advertisement

9+**********+wbg t Solutioms to Exercises om Newtom³s Law of Coolimg S. F. Ellermeyer fi. A thermometer is taken from a room that is XO○C to the outdoors where the temperature is 5○C. After one minute, the thermometer reads fiX○C. Use Newton‘s Law of Cooling to answer the following questions. (a) What will the reading on the thermometer be after one more minute? (b) When will the thermometer read 6 ○C? Solutiom: If f is the thermometer temperature, then Newton‘s Law of Cooling tells us that//.dv=f = h (5 — f) ;]jjjdt f (O) = XO. The solution of this initial value problem is f = 5 ‡ fi5e—ht. We still need to find the value of h. We can do this by using the given information that f (fi) = fiX. In fact, let us pause here to consider the general problem of finding the value of h. We will obtain some facts that can be used in the rest of the problems involving Newton‘s Law of Cooling. Suppose that we have the model df = h (fs — f) dt f (O) = fO f (tfi) = ffi where tfi is some time other than O. Then, from the first two equations in the model, we obtain f = fs ‡ (fO — fs ) e—ht and from the third equation we obtain fs ‡ (fO — fs ) e—htfi = ffi . fi Thus (fO — fs ) e—htfi = ffi — fs which gives us e—htfi = or ff i — fs fO — fs fO — fs ffi — fs . Σ fi f O — fs h = ln . tfi ffi — fs ehtfi = or The latter equation gives us the value of h. However, note that, in most problems that we deal with, it is not really necessary to find the value of h. Since the term e—ht that appears in the solution of Newton‘s Law of Cooling can be written as . Σ tƒtfi e—ht = e—htfi , we really just need (in most situations) to know the value of e—htfi , and this value has been obtained in the work done above. In particular, the solution of Newton‘s Law of Cooling, f = fs ‡ (fO — fs ) e—ht , can be written as or as . Σtƒt f = fs ‡ (fO — fs ) e—htfi fi . f f i — fs f = fs ‡ (fO — fs ) fO — fs Σtƒtfi . Returning now to the problem at hand (with the thermometer), we see that the temperature function for the thermometer is . Σt t . f = 5 ‡ fi5 fi5 Note that this makes sense because this formula gives us . ΣO t f (O) = 5 ‡ fi5 = XO fi5 2 and . f (fi) = 5 ‡ fi5 Σ t fi = fiX. fi5 To find what the thermometer will read two minutes after being taken outside, we compute . XΣ t f (X) = 5 ‡ fi5 = 8.3 fi5 which tells us that the thermometer will read about 8.3○C two minutes after being taken outside. Finally, to determine when the thermometer will read 6○ C, we solve the equation . Σ t t 5 ‡ fi5 = 6. fi5 The step–by–step solution of this equation is . Σt t fi5 = fi fi5 . Σ fi t t = fi5 fi5 . . Σt Σ . Σ fi t = ln ln fi5 . Σ .fi5 Σ t fi t ln = ln fi5 fi5 ln (fi/fi5) t= = 3.5. ln (t/fi5) Thus, the thermometer will reach 6○C after being outside for about 3.5 minutes. Let us remember, in solving the upcoming problems, that the solution of the problem df = h (fs — f ) dt f (O) = fO f (tfi) = ffi 3 (which type of problem is called a boundav4 sa1ue pvob1em because we are given prescribed values of a differential equation at tmo points) can be written as . Σ ff i — fs tƒtfi f = fs ‡ (fO — fs ) . fO — fs 2. At midnight, with the temperature inside your house at tO○5 and the temperature outside at XO○5 , your furnace breaks down. Two hours later, the temperature in your house has fallen to 5O○5 . Assume that the outside temperature remains constant at XO○5 . At what time will the inside temperature of your house reach ŒO○ 5 ? Solutiom: The boundary value problem that models this situation is df = h (XO — f) dt f (O) = tO f (X) = 5O where time O is midnight. The solution of this boundary value problem (from the work done in problem fi above) is . ΣtƒX 3 . f = XO ‡ 5O 5 Note (for the purpose of a reasonableness check) that this formula gives us . ΣOƒX 3 f (O) = XO ‡ 5O = tO 5 and . ΣXƒX 3 f (X) = XO ‡ 5O = 5O. 5 To find when the temperature in the house will reach ŒO○5 , we must solve the equation . ΣtƒX XO ‡ 5O 3 = ŒO. 5 The solution of this equation is . ln (X/5) Σ t=X 3.6. ln (3/5) = Thus, the temperature in the house will reach ŒO○ 5 a little after 3:30 a.m. 4 3. You can find the temperature inside your refrigerator without putting a thermometer inside. Take a can of soda from the refrigerator, let it warm for half an hour, then record its temperature. Let it warm for another half an hour and record its temperature again. Suppose that the readings are f (fi/X) = Œ5○ 5 and f (fi) = 55○ 5 . Assuming that the room temperature is tO○5 , what is the temperature inside the refrigerator? Solutiom: Taking the time one half hour after the soda was removed from the refrigerator to be the ”zero time™ (and stating the given information in an appropriate way), we have the boundary value problem df = h (tO — f ) dt f (O) = Œ5 f (fi/X) = 55 and we know that the solution of this boundary value problem is . ΣXt 3 . f = tO — X5 5 To check this formula for reasonableness, we observe that the formula gives us . ΣX(O) f (O) = tO — X5 3 = Œ5 5 and . ΣXfi2( . Σ fi = tO — X5 = 55. )3 f 5 X The temperature of the refrigerator is the temperature of the can of soda at time t = —fi/X, so we see that the temperature of the refrigerator is . f Σ . ΣX(— fi ) 2 3 fi — = tO — X5 X . 5Σ 5 = tO — X5 3 ○ = X8.3 5 . † 4. In a murder investigation, a corpse was found by a detective at exactly 8 P.M. Being alert, the detective also measured the body temperature and found it to be tO○5 . Two hours later, the detective measured the body temperature again and found it to be 6O○5 . If the room temperature is 5O○5 , and assuming that the body temperature of the person before death was 98.6○5 , at what time did the murder occur? Solutiom: With time O taken to be 8 P.M., we have the boundary value problem df = h (5O — f) dt f (O) = tO f (X) = 6O whose solution is . ΣtƒX fi . f = 5O ‡ XO X We would like to find the value of t for which f (t) = 98.6. Solving the equation . ΣtƒX 5O ‡ XO fi = 98.6 X gives us t=X . ln (Œ8.6/XO) Σ = —X.56. ln (fi/X) It appears that this person was murdered at about †:30 P.M. or so. Here is a graph of the function . ΣtƒX fi f = 5O ‡ XO X over the time interval —X.56 ≤ t ≤ X.56. 6 90 80 body temperature 70 60 -2 -1 0 tim1e in hours 2 †. John and Maria are having dinner and each orders a cup of coffee. John cools his coffee with three tablespoons of cream. They wait ten minutes and then Maria cools her coffee with three tablespoons of cream. The two then begin to drink. Who drinks the hotter coffee? (Assume that adding three tablespoons of cream to coffee immediately cools the coffee by fiO○5 .) Solutiom: Let tO be the time that John adds cream and let tfi be the time (ten minutes after tO) that Maria adds cream. At time tO, John‘s coffee is fiO○5 cooler than Maria‘s coffee. During the ten minute time interval from time tO to time tfi, both John‘s and Maria‘s coffees are cooling (getting closer to room temperature). However, during this ten second time interval, John‘s coffee is cooling more slowly than Maria‘s coffee, and Maria‘s coffee is always warmer than John‘s coffee. At time tfi, there must be less than fiO○5 difference between the coffee temperatures. Thus, when Maria adds cream, it drops her coffee‘s temperature below that of John‘s coffee temperature, so John drinks the warmer coffee. F