Lab # Project II
# Two identical jars of boiling water, labeled A and B, stand beside
# two identical jars of ice water , labeled C and D, in a room whose
# temperature remains constant. The contents of jars A and C are
# immediately mixed. Jars B and D are kept separate for a few minutes
# and then mixed. How will the temperatures of the two mixtures
# compare? One school of thought says that the AC mixture will be
# hotter because jar D, being very hot will lose its heat much more
# rapidly than the cooler AC mixture and the result will be that more
# heat is lost by B alone than by the AC misture. Another says that the
# hot water from A and the cold water from C will act independently,
# just like the separate portions in B and D and therefore the mistures
# will be the same temperature.
#
# Run a number of (theoretical) experiments, varying the temperatures
# of the room and water, the quantities of hot and cold water, and the
# times. Prepare a report on the experiments, complete with an
# explanation of the calculations (the theory) , a separate section of
# results and a third of interpretation of the results relative to the
# original question. In a final section provide a theoretical
# explanation of the results of the experiments.
#
# Alternative Formulation
#
# Ms. Karen Heavin suggests the following alternative statement for the
# lab. You may report on the original, this, or your own version.
#
# There were once three friends, Carl, Paul, and John. In an attempt
# to survive the last two weeks of the semester they had begun a nightly
# ritual of studying into the wee hours of the morning, replacing sleep
# with coffee. During one of their sessions, after consuming about 15
# cups of coffee each, Carl decided to go run around the block to
# control the shaking of his limbs from too much coffee. Just as he
# was about to get up, John inadvertently poured Carl a full cup of
# fresh coffee. Carl said not to worry about it he would drink it when
# he got back. John then offered to put cream in the coffee so it would
# be "just right" when Carl got back from his run. Carl objected
# strenuously to this stating that the creme would cause his coffee to
# be too cold when he got back. Upon hearing this Paul assured Carl his
# coffee would not be any colder than if he left it alone and poured the
# cream in when he got back. John felt Paul was in error and told him
# so.
#
# Anyway, Carl started to prepare for his run but fell asleep as he
# sat on his bed to put on his running shoes. John and Paul argued all
# night over the timing the cream pouring (rather than studying
# calculus).
#
# The next morning Carl found his friends still arguing. However,
# after a nights sleep he could think straight. He drank the cold cup
# of coffee, then quickly put the argument to rest by modeling the
# problem with Newton's law of cooling (or whatever), doing some
# exprimental calculations, looking at some graphs and then explaining
# the results. Then Carl went off to school while John and Paul fell
# asleep.
#
#
#
# GETTING STARTED: One will need to assume that the constant, k , in
# Newton's law of cooling is the same for all aspects of the problem. In
# addition one will need to assume that the temperature of a mixture of
# two containers of water will be the weighted average of the
# temperatures of the two components.
#
# The following takes one through a related problem and contains all of
# the Maple commands and ideas one might need.
#
# Problem : Two of cups soup, the first at 90 C and the second at 100 C
# are put in a room where the temp is maintained at 20 C. The first cup
# cooled from 90 C to 60 C after 10 minutes, at which time the second
# cup is put in a freezer at -5 C. Assume Newton's law of cooling and
# answer the following questions.
#
# (a) Express the temperature of each cup as a function of time, and
# draw their graphs.
#
# (b) How much longer will it take the two cups to reach the same
# temperature?
#
# Solution: Assuming Newton's Law of cooling, the rate at which each
# cup cools is proportional to the difference between the ambient
# temperature and the temperature of the soup in the cup. The
# differential equation for each cup looks like this:
> restart;
> Diff(f(t),t) = k*(f(t)-20);
d
-- f(t) = k (f(t) - 20)
dt
# This is a separable differential equation, whose solution can be
# written in the
# form
> T := 20 + (T0-20)*exp(k*t);
#
T := 20 + (T0 - 20) exp(k t)
# where T = T0 at t=0 and k is determined by the information given
# that the first cup cools to 60 C in 10 minutes (i.e. T=60 at t=10).
#
# So if we denote the temperature of the first cup after t minutes by f
# then
> f := subs(T0=90,T);
f := 20 + 70 exp(k t)
> k :=solve(subs(t=10,f) =60,k);
k := 1/10 ln(4/7)
> f := unapply(f,t);
f := t -> 20 + 70 exp(1/10 ln(4/7) t)
# Assume that this value of k is also valid for the second cup. (Why
# shouldn't it be? They are both soup.) So the second cup's temperature
# is given by
#
> g := subs(T0=100,T);
g := 20 + 80 exp(1/10 ln(4/7) t)
> g := unapply(g,t);
g := t -> 20 + 80 exp(1/10 ln(4/7) t)
#
# for the first 10 minutes. From that time on, the second cup's
# temperature is given by
#
> h := t -> -5 + (itemp - (-5))*exp(k*t);
h := t -> -5 + (itemp + 5) exp(k t)
>
#
# where itemp is the temperature that the 2nd cup would start at in
# ord...