This problem can be modeled using Newton's Law of Cooling, which describes the rate at which an object changes temperature in relation to the surrounding environment. The formula for this is:

$\frac{dT}{dt} = k(T - T_{\text{env}})$

Where:

• $T$ is the temperature of the object (thermometer).
• $T_{\text{env}}$ is the ambient temperature (environment temperature).
• $k$ is a constant that depends on the characteristics of the object.
• $t$ is the time.

Step 1: Set up the problem

• Initial temperature of the thermometer $T_0 = 18^\circ C$ (room temperature).
• Outdoor temperature $T_{\text{env}} = -4^\circ C$.
• After 1 minute, the thermometer reads $T_1 = 11^\circ C$.

We are asked to find the temperature after 2 more minutes, i.e., after 3 minutes in total.

Step 2: Solve the differential equation

The solution to the differential equation is:

$T(t) = T_{\text{env}} + (T_0 - T_{\text{env}}) e^{-kt}$

Where:

• $T(t)$ is the temperature at time $t$.
• $T_{\text{env}}$ is the constant ambient temperature (-4°C).
• $T_0$ is the initial temperature of the thermometer (18°C).
• $k$ is the constant we need to find.

Step 3: Find the constant $k$

At $t = 1$ minute, the thermometer reads 11°C:

$T(1) = -4 + (18 - (-4)) e^{-k \cdot 1}$ $11 = -4 + 22 e^{-k}$ $15 = 22 e^{-k}$ $e^{-k} = \frac{15}{22}$ $-k = \ln\left(\frac{15}{22}\right)$ $k = -\ln\left(\frac{15}{22}\right)$

Now we can compute the value of $k$.

Step 4: Calculate the temperature at $t = 3$ minutes

Once we have $k$, we can use the equation for $T(t)$ to find the temperature at 3 minutes:

$T(3) = -4 + 22 e^{-k \cdot 3}$

I'll calculate this now. math

Given values

T_env = -4 # Ambient temperature in Celsius T_0 = 18 # Initial temperature in Celsius T_1 = 11 # Temperature after 1 minute t_1 = 1 # Time after 1 minute

Calculate k from the equation T(1) = T_env + (T_0 - T_env) * exp(-k * t_1)

k = -math.log((T_1 - T_env) / (T_0 - T_env))

Now calculate the temperature after 3 minutes using the equation T(t) = T_env + (T_0 - T_env) * exp(-k * t)

t_3 = 3 T_3 = T_env + (T_0 - T_env) * math.exp(-k * t_3)

T_3The temperature on the thermometer after 3 minutes will be approximately 2.97°C.

Would you like any further details on this problem? Here are some related questions for you:

1. How would the rate of cooling change if the ambient temperature were higher?
2. How does Newton's Law of Cooling apply to objects cooling in water versus air?
3. What assumptions are made in Newton's Law of Cooling that may not always be true in real-world situations?
4. How would you calculate the time it would take for the thermometer to reach 0°C using this model?
5. How can we use this model to predict the temperature of an object after any given time?

Tip: Newton's Law of Cooling is most accurate when the temperature difference between the object and the environment is large, and it assumes a constant cooling rate.