Answer :
Certainly! Let's walk through the process step-by-step to understand how long it will take for the cake to reach a temperature of [tex]\(75^\circ F\)[/tex].
### Step 1: Understanding Initial Conditions
1. Initial Temperature of the Cake: [tex]\(300^\circ F\)[/tex]
2. Temperature after 10 minutes: [tex]\(200^\circ F\)[/tex]
3. Room Temperature (Ambient Temperature): [tex]\(60^\circ F\)[/tex]
4. Target Temperature: [tex]\(75^\circ F\)[/tex]
### Step 2: Applying Newton's Law of Cooling
Newton's Law of Cooling describes the rate of cooling of an object. The formula is:
[tex]\[
T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) \cdot e^{-kt}
\][/tex]
where:
- [tex]\(T(t)\)[/tex] is the temperature of the object at time [tex]\(t\)[/tex].
- [tex]\(T_{\text{ambient}}\)[/tex] is the ambient (room) temperature.
- [tex]\(T_{\text{initial}}\)[/tex] is the initial temperature of the object.
- [tex]\(k\)[/tex] is a constant that represents the cooling rate.
- [tex]\(t\)[/tex] is the time the object has been cooling.
### Step 3: Calculating the Cooling Constant [tex]\(k\)[/tex]
To find [tex]\(k\)[/tex], we use the information that after 10 minutes, the cake's temperature is [tex]\(200^\circ F\)[/tex].
[tex]\[
200 = 60 + (300 - 60) \cdot e^{-10k}
\][/tex]
This equation is solved for [tex]\(k\)[/tex], yielding:
[tex]\[
k \approx 0.0539
\][/tex]
### Step 4: Finding Time to Reach [tex]\(75^\circ F\)[/tex]
Now, we know the cooling constant [tex]\(k\)[/tex] and want to find [tex]\(t\)[/tex] when the temperature reaches [tex]\(75^\circ F\)[/tex].
[tex]\[
75 = 60 + (300 - 60) \cdot e^{-kt}
\][/tex]
We solve this equation for [tex]\(t\)[/tex], using the previously calculated value of [tex]\(k\)[/tex].
[tex]\[
t \approx 51.44 \text{ minutes}
\][/tex]
Thus, it will take approximately [tex]\(51.44\)[/tex] minutes for the cake to cool down to [tex]\(75^\circ F\)[/tex].
### Step 1: Understanding Initial Conditions
1. Initial Temperature of the Cake: [tex]\(300^\circ F\)[/tex]
2. Temperature after 10 minutes: [tex]\(200^\circ F\)[/tex]
3. Room Temperature (Ambient Temperature): [tex]\(60^\circ F\)[/tex]
4. Target Temperature: [tex]\(75^\circ F\)[/tex]
### Step 2: Applying Newton's Law of Cooling
Newton's Law of Cooling describes the rate of cooling of an object. The formula is:
[tex]\[
T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) \cdot e^{-kt}
\][/tex]
where:
- [tex]\(T(t)\)[/tex] is the temperature of the object at time [tex]\(t\)[/tex].
- [tex]\(T_{\text{ambient}}\)[/tex] is the ambient (room) temperature.
- [tex]\(T_{\text{initial}}\)[/tex] is the initial temperature of the object.
- [tex]\(k\)[/tex] is a constant that represents the cooling rate.
- [tex]\(t\)[/tex] is the time the object has been cooling.
### Step 3: Calculating the Cooling Constant [tex]\(k\)[/tex]
To find [tex]\(k\)[/tex], we use the information that after 10 minutes, the cake's temperature is [tex]\(200^\circ F\)[/tex].
[tex]\[
200 = 60 + (300 - 60) \cdot e^{-10k}
\][/tex]
This equation is solved for [tex]\(k\)[/tex], yielding:
[tex]\[
k \approx 0.0539
\][/tex]
### Step 4: Finding Time to Reach [tex]\(75^\circ F\)[/tex]
Now, we know the cooling constant [tex]\(k\)[/tex] and want to find [tex]\(t\)[/tex] when the temperature reaches [tex]\(75^\circ F\)[/tex].
[tex]\[
75 = 60 + (300 - 60) \cdot e^{-kt}
\][/tex]
We solve this equation for [tex]\(t\)[/tex], using the previously calculated value of [tex]\(k\)[/tex].
[tex]\[
t \approx 51.44 \text{ minutes}
\][/tex]
Thus, it will take approximately [tex]\(51.44\)[/tex] minutes for the cake to cool down to [tex]\(75^\circ F\)[/tex].
