A cup of coffee cools at rate proportional to the difference between the constant room temperature of 20.0°C and the temperature of the coffee. If the temperature of the coffee was 86.1°C 3.0 minutes ago and the current temperature of the coffee is 79.9°C, what will the temperature of the coffee be 29.0 minutes from now?

This is a classic problem of Newton's Law of Cooling, which states that the rate at which an object cools is proportional to the difference between its temperature and the ambient temperature.

The formula for Newton's Law of Cooling is given by:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_{\text{ambient}} \) is the ambient temperature (20.0°C in this case),
- \( T_0 \) is the initial temperature (86.1°C in this case),
- \( k \) is the cooling constant (we'll need to find this),
- \( t \) is the time elapsed in minutes.

### Step 1: Find the cooling constant \( k \)
We are given:
- Initial temperature: \( T_0 = 86.1^\circ C \),
- Current temperature after 3 minutes: \( T(3) = 79.9^\circ C \),
- Ambient temperature: \( T_{\text{ambient}} = 20.0^\circ C \).

Using the formula for Newton's Law of Cooling, we can substitute the values at \( t = 3 \) to find \( k \):

\[
T(3) = 20 + (86.1 - 20) e^{-3k}
\]
\[
79.9 = 20 + 66.1 e^{-3k}
\]
\[
79.9 - 20 = 66.1 e^{-3k}
\]
\[
59.9 = 66.1 e^{-3k}
\]
\[
\frac{59.9}{66.1} = e^{-3k}
\]
\[
e^{-3k} = 0.9067
\]
Now, take the natural logarithm of both sides to solve for \( k \):

\[
-3k = \ln(0.9067)
\]
\[
k = -\frac{\ln(0.9067)}{3}
\]

### Step 2: Calculate the temperature after 29 minutes
Once we find \( k \), we can use the same formula to calculate the temperature after 29 minutes:

\[
T(29) = 20 + (86.1 - 20) e^{-29k}
\]

Let me calculate the value of \( k \) and then use it to find \( T(29) \).The temperature of the coffee 29.0 minutes from now will be approximately \( 45.5^\circ C \).

Would you like more details on how this was calculated or have any questions? 

Here are 5 related questions:
1. How does Newton’s Law of Cooling apply in other contexts like cooling of electronics?
2. How does the cooling constant \( k \) vary with different environments?
3. What happens to the cooling rate as the temperature approaches the ambient temperature?
4. How would the coffee's temperature change if the ambient temperature were higher or lower?
5. How can this cooling model be adjusted for more complex scenarios, like varying room temperature?

**Tip**: Newton’s Law of Cooling assumes ideal conditions, but real-world factors like insulation or wind can affect cooling.

A cup of coffee cools at a rate proportional to the difference between the constant room temperature of 20.0°C and the temperature of the coffee. If the temperature of the coffee was 86.1°C 3.0 minutes ago and the current temperature of the coffee is 79.9°C, what will the temperature of the coffee be 29.0 minutes from now?

Solve a Newton's Law of Cooling problem where a cup of coffee cools from 86.1°C to 79.9°C in 3 minutes. Learn how to find the cooling constant and calculate the coffee temperature 29 minutes later using exponential decay and differential equations.

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