To solve this problem using Newton's Law of Cooling, let's work through each part step-by-step.

(a) Initial Value Problem

Newton's Law of Cooling states that the rate of change of the temperature $T(t)$ of an object is proportional to the difference between the temperature of the object and the ambient temperature $T_{\text{ambient}}$. Mathematically, this is expressed as:

$\frac{dT}{dt} = -k \left( T(t) - T_{\text{ambient}} \right)$

Given:

• The ambient temperature $T_{\text{ambient}} = 68^\circ \text{F}$
• The initial temperature $T(0) = 190^\circ \text{F}$
• The temperature after 2.5 minutes $T(2.5) = 175^\circ \text{F}$

The initial value problem is:

$\frac{dT}{dt} = -k \left( T(t) - 68 \right)$

$T(0) = 190$

(b) Solve the Initial Value Problem

This is a first-order linear differential equation. We solve it using separation of variables:

1. Separate variables: $\frac{dT}{T(t) - 68} = -k \, dt$

2. Integrate both sides: $\int \frac{1}{T(t) - 68} \, dT = -k \int dt$

   This gives: $\ln|T(t) - 68| = -kt + C$

   where $C$ is the constant of integration.

3. Solve for $T(t)$: $T(t) - 68 = e^{-kt + C} = Ce^{-kt}$ where $C = e^C$ is a new constant.

   So, the solution is: $T(t) = 68 + Ce^{-kt}$

4. Apply the initial condition $T(0) = 190$: $190 = 68 + Ce^{0} \implies C = 122$

   Therefore, the solution is: $T(t) = 68 + 122e^{-kt}$

(c) Determine the Value of the Constant $k$

We use the fact that $T(2.5) = 175$ to determine $k$:

$175 = 68 + 122e^{-2.5k}$

$107 = 122e^{-2.5k}$

$\frac{107}{122} = e^{-2.5k}$

$\ln\left(\frac{107}{122}\right) = -2.5k$

$k = -\frac{1}{2.5} \ln\left(\frac{107}{122}\right)$

Let's calculate $k$:

$k = \frac{-1}{2.5} \ln\left(\frac{107}{122}\right) \approx 0.0527 \, \text{per minute}$

(d) Determine When the Coffee Reaches 150 Degrees

We want to find the time $t$ when $T(t) = 150^\circ \text{F}$:

$150 = 68 + 122e^{-0.0527t}$

$82 = 122e^{-0.0527t}$

$\frac{82}{122} = e^{-0.0527t}$

$\ln\left(\frac{82}{122}\right) = -0.0527t$

$t = \frac{\ln\left(\frac{122}{82}\right)}{0.0527} \approx 9.38 \, \text{minutes}$

So, the coffee reaches a temperature of 150 degrees in approximately 9.38 minutes.

Summary:

• (a) Initial Value Problem: $\frac{dT}{dt} = -k \left( T(t) - 68 \right)$ $T(0) = 190$

• (b) Solution: $T(t) = 68 + 122e^{-kt}$

• (c) Constant $k$: $k \approx 0.0527 \, \text{per minute}$

• (d) Time to reach 150 degrees: $t \approx 9.38 \, \text{minutes}$

Would you like more details on any of these steps?

Here are 5 related questions:

1. What happens to the temperature as time approaches infinity?
2. How would the initial value problem change if the ambient temperature were different?
3. What is the physical meaning of the constant $k$ in Newton's law of cooling?
4. Can Newton's law of cooling be applied to other scenarios besides temperature?
5. How would the solution change if the coffee were cooled in a refrigerator at 40 degrees?

Tip: Newton's law of cooling is applicable to a wide range of phenomena, including the cooling of objects, but also to heating, and even radioactive decay, where the rate of change is proportional to the current state.