# 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 cups of 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 # order for it to cool to g(10) degrees C if it were in the freezer from # the start. So itemp satisfies the equation # > eq := -5 + (itemp +5)*exp(k*10) = g(10); eq := - 15/7 + 4/7 itemp = 460/7 > # # Solving for itemp, # > itemp := solve(eq,itemp); itemp := 475/4 > # # Now plot the two temperature curves. # > cup1 := [t,f(t),t=0..20]; cup1 := [t, 20 + 70 exp(1/10 ln(4/7) t), t = 0 .. 20] > cup2 := [t,g(t),t=0..10],[t,h(t),t=10..20]; cup2 := [t, 20 + 80 exp(1/10 ln(4/7) t), t = 0 .. 10], [t, -5 + 495/4 exp(1/10 ln(4/7) t), t = 10 .. 20] > plot( {cup1,cup2},color=black ); > # # The cups are at the same temperature before another 5 minutes pass. # > etime := solve(f(t)=h(t),t); 20 ln(--) 43 etime := 10 ------- ln(4/7) > `Cups are same temp at t=`,evalf(etime),` minutes`; Cups are same temp at t=, 13.67845330, minutes # # At that time the common temperature is # > f(etime); 2260 ---- 43 > evalf(f(etime)),` degrees celsius.`; 52.55813953, degrees celsius. >