A bottle of soda pop at room temperature (72F) is put in a refrigerator (44F). After 30 minutes of cooling, the bottle of soda pop is cooled to 61F. Use the Newton’s Law of Cooling and differentiation, determine 1) The temper of the bottle of soda pop after yet another 30 minute; 2) How long would it take for the soda pop to be cooled to 50F

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the surrounding temperature. Mathematically, it can be expressed as:

\[
\frac{dT}{dt} = -k (T - T_{\text{env}})
\]

where:
- \( T \) is the temperature of the object at time \( t \),
- \( T_{\text{env}} \) is the temperature of the surrounding environment (in this case, 44°F for the refrigerator),
- \( k \) is a positive constant specific to the cooling object and environment.

### Step 1: Setting up the Initial Condition and Solving for \( k \)

Given:
- Initial temperature \( T_0 = 72^\circ F \),
- Surrounding temperature \( T_{\text{env}} = 44^\circ F \),
- Temperature after 30 minutes (\( t = 30 \) minutes): \( T(30) = 61^\circ F \).

The solution to Newton’s Law of Cooling equation can be expressed as:

\[
T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-kt}
\]

Substituting the values we know:

1. **At \( t = 0 \)**:
   \[
   T(0) = T_{\text{env}} + (T_0 - T_{\text{env}}) = 44 + (72 - 44) = 72^\circ F
   \]

2. **At \( t = 30 \)**:
   \[
   T(30) = 44 + (72 - 44)e^{-30k} = 61
   \]

   Simplifying this equation to solve for \( k \):

   \[
   61 = 44 + 28 e^{-30k}
   \]
   \[
   17 = 28 e^{-30k}
   \]
   \[
   e^{-30k} = \frac{17}{28}
   \]
   \[
   -30k = \ln\left(\frac{17}{28}\right)
   \]
   \[
   k = -\frac{1}{30} \ln\left(\frac{17}{28}\right)
   \]

   Calculating this value:

   \[
   k \approx 0.01566
   \]

### Step 2: Finding the Temperature after Another 30 Minutes (Total 60 Minutes)

Now, we can substitute \( t = 60 \) minutes into the equation to find \( T(60) \):

\[
T(60) = 44 + (72 - 44)e^{-0.01566 \times 60}
\]

Calculating this:

1. Find \( e^{-0.01566 \times 60} \):
   \[
   e^{-0.01566 \times 60} \approx 0.4066
   \]

2. Substitute back:
   \[
   T(60) = 44 + 28 \times 0.4066
   \]
   \[
   T(60) \approx 44 + 11.4 = 55.4^\circ F
   \]

So, **the temperature after another 30 minutes (total 60 minutes) will be approximately \( 55.4^\circ F \)**.

### Step 3: Finding the Time to Reach 50°F

Now we need to find \( t \) when \( T(t) = 50 \):

\[
50 = 44 + (72 - 44)e^{-0.01566t}
\]
\[
6 = 28 e^{-0.01566t}
\]
\[
e^{-0.01566t} = \frac{6}{28}
\]
\[
-0.01566t = \ln\left(\frac{6}{28}\right)
\]
\[
t = \frac{\ln\left(\frac{6}{28}\right)}{-0.01566}
\]

Calculating this:

1. \( \ln\left(\frac{6}{28}\right) \approx -1.5404 \)
2. Then, \( t \approx \frac{-1.5404}{-0.01566} \approx 98.35 \) minutes.

So, **it would take approximately 98.35 minutes for the soda pop to cool to 50°F**.

---

Would you like further details or have questions about any of the steps? 

Here are some related questions:
1. How does Newton's Law of Cooling apply to heating as well as cooling?
2. How would the cooling rate change if the refrigerator temperature were lower?
3. How could this approach be adapted for objects with non-uniform temperatures?
4. What are real-world factors that might cause deviations from this theoretical model?
5. How could we use integration to derive the general solution for Newton’s Law of Cooling?

**Tip:** Remember that the constant \( k \) depends on both the properties of the object (like material and mass) and the conditions of the surroundings.

A bottle of soda pop at room temperature (72F) is put in a refrigerator (44F). After 30 minutes of cooling, the bottle of soda pop is cooled to 61F. Use Newton’s Law of Cooling and differentiation to determine 1) The temperature of the bottle of soda pop after another 30 minutes; 2) How long it would take for the soda pop to cool to 50F.

Explore how to apply Newton's Law of Cooling to determine the temperature change of a soda bottle as it cools in a refrigerator. Given an initial temperature of 72°F, calculate the temperature after 60 minutes and the time needed to reach 50°F, with step-by-step solutions involving differential equations. Suitable for high school calculus students.

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