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To solve this, we need to apply **Newton's Law of Cooling**, which states that the rate of change of the temperature of an object is proportional to the difference between its current temperature and the ambient temperature.

The equation is:

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

Where:
- \(T(t)\) is the temperature of the object at time \(t\),
- \(T_{\text{env}}\) is the ambient temperature (here 25°C),
- \(T_0\) is the initial temperature of the object (200°C),
- \(k\) is the cooling constant, and
- \(t\) is the time.

We are given:
- Initial temperature \(T_0 = 200°C\),
- After 5 minutes, the temperature drops to \(T(5) = 150°C\),
- Ambient temperature \(T_{\text{env}} = 25°C\).

We first need to calculate \(k\) using the information after 5 minutes:

\[
150 = 25 + (200 - 25) e^{-5k}
\]

Simplifying:

\[
150 - 25 = 175 e^{-5k}
\]
\[
125 = 175 e^{-5k}
\]
\[
\frac{125}{175} = e^{-5k}
\]
\[
\frac{5}{7} = e^{-5k}
\]

Taking the natural logarithm on both sides:

\[
\ln\left(\frac{5}{7}\right) = -5k
\]

Solving for \(k\):

\[
k = -\frac{1}{5} \ln\left(\frac{5}{7}\right)
\]

Now, we calculate how long it will take to reach 100°C. We set \(T(t) = 100°C\) and solve for \(t\):

\[
100 = 25 + (200 - 25) e^{-kt}
\]
\[
100 - 25 = 175 e^{-kt}
\]
\[
75 = 175 e^{-kt}
\]
\[
\frac{75}{175} = e^{-kt}
\]
\[
\frac{3}{7} = e^{-kt}
\]

Taking the natural logarithm:

\[
\ln\left(\frac{3}{7}\right) = -kt
\]

Substitute the value of \(k\) from earlier to find \(t\).

Let me calculate the precise time now.The pie will take approximately **13 minutes** to cool down to 100°C.

Do you have any questions or would you like further details on the calculation?

Here are five related questions for further practice:
1. What would the cooling constant \(k\) be if the ambient temperature were 30°C instead?
2. How long would it take for the pie to cool to 50°C under the same conditions?
3. If the initial temperature was 180°C, how would the time to cool to 100°C change?
4. How does changing the ambient temperature affect the cooling time?
5. What if the pie cooled to 175°C in 5 minutes instead? How would this impact the time to 100°C?

**Tip**: When working with exponential functions like Newton's Law of Cooling, logarithms are essential for solving time-dependent problems.

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 problem demonstrates how to use Newton's Law of Cooling to calculate the time it takes for a pie to cool from 200°C to 100°C on a kitchen counter where the ambient temperature is 25°C. The solution involves solving an exponential decay equation, making it ideal for high school students in Grades 11-12.

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