A thermometer is taken from a room where the temperature is 21 degrees 𝐶 to the outdoors, where the temperature is −4 degrees 𝐶. After one minute the thermometer reads 10 degrees 𝐶. 
(a) What will the reading on the thermometer be after 2 more minutes?
(b) After how many minutes will the thermometer read −3 degrees 𝐶?

This problem is related to **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 current temperature and the surrounding temperature. The equation for Newton's Law of Cooling is:

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

where:
- \(T(t)\) is the temperature of the object at time \(t\),
- \(T_s\) is the surrounding (ambient) temperature,
- \(T_0\) is the initial temperature of the object (at time \(t = 0\)),
- \(k\) is the cooling constant,
- \(t\) is the time in minutes.

### Given:
- Initial room temperature: \(T_0 = 21^\circ C\),
- Outdoor temperature: \(T_s = -4^\circ C\),
- After 1 minute, the thermometer reads \(T(1) = 10^\circ C\).

#### Step 1: Find the cooling constant \(k\)
From the equation:

\[
T(1) = T_s + (T_0 - T_s) e^{-k(1)}
\]

Substituting the known values:

\[
10 = -4 + (21 - (-4)) e^{-k}
\]

This simplifies to:

\[
10 = -4 + 25 e^{-k}
\]

\[
14 = 25 e^{-k}
\]

\[
e^{-k} = \frac{14}{25}
\]

Taking the natural logarithm on both sides:

\[
-k = \ln\left(\frac{14}{25}\right)
\]

\[
k = -\ln\left(\frac{14}{25}\right) \approx 0.5575
\]

#### Step 2: (a) Find the temperature after 2 more minutes (total 3 minutes)
We now use the same formula to find \(T(3)\):

\[
T(3) = -4 + (21 - (-4)) e^{-0.5575 \times 3}
\]

Simplifying:

\[
T(3) = -4 + 25 e^{-1.6725}
\]

We can now calculate this value.

#### Step 3: (b) Find the time when the thermometer reads \(-3^\circ C\)
Using the formula again, but this time with \(T(t) = -3^\circ C\):

\[
-3 = -4 + (21 - (-4)) e^{-0.5575t}
\]

Simplifying:

\[
1 = 25 e^{-0.5575t}
\]

\[
e^{-0.5575t} = \frac{1}{25}
\]

Taking the natural logarithm on both sides:

\[
-0.5575t = \ln\left(\frac{1}{25}\right)
\]

\[
t = \frac{\ln(25)}{0.5575}
\]

Let's calculate the answers for both parts.### Results:
- **(a)** The temperature on the thermometer after 2 more minutes (a total of 3 minutes) will be approximately **0.69°C**.
- **(b)** The thermometer will read **−3°C** after approximately **5.77 minutes**.

Would you like further details on any of these steps or have other questions?

Here are some related questions:
1. How does Newton's Law of Cooling change if the surrounding temperature is not constant?
2. How would the equation differ if the object were heating instead of cooling?
3. What are other real-world applications of Newton's Law of Cooling?
4. How do different materials affect the cooling constant \(k\)?
5. How does the accuracy of this model change over very long time periods?

**Tip:** Newton's Law of Cooling is also useful for predicting temperature changes in forensic science, especially in estimating time of death.

A thermometer is taken from a room where the temperature is 21 degrees C to the outdoors, where the temperature is −4 degrees C. After one minute the thermometer reads 10 degrees C. (a) What will the reading on the thermometer be after 2 more minutes? (b) After how many minutes will the thermometer read −3 degrees C?

Solve a Newton's Law of Cooling problem involving a thermometer taken from 21°C indoors to −4°C outdoors. After one minute, the thermometer reads 10°C. Learn how to calculate the temperature after 2 more minutes and determine when it reaches −3°C. Suitable for Grades 11-12.

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