Solutions to Exercises on Newton’s Law of Cooling
S. F. Ellermeyer
1. A thermometer is taken from a room that is 20 C to the outdoors where
the temperature is 5 C. After one minute, the thermometer reads 12 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 (5
dt
T (0) = 20.
T)
The solution of this initial value problem is
T = 5 + 15e
kt
.
We still need to …nd the value of k. We can do this by
information that T (1) = 12. In fact, let us pause here
general problem of …nding the value of k. We will obtain
can be used in the rest of the problems involving Newton’s
Suppose that we have the model
dT
= k (Ts
dt
T (0) = T0
T (t1 ) = T1
using the given
to consider the
some facts that
Law of Cooling.
T)
where t1 is some time other than 0. Then, from the …rst two equations in
the model, we obtain
T = Ts + (T0 Ts ) e kt
and from the third equation we obtain
Ts + (T0
Ts ) e
1
kt1
= T1 .
Thus
(T0
kt1
Ts ) e
= T1
Ts
which gives us
=
T1
T0
Ts
Ts
ekt1 =
T0
T1
Ts
Ts
e
kt1
or
or
k=
1
ln
t1
T0
T1
Ts
Ts
.
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
t=t1
e kt = e kt1
,
we really just need (in most situations) to know the value of e kt1 , and this
value has been obtained in the work done above. In particular, the solution
of Newton’s Law of Cooling,
T = Ts + (T0
can be written as
T = Ts + (T0
kt
Ts ) e
kt1 t=t1
Ts ) e
or as
T = Ts + (T0
Ts )
,
T1
T0
Ts
Ts
t=t1
.
Returning now to the problem at hand (with the thermometer), we see that
the temperature function for the thermometer is
T = 5 + 15
7
15
t
.
Note that this makes sense because this formula gives us
T (0) = 5 + 15
2
7
15
0
= 20
and
T (1) = 5 + 15
1
7
15
= 12.
To …nd what the thermometer will read two minutes after being taken outside, we compute
2
7
8:3
T (2) = 5 + 15
15
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
t
7
5 + 15
= 6.
15
The step–by–step solution of this equation is
15
ln
7
15
t
7
15
t
7
15
t ln
=1
=
t
7
15
!
1
15
= ln
1
15
1
15
ln (1=15)
t=
ln (7=15)
= ln
3:5.
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
dT
= k (Ts
dt
T (0) = T0
T (t1 ) = T1
3
T)
(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=t1
T1 Ts
T = Ts + (T0 Ts )
.
T0 Ts
2. At midnight, with the temperature inside your house at 70 F and the
temperature outside at 20 F , your furnace breaks down. Two hours
later, the temperature in your house has fallen to 50 F . Assume that
the outside temperature remains constant at 20 F . At what time will
the inside temperature of your house reach 40 F ?
Solution: The boundary value problem that models this situation is
dT
= k (20
dt
T (0) = 70
T (2) = 50
T)
where time 0 is midnight. The solution of this boundary value problem (from
the work done in problem 1 above) is
T = 20 + 50
3
5
t=2
.
Note (for the purpose of a reasonableness check) that this formula gives us
T (0) = 20 + 50
and
3
5
0=2
= 70
2=2
3
= 50.
5
To …nd when the temperature in the house will reach 40 F , we must solve
the equation
t=2
3
20 + 50
= 40.
5
The solution of this equation is
T (2) = 20 + 50
t=2
ln (2=5)
ln (3=5)
3:6.
Thus, the temperature in the house will reach 40 F a little after 3:30 a.m.
4
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 (1=2) = 45 F and T (1) = 55 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
dt
T (0) = 45
T (1=2) = 55
T)
and we know that the solution of this boundary value problem is
T = 70
3
5
25
2t
.
To check this formula for reasonableness, we observe that the formula gives
us
2(0)
3
T (0) = 70 25
= 45
5
and
1
2
T
= 70
25
3
5
2( 12 )
= 55.
The temperature of the refrigerator is the temperature of the can of soda at
time t = 1=2, so we see that the temperature of the refrigerator is
T
1
2
= 70
25
= 70
25
28:3 F .
5
3
5
5
3
2(
1
2
)
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 70 F . Two hours later, the detective measured
the body temperature again and found it to be 60 F . If the room
temperature is 50 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 P.M., we have the boundary value
problem
dT
= k (50
dt
T (0) = 70
T (2) = 60
T)
whose solution is
T = 50 + 20
1
2
t=2
.
We would like to …nd the value of t for which T (t) = 98:6. Solving the
equation
t=2
1
= 98:6
50 + 20
2
gives us
ln (48:6=20)
ln (1=2)
t=2
2:56.
It appears that this person was murdered at about 5:30 P.M. or so.
Here is a graph of the function
T = 50 + 20
over the time interval
2:56
t
2:56.
6
1
2
t=2
90
80
body temperature
70
60
-2
-1
0
1 in hours
time
2
5. 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 10 F .)
Solution: Let t0 be the time that John adds cream and let t1 be the time
(ten minutes after t0 ) that Maria adds cream.
At time t0 , John’s co¤ee is 10 F cooler than Maria’s co¤ee. During the
ten minute time interval from time t0 to time t1 , 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 t1 ,
there must be less than 10 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