Initial temp 420 degrees after 11 minutes cooled to 297 degrees ambient temp is 82 degrees what is temp after 23 minutes of a rock spewed out of fire

This problem is modeled by **Newton's Law of Cooling**, which describes the rate of cooling of an object. The formula 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 (surrounding) temperature,
- \(T_0\) is the initial temperature of the object,
- \(k\) is the cooling constant (which we need to find),
- \(t\) is the time elapsed.

### Step 1: Find the cooling constant \(k\)
You are given:
- Initial temperature \(T_0 = 420^\circ F\),
- Temperature after 11 minutes, \(T(11) = 297^\circ F\),
- Ambient temperature \(T_{\text{ambient}} = 82^\circ F\).

Using the information at \(t = 11\):

\[
297 = 82 + (420 - 82)e^{-11k}
\]

Simplifying:

\[
297 = 82 + 338e^{-11k}
\]

Subtract 82 from both sides:

\[
215 = 338e^{-11k}
\]

Now solve for \(e^{-11k}\):

\[
\frac{215}{338} = e^{-11k}
\]

Take the natural logarithm of both sides to solve for \(k\):

\[
\ln\left(\frac{215}{338}\right) = -11k
\]

Now solve for \(k\):

\[
k = -\frac{1}{11} \ln\left(\frac{215}{338}\right)
\]

### Step 2: Find the temperature at \(t = 23\) minutes
Once we calculate \(k\), we can substitute it into the original equation to find the temperature at 23 minutes.

Let me calculate this step by step.The cooling constant \(k\) is approximately \(0.0411\).

Using this constant, the temperature of the rock after 23 minutes is approximately \(213.25^\circ F\).

Would you like more details or have any questions? 

Here are 5 related questions:
1. How would the temperature change if the ambient temperature were higher or lower?
2. What happens to the temperature of the rock as time approaches infinity?
3. Can we use this formula to predict the cooling rate of other materials?
4. How does the cooling constant \(k\) change with different materials?
5. How could you experimentally determine the cooling constant for different objects?

**Tip:** Newton's Law of Cooling assumes a uniform environment, so real-world factors like wind or varying ambient temperatures could slightly affect predictions.

Learn how to calculate the temperature of a rock after 23 minutes using Newton's Law of Cooling. The problem starts with an initial temperature of 420°F, cooling to 297°F after 11 minutes in an ambient temperature of 82°F. Explore step-by-step calculations, including solving for the cooling constant and applying it to find the final temperature.

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