# Solutions to Exercises on Newton s Law of Cooling S. F. Ellermeyer

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1 Solutions to Exercises on Newton s Law of Cooling S F Ellermeyer A thermometer is taken from a room that is 0 C to the outdoors where the temperature is C After one minute, the thermometer reads 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? Solution: If T is the thermometer temperature, then Newton s Law of Cooling tells us that dt = k ( T ) T (0) = 0 The solution of this initial value problem is T = + e kt We still need to nd the value of k We can do this by using the given information that T () = In fact, let us pause here to consider the general problem of nding the value of k 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 dt = k ( T ) T (0) = T 0 T (t ) = T where t is some time other than 0 Then, from the rst two equations in the model, we obtain T = + (T 0 ) e kt and from the third equation we obtain + (T 0 ) e kt = T

2 Thus which gives us or or (T 0 ) e kt = T e kt = T T 0 e kt = T 0 T k = T0 ln t The latter equation gives us the value of k However, note that, in most problems that we deal with, it is not really necessary to nd the value of k Since the term e kt that appears in the solution of Newton s Law of Cooling can be written as e kt = e kt t=t, we really just need (in most situations) to know the value of e kt, and this value has been obtained in the work done above In particular, the solution of Newton s Law of Cooling, can be written as or as T T = + (T 0 ) e kt, T = + (T 0 ) e kt t=t T T = + (T 0 ) T 0 t=t Returning now to the problem at hand (with the thermometer), we see that the temperature function for the thermometer is T = + Note that this makes sense because this formula gives us 0 7 T (0) = + = 0

3 and 7 T () = + = To nd what the thermometer will read two minutes after being taken outside, we compute 7 T () = + 8:3 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 + = 6 The step by step solution of this equation is = = ln! t 7 = ln 7 t ln = ln ln (=) t = ln (7=) 3: Thus, the thermometer will reach 6 C after being outside for about 3: minutes Let us remember, in solving the upcoming problems, that the solution of the problem dt = k ( T ) T (0) = T 0 T (t ) = T 3

4 (which type of problem is called a boundary value problem because we are given prescribed values of a di erential equation at two points) can be written as t=t T T = + (T 0 ) At midnight, with the temperature inside your house a0 F and the temperature outside at 0 F, your furnace breaks down Two hours later, the temperature in your house has fallen to 0 F Assume that the outside temperature remains constant at 0 F At what time will the inside temperature of your house reach 40 F? Solution: The boundary value problem that models this situation is T 0 dt = k (0 T ) T (0) = 70 T () = 0 where time 0 is midnight The solution of this boundary value problem (from the work done in problem above) is t= 3 T = Note (for the purpose of a reasonableness check) that this formula gives us 0= 3 T (0) = = 70 and = 3 T () = = 0 To nd when the temperature in the house will reach 40 F, we must solve the equation t= = 40 The solution of this equation is ln (=) t = 3:6 ln (3=) Thus, the temperature in the house will reach 40 F a little after 3:30 am 4

5 3 You can nd 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 T (=) = 4 F and T () = F Assuming that the room temperature is 70 F, what is the temperature inside the refrigerator? Solution: 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 dt = k (70 T ) T (0) = 4 T (=) = and we know that the solution of this boundary value problem is T = 70 t 3 To check this formula for reasonableness, we observe that the formula gives us (0) 3 T (0) = 70 = 4 and T 3 ( ) = 70 = The temperature of the refrigerator is the temperature of the can of soda at time t = =, so we see that the temperature of the refrigerator is T 3 ( ) = 70 = :3 F

6 4 In a murder investigation, a corpse was found by a detective at exactly 8 PM Being alert, the detective also measured the body temperature and found it to be 70 F Two hours later, the detective measured the body temperature again and found it to be 60 F If the room temperature is 0 F, and assuming that the body temperature of the person before death was 98:6 F, at what time did the murder occur? Solution: With time 0 taken to be 8 PM, we have the boundary value problem whose solution is dt = k (0 T ) T (0) = 70 T () = 60 T = t= We would like to nd the value of t for which T (t) = 98:6 equation t= = 98:6 gives us ln (48:6=0) t = :6 ln (=) It appears that this person was murdered at about :30 PM or so Here is a graph of the function T = over the time interval :6 t :6 t= Solving the 6

7 90 80 body temperature time in hours John and Maria are having dinner and each orders a cup of co ee John cools his co ee with three tablespoons of cream They wait ten minutes and then Maria cools her co ee with three tablespoons of cream The two then begin to drink Who drinks the hotter co ee? (Assume that adding three tablespoons of cream to co ee immediately cools the co ee by 0 F ) Solution: Let t 0 be the time that John adds cream and let t be the time (ten minutes after t 0 ) that Maria adds cream At time t 0, John s co ee is 0 F cooler than Maria s co ee During the ten minute time interval from time t 0 to time t, both John s and Maria s co ees are cooling (getting closer to room temperature) However, during this ten second time interval, John s co ee is cooling more slowly than Maria s co ee, and Maria s co ee is always warmer than John s co ee At time t, there must be less than 0 F di erence between the co ee temperatures Thus, when Maria adds cream, it drops her co ee s temperature below that of John s co ee temperature, so John drinks the warmer co ee 7

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