This problem involves the Newton's Law of Cooling, which describes the rate of temperature change of an object based on the difference between the object's temperature and its surrounding temperature. The formula for Newton's Law of Cooling is:

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

Where:

• $T(t)$ is the temperature of the object at time $t$,
• $T_{\text{outside}}$ is the constant outside temperature,
• $T_{\text{initial}}$ is the initial temperature of the object,
• $k$ is the rate constant, and
• $t$ is the time elapsed.

Part 1: Temperature at 8:00 AM

Given data:

• Initial temperature inside the lecture hall, $T_{\text{initial}} = 41^\circ$F.
• Outside temperature, $T_{\text{outside}} = 41^\circ$F.
• Furnace temperature, $T_{\text{furnace}} = 72^\circ$F.
• The time constant for the building is $\frac{1}{K} = 33$ hours, so $K = \frac{1}{33}$.
• The time constant for the building and heating system together is $\frac{1}{K_1} = 12$ hours, so $K_1 = \frac{1}{12}$.

Since the furnace was turned on at 7:00 AM, we need to find the temperature at 8:00 AM, which is after 1 hour.

Part 2: When the temperature reaches 68°F

Now we need to determine when the temperature inside the lecture hall will reach $68^\circ$F.

Let's start by applying the formula for both parts.

I'll calculate the temperature at 8:00 AM first and then determine when it will reach 68°F.

Let me compute this now. math

Given values

T_outside = 41 # degrees F (outside temperature) T_initial = 41 # degrees F (initial temperature inside) T_furnace = 72 # degrees F (furnace temperature) K1 = 1 / 12 # rate constant for the heating system (in hr^-1)

Time elapsed: 1 hour from 7:00 AM to 8:00 AM

t = 1 # time in hours

Applying the formula for Newton's Law of Cooling

T_8am = T_outside + (T_furnace - T_outside) * math.exp(-K1 * t) T_8am = round(T_8am, 1)

Now, calculate when the temperature will reach 68°F

target_temp = 68 # degrees F

Rearrange the formula to solve for time (t):

target_temp = T_outside + (T_furnace - T_outside) * exp(-K1 * t)

target_temp - T_outside = (T_furnace - T_outside) * exp(-K1 * t)

(target_temp - T_outside) / (T_furnace - T_outside) = exp(-K1 * t)

ln((target_temp - T_outside) / (T_furnace - T_outside)) = -K1 * t

t = -ln((target_temp - T_outside) / (T_furnace - T_outside)) / K1

t_target = -math.log((target_temp - T_outside) / (T_furnace - T_outside)) / K1 t_target = round(t_target, 1)

T_8am, t_targetAt 8:00 AM, the temperature inside the lecture hall will be approximately 69.5°F.

To reach a temperature of 68°F, it will take approximately 1.7 hours from when the furnace was turned on, which is around 8:42 AM.

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If you'd like more details or further explanations, feel free to ask! Here are some related questions:

1. How does the time constant $K_1$ affect the heating rate?
2. What would happen if the furnace temperature were set higher, say at 75°F?
3. How would the result change if the outside temperature were different, for example, 30°F?
4. How can we modify the formula if the temperature inside the building also loses heat to the outside (i.e., a cooling factor)?
5. What is the general behavior of temperature change in such heating systems over time?

Tip: When working with Newton's Law of Cooling, remember that the cooling rate depends on the difference between the object's temperature and the surrounding temperature. The larger the difference, the faster the change.