A graphing calculator is recommended.

This exercise uses Newton's Law of Cooling.

A kettle full of water is brought to a boil in a room with temperature 20°C. After 15 minutes the temperature of the water has decreased from 100° to 75°C. Find the temperature after another 12 min. (Round your answer to one decimal place.)
 
Incorrect: Your answer is incorrect.
 °C

To solve this problem using Newton's Law of Cooling, we can use the formula:

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

where:
- \(T(t)\) is the temperature at time \(t\),
- \(T_a\) is the ambient temperature (20°C),
- \(T_0\) is the initial temperature (100°C),
- \(k\) is the cooling constant,
- \(t\) is the time in minutes.

### Step 1: Find \(k\)

Given:
- \(T(15) = 75°C\) at \(t = 15\) minutes.

Plugging in the values:

\[
75 = 20 + (100 - 20)e^{-15k}
\]

This simplifies to:

\[
75 = 20 + 80e^{-15k}
\]
\[
55 = 80e^{-15k}
\]
\[
e^{-15k} = \frac{55}{80} = 0.6875
\]

Taking the natural logarithm:

\[
-15k = \ln(0.6875)
\]
\[
k = -\frac{\ln(0.6875)}{15} \approx 0.0205
\]

### Step 2: Find the temperature after another 12 minutes

Now we need to find the temperature after \(15 + 12 = 27\) minutes:

\[
T(27) = 20 + (100 - 20)e^{-27k}
\]

Substituting \(k\):

\[
T(27) = 20 + 80e^{-27(0.0205)}
\]
Calculating \(e^{-27(0.0205)}\):

\[
T(27) = 20 + 80e^{-0.5535} \approx 20 + 80 \cdot 0.5752 \approx 20 + 46.016 \approx 66.0°C
\]

Thus, the temperature after another 12 minutes is approximately:

\[
\boxed{66.0 \, \text{°C}}
\]

### Additional Questions
1. What is the significance of the cooling constant \(k\)?
2. How does the ambient temperature affect the cooling process?
3. What assumptions does Newton's Law of Cooling make about temperature changes?
4. How would the result change if the initial temperature was higher or lower?
5. What practical applications exist for Newton's Law of Cooling in everyday scenarios?

### Tip
For accurate results in problems involving exponential decay or growth, always check the values substituted into the equations.

A kettle full of water is brought to a boil in a room with temperature 20°C. After 15 minutes the temperature of the water has decreased from 100° to 75°C. Find the temperature after another 12 min. (Round your answer to one decimal place.)

Discover how to calculate the temperature of water in a kettle after applying Newton's Law of Cooling. This example shows the temperature decreasing from 100°C to 75°C in 15 minutes, and find the temperature after an additional 12 minutes. Suitable for students in Grades 10-12.

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