A thermometer reading 6degreesC is brought into a room with a constant temperature of 38 degrees C. If the thermometer reads 17degreesC after 2 ​minutes, what will it read after being in the room for 4 ​minutes? For 9 ​minutes?
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Part 1
After 4 ​minutes, the thermometer will read

enter your response heredegreesC.
​(Do not round until the final answer. Then round to the nearest hundredth as​ needed.)

To solve this problem, we will use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its 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 (room) temperature,
- \( T_0 \) is the initial temperature of the object,
- \( k \) is the cooling rate constant,
- \( t \) is the time elapsed.

### Given:
- Initial temperature, \( T_0 = 6^\circ \text{C} \)
- Ambient temperature, \( T_{\text{ambient}} = 38^\circ \text{C} \)
- Temperature after 2 minutes, \( T(2) = 17^\circ \text{C} \)

### Step 1: Find the cooling rate constant \( k \)
Substitute the given values into the equation for \( T(2) \):

\[
17 = 38 + (6 - 38) e^{-2k}
\]

Simplify the equation:

\[
17 = 38 - 32 e^{-2k}
\]

\[
32 e^{-2k} = 21
\]

\[
e^{-2k} = \frac{21}{32}
\]

Take the natural logarithm of both sides:

\[
-2k = \ln\left(\frac{21}{32}\right)
\]

\[
k = -\frac{1}{2} \ln\left(\frac{21}{32}\right)
\]

Calculating \( k \):

\[
k = -\frac{1}{2} \ln\left(\frac{21}{32}\right) \approx 0.207
\]

### Step 2: Find the temperature after 4 minutes
Now, use the formula with \( t = 4 \):

\[
T(4) = 38 + (6 - 38) e^{-0.207 \cdot 4}
\]

\[
T(4) = 38 - 32 e^{-0.828}
\]

Calculating \( e^{-0.828} \approx 0.437 \):

\[
T(4) = 38 - 32 \times 0.437
\]

\[
T(4) = 38 - 13.984 \approx 24.02
\]

So, after 4 minutes, the thermometer will read approximately **24.02°C**.

### Step 3: Find the temperature after 9 minutes
Repeat the process for \( t = 9 \):

\[
T(9) = 38 + (6 - 38) e^{-0.207 \cdot 9}
\]

\[
T(9) = 38 - 32 e^{-1.863}
\]

Calculating \( e^{-1.863} \approx 0.155 \):

\[
T(9) = 38 - 32 \times 0.155
\]

\[
T(9) = 38 - 4.96 \approx 33.04
\]

So, after 9 minutes, the thermometer will read approximately **33.04°C**.

Would you like more details or have any questions?

### Related Questions:
1. What would the thermometer read after 5 minutes?
2. How would the cooling rate change if the ambient temperature was different?
3. What is the physical significance of the cooling rate constant \( k \)?
4. How does Newton's Law of Cooling apply to real-world situations?
5. Can Newton's Law of Cooling be used to determine the time of death in forensic science?

**Tip:** When solving exponential equations like these, understanding logarithmic functions and their properties is essential for simplifying calculations effectively.

A thermometer reading 6 degrees C is brought into a room with a constant temperature of 38 degrees C. If the thermometer reads 17 degrees C after 2 minutes, what will it read after being in the room for 4 minutes? For 9 minutes?

Learn how to solve a real-world application of Newton's Law of Cooling. A thermometer's temperature changes over time as it is brought into a warmer room. This problem involves finding the thermometer's temperature after 4 and 9 minutes, using exponential decay principles. Ideal for students in Grades 11-12.

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