It is a well-know fact that Mr. G loves his coffee. It is
also a well-known that Mr. G does not guzzle his
coffee. In fact, he will linger over a cup of coffee for a
long time, but he will stop drinking his coffee if the
temperature falls below 120 degrees Fahrenheit. (Iced
coffee is an abomination to Mr. G.)
Mr. G is well familiar with Newton’s Law of Cooling. Newton discovered that the rate at
which the temperature of an object (including coffee in a cup) cools is directly
proportional to the difference between the temperature of the object and the
surrounding temperature.
In calculus terms this means:
dT
dt
= − k T Ts
( ) , where T(t) is the temperature of the
object at any time, t; k is the constant of proportionality; and Ts is the surrounding
temperature (which in our case will be a constant).
On his way to school one morning, Mr. G stops at his favorite Joe McBuck’s Coffee
Shoppe in South Riding and purchases his usual twenty (venti) ounce coffee. He
inquires of the helpful barista and finds that the temperature at which Joe McBuck’s
serves coffee is 185 degrees Fahrenheit. (So, at time t = 0, that is the temperature of
the coffee.) While waiting for the barista to pour his coffee, Mr. G checks the thermostat
of the shop discreetly and finds the temperature of the shop is precisely 70 degrees.
Mr. G pays for his coffee, grabs a few napkins, and settles down at a table by the
window. He carefully pulls out his handy pocket thermometer and notes that the
temperature of the coffee is now 182 degrees. He checks his watch and notes that
precisely 1 minute and 30 seconds has gone by since the barista poured his coffee.
Here are your questions:
Question 1: Solve the differential equation to get the function T(t) that gives the
temperature of the coffee at any time, t. (Show all work.)

Question 2: Having prepared his lesson plans for the day and made all his copies the
night before, Mr. G can linger in the coffee shop for a total of 30 minutes (after receiving
his coffee) before heading to school (that’s 28 minutes and 30 seconds after he sat
down). What is the temperature of his coffee when he leaves the coffee shop?

Question 3: Mr. G leaves the coffee shop after 30 minutes and makes the 5 minute
drive to school. Miraculously, the temperature outside and at school is a constant 70
degrees, just like it is in the coffee shop (i.e., the surrounding temperature is the same
everywhere). How long after he arrives at school does Mr. G have to drink his coffee
before it gets too cool for him to drink?

Question 4: At what time does the coffee reach EXACTLY room temperature?

Question 1
~~~~~~~~
NB: i will do it my way since this
dT
dt
= − k T Ts
is not making sense.
~~~~~~~~~~~~~~~~~~~~~~

dT/dt = k(T - Ts) = (T - 70)
dT /(T - 70) = k dt
integrate
ln(T - 70) = kt + C
T - 70 = e^(kt) e^C
initial conditions:
T(0) = 185
T(1.5) = 182

185 - 70 = e^C
e^C = 115
T - 70 = e^(kt) e^C
T = 70 + 115e^(kt)

182 = 70 + 115e^(1.5k)
k = 2/3 ln(112/115)
k = ln[ (112/115)^(2/3) ]

T = 70 + 115e^( ln[ (112/115)^(2t / 3) ]
T(t) = 70 + 115[ (112/115)^(2t / 3) ]

Question 2:
~~~~~~~~
after 30 mins
T(t) = 70 + 115[ (112/115)^(2t / 3) ]
T(30) = 70 + 115[ (112/115)^(20) ]
T(30) = 137.8 degrees fahrenheit

Question 3:
~~~~~~~~
when he gets to school would be 35 minutes after he was served;
coffee gets too cold to drink when temp falls below 120 degrees
T(t) = 70 + 115[ (112/115)^(2t / 3) ]
120 = 70 + 115[ (112/115)^(2t / 3) ]

t = 3/2 [ ln(112/115) / ln( 50/115) ]
t = 47.3 minutes

will drink coffee when he gets to school for ( 47.3 - 35 = 12.3 minutes )
(approximately 12 minutes before it gets too cold)

Question 4:
~~~~~~~~
T(t) = 70 + 115[ (112/115)^(2t / 3) ]
70 = 70 + 115[ (112/115)^(2t / 3) ]
t = undefined