Calculus Word Problem?

In this question we use Newton’s Law of heating to find a temperature function.
Suppose that a container of milk is removed from the refrigerator and
placed on a table. The temperature of the room is constantly E and we let
T(t) denote the temperature (in degrees Celsius) of the milk after t minutes.

1.Write an equation which corresponds to the description: “The rate of
change of the temperature of the milk is proportional to the difference
between the room temperature and the milk temperature”. Introduce
new constant(s) as needed.
2.Define a new function of time S by the rule S(t) = E − T(t). How do dS/dt and dT/dt relate?
3. Use 1. and 2. to determine a differential equation (not using T) which
is satisfied by S.
4. What is the general form of S?
5. What is the general form of T?
6. Assume that E = 20◦ C and suppose that the initial temperature of the
milk is 5◦ C and the temperature of the milk after 10 minutes is 10◦ C
degrees. What is the function T(t)? (keep constants to 4 decimal places)

1 Answer
Nov 23, 2016

# T=20-15e^(-0.0406t) #

Explanation:

Room Temp = #E# (const)
Milk Temp = #T(t)#

1. The Equation

# "rate of change of " T prop " Room Temp " - "Milk Temp" #
# :. (dT)/dt prop (E-T) #
# :. (dT)/dt =k(E-T) #

2. The Relationship
# S(t) = E-T(t) #
# :. (dS)/dt = 0 - (dT)/dt #
# :. (dS)/dt = - (dT)/dt #

3. The DE
Subs result from (2) into result from from (1) we get:
# -(dS)/dt = kS #
# :. (dS)/dt = -kS #

4. Solve for S
The DE from (3) is a First Order Separable DE, and we can collect terms as follows;
# (dS)/dt = -kS => 1/S (dS)/dt = -k#

Separating the variables we get:
# int 1/S dS = int -k dt#

And Integrating gives us:
# lnS = -kt + C#
# :. S = e^(-kt + C)#
# :. S = e^(-kt)e^C#
# :. S = Ae^(-kt)#

5. Solve for T
# S = Ae^(-kt)# and #S(t) = E-T(t)#
# :. T=E-Ae^(-kt) #

6. Apply Initial Conditions
# T=20-Ae^(-kt) #
# :. T(0)=5 => 5=20-Ae^0#
# :. A=20-5#
# :. A=15#

So we have:
# T=20-15e^(-kt) #
# :. T(10) = 10 => 10=20-15e^(-10k) #
# :. 15e^(-10k) = 10 #
# :. e^(-10k) = 2/3 #
# :. -10k = ln(2/3) #
# :. k = -1/10ln(2/3) #
# :. k = 0.0406 # (4dp)

And so:
# T=20-15e^(-0.0406t) #