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Answer :
We can model the cooling of the object using Newton's Law of Cooling, which states that the temperature [tex]$T(t)$[/tex] at time [tex]$t$[/tex] is given by
[tex]$$
T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) e^{kt},
$$[/tex]
where
- [tex]$T_{\text{ambient}}$[/tex] is the ambient (room) temperature,
- [tex]$T_{\text{initial}}$[/tex] is the initial temperature of the object, and
- [tex]$k$[/tex] is the constant that describes the rate of cooling.
In this problem:
- The ambient temperature is [tex]$T_{\text{ambient}} = 50^\circ\text{F}$[/tex],
- The initial temperature is [tex]$T_{\text{initial}} = 200^\circ\text{F}$[/tex], so the difference is [tex]$200 - 50 = 150^\circ\text{F}$[/tex].
Thus, the equation becomes
[tex]$$
T(t) = 50 + 150 \, e^{kt}.
$$[/tex]
It is also given that after [tex]$t = 8$[/tex] minutes the temperature of the object is [tex]$100^\circ\text{F}$[/tex]. Substitute [tex]$t = 8$[/tex] and [tex]$T(8) = 100$[/tex] into the equation:
[tex]$$
100 = 50 + 150 \, e^{8k}.
$$[/tex]
Subtract [tex]$50$[/tex] from both sides:
[tex]$$
50 = 150 \, e^{8k}.
$$[/tex]
Divide both sides by [tex]$150$[/tex]:
[tex]$$
e^{8k} = \frac{50}{150} = \frac{1}{3}.
$$[/tex]
Now take the natural logarithm on both sides:
[tex]$$
8k = \ln\left(\frac{1}{3}\right).
$$[/tex]
Solve for [tex]$k$[/tex]:
[tex]$$
k = \frac{\ln\left(\frac{1}{3}\right)}{8}.
$$[/tex]
Evaluating [tex]$k$[/tex], we find it is approximately
[tex]$$
k \approx -0.137,
$$[/tex]
rounded to the nearest thousandth.
Thus, the equation that models the temperature of the object as a function of time is
[tex]$$
T(t) = 50 + 150 \, e^{-0.137t}.
$$[/tex]
This completes the step-by-step derivation of the cooling model.
[tex]$$
T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) e^{kt},
$$[/tex]
where
- [tex]$T_{\text{ambient}}$[/tex] is the ambient (room) temperature,
- [tex]$T_{\text{initial}}$[/tex] is the initial temperature of the object, and
- [tex]$k$[/tex] is the constant that describes the rate of cooling.
In this problem:
- The ambient temperature is [tex]$T_{\text{ambient}} = 50^\circ\text{F}$[/tex],
- The initial temperature is [tex]$T_{\text{initial}} = 200^\circ\text{F}$[/tex], so the difference is [tex]$200 - 50 = 150^\circ\text{F}$[/tex].
Thus, the equation becomes
[tex]$$
T(t) = 50 + 150 \, e^{kt}.
$$[/tex]
It is also given that after [tex]$t = 8$[/tex] minutes the temperature of the object is [tex]$100^\circ\text{F}$[/tex]. Substitute [tex]$t = 8$[/tex] and [tex]$T(8) = 100$[/tex] into the equation:
[tex]$$
100 = 50 + 150 \, e^{8k}.
$$[/tex]
Subtract [tex]$50$[/tex] from both sides:
[tex]$$
50 = 150 \, e^{8k}.
$$[/tex]
Divide both sides by [tex]$150$[/tex]:
[tex]$$
e^{8k} = \frac{50}{150} = \frac{1}{3}.
$$[/tex]
Now take the natural logarithm on both sides:
[tex]$$
8k = \ln\left(\frac{1}{3}\right).
$$[/tex]
Solve for [tex]$k$[/tex]:
[tex]$$
k = \frac{\ln\left(\frac{1}{3}\right)}{8}.
$$[/tex]
Evaluating [tex]$k$[/tex], we find it is approximately
[tex]$$
k \approx -0.137,
$$[/tex]
rounded to the nearest thousandth.
Thus, the equation that models the temperature of the object as a function of time is
[tex]$$
T(t) = 50 + 150 \, e^{-0.137t}.
$$[/tex]
This completes the step-by-step derivation of the cooling model.
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