to find the time when the temperature reaches –269.5º C.
269.5 26.66 298.16 0.000168^ 0.000168 ln 296.16 40.06 minutes
− = − e^ −^^ t ⇒ − t = ^ − ⇒ t = (^) −
