This exam was adminstered in January 2024 .
June 2024 Algebra 2, Part IV
A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.
37. Megan is performing an experiment in a lab where the air temperature is a constant 73°F and the liquid is 237°F. One and a half hours later, the temperature of the liquid is 112°F. Newton’s law of cooling states T(t) = T_{a} + (T_{0}  T_{a})e^{kt} where:
T(t): temperature, °F, of the liquid at t hours
T_{a}: air temperature
T_{0}: initial temperature of the liquid
k: constant
Determine the value of k, to the nearest thousandth, for this liquid.
Determine the temperature of the liquid using your value for k, to the nearest degree, after two and a half hours.
Megan needs the temperature of the liquid to be 80°F to perform the next step in her experiment. Use your value for k to determine, to the nearest tenth of an hour, how much time she must wait since she first began the experiment.
Answer:
In the first part, to solve for k, you have to substitute all the other variables and then isolate k.
112 = 73 + (237  73)e^{1.5k}
39 = 164e^{1.5k}
39/164 = e^{1.5k}
ln(39/164) = 1.5k
1.4363 = 1.5k
0.9575 = k
k = 0.958 to the nearest onethousandth of an hour.
For the second part, substitute t = 2.5 hours.
T(2.5) = 73 + (237  73)e^{2.5(.958)}
T(2.5) = 87.9523 = 88 degrees.
For the final part, set T(t) = 80 and solve for t.
80 = 73 + (237  73)e^{.958t}
7 = 164e^{.958t}
7/164 = e^{.958t}
ln(7/164) = .958t
3.1539 = .958t
3.2922 = t
It would take about 3.3 hours.
End of Exam
How did you do?
Questions, comments and corrections welcome.
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