# nanopdf.com solutions-to-problems-on-newtons-law-of-cooling (1)-converted ```9+**********+wbg t
Solutioms to Exercises om Newtom&sup3;s 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
```