When a vegetable soup is removed from an oven, its temperature is measured at
300°F. Three minutes later its temperature is 200° F. If the room temperature is
70° F., then find the temperature of the vegetable soup after 20 minutes? (Assume Newton's Law of Cooling)

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the temperature of the object and the ambient temperature. The formula is given by:

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

Where:
- \(T(t)\) is the temperature of the object at time \(t\),
- \(T_{\text{room}}\) is the room temperature (70°F in this case),
- \(T_0\) is the initial temperature of the object (300°F),
- \(k\) is a constant related to the rate of cooling,
- \(t\) is the time in minutes.

### Step 1: Set up the problem
We know the following:
- Initial temperature of the soup, \(T_0 = 300°F\),
- Temperature after 3 minutes, \(T(3) = 200°F\),
- Room temperature, \(T_{\text{room}} = 70°F\).

First, let's use this information to find the cooling constant \(k\).

### Step 2: Find the constant \(k\)
Using the formula for \(t = 3\) and \(T(3) = 200°F\):

\[
200 = 70 + (300 - 70) e^{-3k}
\]

Simplifying:

\[
200 = 70 + 230 e^{-3k}
\]

\[
130 = 230 e^{-3k}
\]

\[
\frac{130}{230} = e^{-3k}
\]

\[
\frac{13}{23} = e^{-3k}
\]

Take the natural logarithm of both sides:

\[
\ln\left(\frac{13}{23}\right) = -3k
\]

\[
k = -\frac{1}{3} \ln\left(\frac{13}{23}\right)
\]

Now, let's calculate the value of \(k\).

### Step 3: Calculate the temperature after 20 minutes
Once we have \(k\), we can substitute it back into the formula to find \(T(20)\).The temperature of the vegetable soup after 20 minutes will be approximately \( 75.13^\circ F \).

Would you like more details on the calculations or have any questions?

Here are 5 related questions you might find helpful:
1. How does changing the room temperature affect the cooling process?
2. How do we derive Newton's Law of Cooling from differential equations?
3. What other factors, besides room temperature, can influence the cooling rate?
4. How does the soup's temperature behave after a very long period?
5. Can Newton's Law of Cooling apply to heating processes as well?

**Tip:** The cooling constant \(k\) is specific to the object and environment. If the material of the object changes, \(k\) will also change, affecting the cooling rate.

When a vegetable soup is removed from an oven, its temperature is measured at 300°F. Three minutes later its temperature is 200°F. If the room temperature is 70°F, then find the temperature of the vegetable soup after 20 minutes? (Assume Newton's Law of Cooling)

Solve a cooling problem using Newton's Law of Cooling: a vegetable soup cools from 300°F to 200°F in 3 minutes, with room temperature at 70°F. Learn how to find the soup's temperature after 20 minutes with step-by-step calculations and key formulas. Suitable for high school students.

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