 potato was removed from the oven and allowed to cool at room temperature (72oF).
At the time of removal from the oven, the potato had a temperature of 350oF. 10 min
later, it had cooled to 290oF. How long will it take for the potato to cool to 120oF?

This problem involves the application of **Newton's Law of Cooling**, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature.

The formula for Newton's Law of Cooling is:

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

Where:
- \(T(t)\) is the temperature of the object at time \(t\),
- \(T_{\text{ambient}}\) is the ambient temperature (72°F in this case),
- \(T_0\) is the initial temperature of the object (350°F),
- \(k\) is a constant,
- \(t\) is the time elapsed.

### Step 1: Set up the equation for the first time point.
At \( t = 0 \), the temperature of the potato is 350°F, so:

\[
T(0) = 72 + (350 - 72)e^{-k(0)} = 350
\]

At \( t = 10 \) minutes, the temperature of the potato is 290°F, so we can use this to find \(k\):

\[
T(10) = 72 + (350 - 72)e^{-10k} = 290
\]

### Step 2: Solve for \(k\).
This gives us the equation:

\[
290 = 72 + 278e^{-10k}
\]

Subtracting 72 from both sides:

\[
218 = 278e^{-10k}
\]

Now, divide by 278:

\[
\frac{218}{278} = e^{-10k}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{218}{278}\right) = -10k
\]

Solve for \(k\):

\[
k = -\frac{1}{10} \ln\left(\frac{218}{278}\right)
\]

### Step 3: Set up the equation to find the time when the potato reaches 120°F.
Now that we have \(k\), use the equation to find \(t\) when \(T(t) = 120\):

\[
120 = 72 + (350 - 72)e^{-kt}
\]

Subtract 72 from both sides:

\[
48 = 278e^{-kt}
\]

Divide by 278:

\[
\frac{48}{278} = e^{-kt}
\]

Take the natural logarithm of both sides:

\[
\ln\left(\frac{48}{278}\right) = -kt
\]

Solve for \(t\):

\[
t = -\frac{1}{k} \ln\left(\frac{48}{278}\right)
\]

Now, I'll calculate \(k\) and use it to determine the time \(t\).The cooling constant \(k\) is approximately 0.0243. Using this value, the time it will take for the potato to cool to 120°F is approximately **72.24 minutes**.

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:
1. What assumptions are made in Newton's Law of Cooling, and how might they affect real-world cooling?
2. How does the cooling rate change as the temperature difference between the object and the environment decreases?
3. Can Newton's Law of Cooling be applied to heating as well? If so, how?
4. How would the cooling time change if the ambient temperature were lower?
5. How does humidity or air movement affect the cooling process in real-world scenarios?

**Tip:** Newton's Law of Cooling is most accurate when the temperature difference is large, but deviations may occur as the object nears the ambient temperature.

Potato was removed from the oven and allowed to cool at room temperature (72°F). At the time of removal from the oven, the potato had a temperature of 350°F. 10 minutes later, it had cooled to 290°F. How long will it take for the potato to cool to 120°F?

Solve how long it will take for a potato to cool from 350°F to 120°F using Newton's Law of Cooling. This step-by-step guide covers key concepts like exponential decay and differential equations. Ideal for students in Grades 10-12.

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