Answer :
We start with Newton’s Law of Cooling, which tells us that the temperature of an object over time is given by
[tex]$$
T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) e^{kt},
$$[/tex]
where
- [tex]$T_{\text{initial}}$[/tex] is the initial temperature of the object,
- [tex]$T_{\text{room}}$[/tex] is the ambient room temperature,
- [tex]$k$[/tex] is a constant, and
- [tex]$t$[/tex] is the time in minutes.
In this problem:
- The object’s initial temperature is [tex]$200^\circ$[/tex]F,
- The room temperature is [tex]$50^\circ$[/tex]F, and
- After [tex]$8$[/tex] minutes, the object's temperature is [tex]$100^\circ$[/tex]F.
1. First, substitute these values into the equation. The difference between the initial temperature and the room temperature is
[tex]$$
\Delta T = T_{\text{initial}} - T_{\text{room}} = 200 - 50 = 150.
$$[/tex]
This allows us to write the temperature function as
[tex]$$
T(t) = 50 + 150e^{kt}.
$$[/tex]
2. Next, use the fact that [tex]$T(8) = 100$[/tex] to solve for [tex]$k$[/tex]. Substitute [tex]$t = 8$[/tex] and [tex]$T(8) = 100$[/tex]:
[tex]$$
100 = 50 + 150e^{8k}.
$$[/tex]
3. Subtract [tex]$50$[/tex] from both sides:
[tex]$$
100 - 50 = 150e^{8k},
$$[/tex]
which simplifies to
[tex]$$
50 = 150e^{8k}.
$$[/tex]
4. Divide both sides by [tex]$150$[/tex]:
[tex]$$
\frac{50}{150} = e^{8k},
$$[/tex]
so
[tex]$$
e^{8k} = \frac{1}{3}.
$$[/tex]
5. Take the natural logarithm of both sides to solve for [tex]$k$[/tex]:
[tex]$$
8k = \ln\left(\frac{1}{3}\right).
$$[/tex]
Thus,
[tex]$$
k = \frac{\ln\left(\frac{1}{3}\right)}{8}.
$$[/tex]
6. Evaluating the natural logarithm (to the necessary accuracy) gives
[tex]$$
k \approx -0.137.
$$[/tex]
7. Substitute the value of [tex]$k$[/tex] back into the equation for [tex]$T(t)$[/tex]:
[tex]$$
T(t) = 50 + 150e^{-0.137t}.
$$[/tex]
This is the equation that models the temperature of the object as a function of time, with the constant rounded to the nearest thousandth.
Thus, the final answer is:
[tex]$$
\boxed{T(t) = 50 + 150e^{-0.137t}}.
$$[/tex]
[tex]$$
T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) e^{kt},
$$[/tex]
where
- [tex]$T_{\text{initial}}$[/tex] is the initial temperature of the object,
- [tex]$T_{\text{room}}$[/tex] is the ambient room temperature,
- [tex]$k$[/tex] is a constant, and
- [tex]$t$[/tex] is the time in minutes.
In this problem:
- The object’s initial temperature is [tex]$200^\circ$[/tex]F,
- The room temperature is [tex]$50^\circ$[/tex]F, and
- After [tex]$8$[/tex] minutes, the object's temperature is [tex]$100^\circ$[/tex]F.
1. First, substitute these values into the equation. The difference between the initial temperature and the room temperature is
[tex]$$
\Delta T = T_{\text{initial}} - T_{\text{room}} = 200 - 50 = 150.
$$[/tex]
This allows us to write the temperature function as
[tex]$$
T(t) = 50 + 150e^{kt}.
$$[/tex]
2. Next, use the fact that [tex]$T(8) = 100$[/tex] to solve for [tex]$k$[/tex]. Substitute [tex]$t = 8$[/tex] and [tex]$T(8) = 100$[/tex]:
[tex]$$
100 = 50 + 150e^{8k}.
$$[/tex]
3. Subtract [tex]$50$[/tex] from both sides:
[tex]$$
100 - 50 = 150e^{8k},
$$[/tex]
which simplifies to
[tex]$$
50 = 150e^{8k}.
$$[/tex]
4. Divide both sides by [tex]$150$[/tex]:
[tex]$$
\frac{50}{150} = e^{8k},
$$[/tex]
so
[tex]$$
e^{8k} = \frac{1}{3}.
$$[/tex]
5. Take the natural logarithm of both sides to solve for [tex]$k$[/tex]:
[tex]$$
8k = \ln\left(\frac{1}{3}\right).
$$[/tex]
Thus,
[tex]$$
k = \frac{\ln\left(\frac{1}{3}\right)}{8}.
$$[/tex]
6. Evaluating the natural logarithm (to the necessary accuracy) gives
[tex]$$
k \approx -0.137.
$$[/tex]
7. Substitute the value of [tex]$k$[/tex] back into the equation for [tex]$T(t)$[/tex]:
[tex]$$
T(t) = 50 + 150e^{-0.137t}.
$$[/tex]
This is the equation that models the temperature of the object as a function of time, with the constant rounded to the nearest thousandth.
Thus, the final answer is:
[tex]$$
\boxed{T(t) = 50 + 150e^{-0.137t}}.
$$[/tex]
