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Answer :
To solve this problem, we will use Newton's Law of Cooling, which describes the rate at which an exposed body changes temperature through radiation. The formula is:
[tex]\[
\frac{T(t) - T_{\text{room}}}{T_0 - T_{\text{room}}} = e^{-kt}
\][/tex]
where:
- [tex]\(T(t)\)[/tex] is the temperature of the object at time [tex]\(t\)[/tex],
- [tex]\(T_{\text{room}}\)[/tex] is the ambient or room temperature,
- [tex]\(T_0\)[/tex] is the initial temperature of the object,
- [tex]\(k\)[/tex] is the cooling constant,
- [tex]\(t\)[/tex] is the time elapsed.
### Step-by-Step Solution:
1. Identify the temperatures and time given:
- Initial temperature of the cake, [tex]\(T_0 = 300^\circ F\)[/tex].
- Temperature after 10 minutes, [tex]\(T(10) = 200^\circ F\)[/tex].
- Room temperature, [tex]\(T_{\text{room}} = 60^\circ F\)[/tex].
2. Apply Newton's Law of Cooling for the first 10 minutes:
We can find the cooling constant [tex]\(k\)[/tex] by plugging the values into the formula for [tex]\(t = 10\)[/tex] minutes:
[tex]\[
\frac{200 - 60}{300 - 60} = e^{-10k}
\][/tex]
Solving this equation gives the cooling constant [tex]\(k\)[/tex].
3. Calculate the cooling constant [tex]\(k\)[/tex]:
- [tex]\(\frac{140}{240} = e^{-10k}\)[/tex]
- Simplifying gives approximately [tex]\(0.5833 = e^{-10k}\)[/tex].
4. Use the calculated cooling constant to find the time required to reach [tex]\(75^\circ F\)[/tex]:
Now, we use the same formula to find when the cake will reach [tex]\(75^\circ F\)[/tex]:
[tex]\[
\frac{75 - 60}{300 - 60} = e^{-kt'}
\][/tex]
Solving for [tex]\(t'\)[/tex] (the time needed to reach [tex]\(75^\circ F\)[/tex]):
5. Finding [tex]\(t'\)[/tex]:
- Substitute the cooling constant found earlier into this equation:
- [tex]\(\frac{15}{240} = e^{-kt'}\)[/tex]
- Using the same constant, solve for [tex]\(t'\)[/tex], which gives approximately [tex]\(t' = 5.14\)[/tex] minutes.
Therefore, it takes about 5.14 minutes for the cake to cool from 300°F to 75°F in a room at 60°F, after the initial cooling to 200°F in the first 10 minutes.
[tex]\[
\frac{T(t) - T_{\text{room}}}{T_0 - T_{\text{room}}} = e^{-kt}
\][/tex]
where:
- [tex]\(T(t)\)[/tex] is the temperature of the object at time [tex]\(t\)[/tex],
- [tex]\(T_{\text{room}}\)[/tex] is the ambient or room temperature,
- [tex]\(T_0\)[/tex] is the initial temperature of the object,
- [tex]\(k\)[/tex] is the cooling constant,
- [tex]\(t\)[/tex] is the time elapsed.
### Step-by-Step Solution:
1. Identify the temperatures and time given:
- Initial temperature of the cake, [tex]\(T_0 = 300^\circ F\)[/tex].
- Temperature after 10 minutes, [tex]\(T(10) = 200^\circ F\)[/tex].
- Room temperature, [tex]\(T_{\text{room}} = 60^\circ F\)[/tex].
2. Apply Newton's Law of Cooling for the first 10 minutes:
We can find the cooling constant [tex]\(k\)[/tex] by plugging the values into the formula for [tex]\(t = 10\)[/tex] minutes:
[tex]\[
\frac{200 - 60}{300 - 60} = e^{-10k}
\][/tex]
Solving this equation gives the cooling constant [tex]\(k\)[/tex].
3. Calculate the cooling constant [tex]\(k\)[/tex]:
- [tex]\(\frac{140}{240} = e^{-10k}\)[/tex]
- Simplifying gives approximately [tex]\(0.5833 = e^{-10k}\)[/tex].
4. Use the calculated cooling constant to find the time required to reach [tex]\(75^\circ F\)[/tex]:
Now, we use the same formula to find when the cake will reach [tex]\(75^\circ F\)[/tex]:
[tex]\[
\frac{75 - 60}{300 - 60} = e^{-kt'}
\][/tex]
Solving for [tex]\(t'\)[/tex] (the time needed to reach [tex]\(75^\circ F\)[/tex]):
5. Finding [tex]\(t'\)[/tex]:
- Substitute the cooling constant found earlier into this equation:
- [tex]\(\frac{15}{240} = e^{-kt'}\)[/tex]
- Using the same constant, solve for [tex]\(t'\)[/tex], which gives approximately [tex]\(t' = 5.14\)[/tex] minutes.
Therefore, it takes about 5.14 minutes for the cake to cool from 300°F to 75°F in a room at 60°F, after the initial cooling to 200°F in the first 10 minutes.
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