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Answer :
We begin by using Newton's Law of Cooling, which states that the temperature of an object at time [tex]$t$[/tex] can be modeled by
[tex]$$
T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}})e^{kt},
$$[/tex]
where
- [tex]$T_0$[/tex] is the initial temperature of the object,
- [tex]$T_{\text{ambient}}$[/tex] is the ambient (room) temperature,
- and [tex]$k$[/tex] is a constant related to the cooling (or heating) rate.
In this problem, we have:
- [tex]$T_0 = 190^\circ\text{F}$[/tex],
- [tex]$T_{\text{ambient}} = 65^\circ\text{F}$[/tex].
Thus, the difference between the initial temperature and the ambient temperature is
[tex]$$
A = T_0 - T_{\text{ambient}} = 190 - 65 = 125.
$$[/tex]
So the temperature function becomes
[tex]$$
T(t) = 65 + 125 \, e^{kt}.
$$[/tex]
We are also given that after [tex]$2$[/tex] minutes the temperature of the object drops to [tex]$150^\circ\text{F}$[/tex]. Therefore, we have
[tex]$$
T(2) = 150.
$$[/tex]
Substitute [tex]$t = 2$[/tex] into the function:
[tex]$$
150 = 65 + 125\, e^{2k}.
$$[/tex]
Subtract [tex]$65$[/tex] from both sides to isolate the exponential term:
[tex]$$
150 - 65 = 125\, e^{2k} \quad \Longrightarrow \quad 85 = 125\, e^{2k}.
$$[/tex]
Now, solve for [tex]$e^{2k}$[/tex]:
[tex]$$
e^{2k} = \frac{85}{125} = \frac{17}{25}.
$$[/tex]
Taking the natural logarithm on both sides gives
[tex]$$
2k = \ln\left(\frac{17}{25}\right).
$$[/tex]
Thus, the constant [tex]$k$[/tex] is
[tex]$$
k = \frac{1}{2} \ln\left(\frac{17}{25}\right).
$$[/tex]
Evaluating the numerical value, we get approximately
[tex]$$
k \approx -0.1928312404059923.
$$[/tex]
Substituting back into the temperature function, the model for the temperature as a function of time is
[tex]$$
T(t) = 125\, e^{-0.1928312404059923\, t} + 65.
$$[/tex]
This is the equation that models the temperature of the object as a function of time.
[tex]$$
T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}})e^{kt},
$$[/tex]
where
- [tex]$T_0$[/tex] is the initial temperature of the object,
- [tex]$T_{\text{ambient}}$[/tex] is the ambient (room) temperature,
- and [tex]$k$[/tex] is a constant related to the cooling (or heating) rate.
In this problem, we have:
- [tex]$T_0 = 190^\circ\text{F}$[/tex],
- [tex]$T_{\text{ambient}} = 65^\circ\text{F}$[/tex].
Thus, the difference between the initial temperature and the ambient temperature is
[tex]$$
A = T_0 - T_{\text{ambient}} = 190 - 65 = 125.
$$[/tex]
So the temperature function becomes
[tex]$$
T(t) = 65 + 125 \, e^{kt}.
$$[/tex]
We are also given that after [tex]$2$[/tex] minutes the temperature of the object drops to [tex]$150^\circ\text{F}$[/tex]. Therefore, we have
[tex]$$
T(2) = 150.
$$[/tex]
Substitute [tex]$t = 2$[/tex] into the function:
[tex]$$
150 = 65 + 125\, e^{2k}.
$$[/tex]
Subtract [tex]$65$[/tex] from both sides to isolate the exponential term:
[tex]$$
150 - 65 = 125\, e^{2k} \quad \Longrightarrow \quad 85 = 125\, e^{2k}.
$$[/tex]
Now, solve for [tex]$e^{2k}$[/tex]:
[tex]$$
e^{2k} = \frac{85}{125} = \frac{17}{25}.
$$[/tex]
Taking the natural logarithm on both sides gives
[tex]$$
2k = \ln\left(\frac{17}{25}\right).
$$[/tex]
Thus, the constant [tex]$k$[/tex] is
[tex]$$
k = \frac{1}{2} \ln\left(\frac{17}{25}\right).
$$[/tex]
Evaluating the numerical value, we get approximately
[tex]$$
k \approx -0.1928312404059923.
$$[/tex]
Substituting back into the temperature function, the model for the temperature as a function of time is
[tex]$$
T(t) = 125\, e^{-0.1928312404059923\, t} + 65.
$$[/tex]
This is the equation that models the temperature of the object as a function of time.
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