Question

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)

Answer 1

`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)`