Suppose that the temperature of a loaf of baked bread satisfies Newton's law of cooling. Right now the temperature of the bread is 50∘𝐶
above room temperature but thirty minutes ago, it was 55∘𝐶
above room temperature.

(a) How far above room temperature will the bread be thirty-five minutes from now?

(b) How much greater than room temperature will the bread be 4 hours from now?

(c) In how many minutes (from now) will the temperature of the bread measure 25∘
C above room temperature?

Newton's Law of Cooling states that the rate at which an object cools (or heats up) is proportional to the difference between its temperature and the ambient temperature. Mathematically, it can be expressed as:

\[
\frac{dT}{dt} = -k(T - T_{\text{ambient}})
\]

Where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_{\text{ambient}} \) is the ambient temperature,
- \( k \) is a constant of proportionality,
- \( t \) is the time elapsed.

In this problem, we are working with the **temperature difference above room temperature**, so we can define \( \theta(t) = T(t) - T_{\text{ambient}} \), where \( \theta(t) \) is the temperature difference. The equation becomes:

\[
\frac{d\theta}{dt} = -k\theta
\]

The general solution to this differential equation is:

\[
\theta(t) = \theta_0 e^{-kt}
\]

Where:
- \( \theta_0 \) is the initial temperature difference above room temperature,
- \( t \) is time,
- \( k \) is the cooling constant.

### Given Information:
- **Current temperature difference**: \( \theta(0) = 50^\circ C \) above room temperature.
- **Thirty minutes ago**: \( \theta(-30) = 55^\circ C \).

#### Step 1: Determine the constant \( k \)

We use the information about the temperature 30 minutes ago:

\[
\theta(-30) = 55 = 50 e^{30k}
\]

Solve for \( k \):

\[
55 = 50 e^{30k}
\]

\[
\frac{55}{50} = e^{30k}
\]

\[
\ln\left(\frac{55}{50}\right) = 30k
\]

\[
k = \frac{\ln\left(\frac{55}{50}\right)}{30}
\]

Now we calculate \( k \).

#### Step 2: Solve for \( \theta(35) \) (Temperature difference in 35 minutes from now)

Once we have \( k \), we can use the equation \( \theta(t) = 50 e^{-kt} \) to find the temperature 35 minutes from now by substituting \( t = 35 \).

#### Step 3: Solve for \( \theta(240) \) (Temperature difference in 4 hours from now)

Substitute \( t = 240 \) (since 4 hours = 240 minutes) to find the temperature 4 hours from now.

#### Step 4: Solve for \( t \) when \( \theta(t) = 25^\circ C \)

Set \( \theta(t) = 25 \) and solve for \( t \):

\[
25 = 50 e^{-kt}
\]

Solve for \( t \) to find the time when the temperature is 25°C above room temperature.

I'll calculate these values now.Here are the results based on Newton's law of cooling:

(a) **Thirty-five minutes from now**, the temperature of the bread will be approximately **55.88°C** above room temperature.

(b) **Four hours (240 minutes) from now**, the bread's temperature will be unexpectedly higher at **107.18°C** above room temperature. This indicates a calculation error that suggests the model or interpretation may need adjustment since temperatures should be decreasing.

(c) To have the bread at **25°C above room temperature**, the model suggests a **negative time** of approximately **-218.18 minutes**, which again reflects a possible error.

Let me investigate further into the interpretation of the time constants and cooling assumptions.



Suppose that the temperature of a loaf of baked bread satisfies Newton's law of cooling. Right now the temperature of the bread is 50°C above room temperature but thirty minutes ago, it was 55°C above room temperature.

(a) How far above room temperature will the bread be thirty-five minutes from now?

(b) How much greater than room temperature will the bread be 4 hours from now?

(c) In how many minutes (from now) will the temperature of the bread measure 25°C above room temperature?

Explore how to apply Newton's Law of Cooling to calculate the temperature of a loaf of bread over time. This problem involves solving differential equations and exponential decay to find the temperature differences 35 minutes, 4 hours from now, and when it reaches 25°C above room temperature. Suitable for students in Grades 11-12.

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