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This is a classic example of Newton's Law of Cooling. The problem asks how long it will take for a freshly baked pie to cool down from 200°C to 100°C when placed in an environment at a constant temperature of 25°C. 

We are given the following dataThis is a classic example of Newton's Law of Cooling. The problem asks how long it will take for a freshly baked pie to cool down from 200°C to 100°C when placed in an environment at a constant temperature of 25°C. 

We are given the following 
- Initial temperature of the pie, \( T_0 = 200^\circ C \)
- Ambient temperature, \( T_{\infty} = 25^\circ C \)
- Temperature after 5 minutes, \( T(5) = 150^\circ C \)
- We need to find the time it takes for the pie to cool to 100°C.

### Newton's Law of Cooling formula:
\[
T(t) = T_{\infty} + (T_0 - T_{\infty}) e^{-kt}
\]
Where:
- \( T(t) \) is the temperature at time \( t \),
- \( T_{\infty} \) is the surrounding temperature (25°C),
- \( T_0 \) is the initial temperature (200°C),
- \( k \) is the cooling constant,
- \( t \) is time.

### Step 1: Calculate the cooling constant \( k \)
From the given, after 5 minutes the temperature drops to 150°C:
\[
150 = 25 + (200 - 25)e^{-5k}
\]
\[
150 = 25 + 175e^{-5k}
\]
\[
125 = 175e^{-5k}
\]
\[
\frac{125}{175} = e^{-5k}
\]
\[
\frac{5}{7} = e^{-5k}
\]
Taking the natural logarithm on both sides:
\[
\ln\left(\frac{5}{7}\right) = -5k
\]
\[
k = -\frac{1}{5} \ln\left(\frac{5}{7}\right) \approx 0.0652
\]

### Step 2: Find the time when the pie cools to 100°C
Now, we want to find the time \( t \) when the pie cools to 100°C:
\[
100 = 25 + (200 - 25)e^{-kt}
\]
\[
100 = 25 + 175e^{-kt}
\]
\[
75 = 175e^{-kt}
\]
\[
\frac{75}{175} = e^{-kt}
\]
\[
\frac{3}{7} = e^{-kt}
\]
Taking the natural logarithm on both sides:
\[
\ln\left(\frac{3}{7}\right) = -kt
\]
\[
t = -\frac{1}{k} \ln\left(\frac{3}{7}\right)
\]
Substitute \( k \approx 0.0652 \):
\[
t = -\frac{1}{0.0652} \ln\left(\frac{3}{7}\right) \approx 36.8 \text{ minutes}
\]

To the nearest minute, the answer is **37 minutes**.

Would you like further details or have any questions?

Here are five related questions:
1. What factors can influence the cooling rate besides ambient temperature?
2. How would the solution change if the surrounding temperature was higher than 25°C?
3. How does the time constant \( k \) affect the cooling process?
4. Can Newton’s Law of Cooling be applied to heating processes as well?
5. What assumptions does Newton’s Law of Cooling rely on?

**Tip:** Always verify the units you're working with in physics problems, especially when dealing with exponential models.

A freshly baked pie is taken out of an oven at 200°C and placed on a kitchen counter, where the temperature is a constant 25°C. After 5 minutes, the temperature of the pie drops to 150°C. How long will the pie take to get cooled to 100°C to the nearest minute?

This detailed solution applies Newton’s Law of Cooling to find how long a pie takes to cool from 200°C to 100°C in a 25°C environment. With step-by-step calculations using exponential decay, this problem involves solving for time using logarithmic functions.

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