Answer :
We begin with Newton's Law of Cooling, which states that the temperature of an object at time [tex]$t$[/tex] can be modeled by
[tex]$$
T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \, e^{kt}.
$$[/tex]
In this problem, the given values are:
- The initial temperature of the object is [tex]$T_{\text{initial}} = 190^\circ\text{F}$[/tex].
- The ambient (room) temperature is [tex]$T_{\text{room}} = 65^\circ\text{F}$[/tex].
- After [tex]$t=2$[/tex] minutes, the temperature is [tex]$T(2)=150^\circ\text{F}$[/tex].
First, we calculate the temperature difference between the object and the room at time [tex]$t=0$[/tex]:
[tex]$$
\Delta T = T_{\text{initial}} - T_{\text{room}} = 190 - 65 = 125.
$$[/tex]
Thus, the temperature function takes the form
[tex]$$
T(t) = 65 + 125 \, e^{kt}.
$$[/tex]
At [tex]$t=2$[/tex] minutes, the object's temperature is [tex]$150^\circ\text{F}$[/tex], so we set up the equation
[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]
Divide both sides by [tex]$125$[/tex]:
[tex]$$
e^{2k} = \frac{85}{125} = 0.68.
$$[/tex]
Next, take the natural logarithm of both sides to solve for [tex]$k$[/tex]:
[tex]$$
2k = \ln(0.68).
$$[/tex]
So, the constant [tex]$k$[/tex] is
[tex]$$
k = \frac{1}{2} \ln(0.68).
$$[/tex]
Evaluating this (and rounding to the nearest thousandth) gives
[tex]$$
k \approx -0.193.
$$[/tex]
Now, substitute the value of [tex]$k$[/tex] back into the temperature function:
[tex]$$
T(t)=65+125\, e^{-0.193t}.
$$[/tex]
Thus, the equation that models the temperature of the object as a function of time is
[tex]$$
\boxed{T(t)=65+125\, e^{-0.193t}}.
$$[/tex]
[tex]$$
T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \, e^{kt}.
$$[/tex]
In this problem, the given values are:
- The initial temperature of the object is [tex]$T_{\text{initial}} = 190^\circ\text{F}$[/tex].
- The ambient (room) temperature is [tex]$T_{\text{room}} = 65^\circ\text{F}$[/tex].
- After [tex]$t=2$[/tex] minutes, the temperature is [tex]$T(2)=150^\circ\text{F}$[/tex].
First, we calculate the temperature difference between the object and the room at time [tex]$t=0$[/tex]:
[tex]$$
\Delta T = T_{\text{initial}} - T_{\text{room}} = 190 - 65 = 125.
$$[/tex]
Thus, the temperature function takes the form
[tex]$$
T(t) = 65 + 125 \, e^{kt}.
$$[/tex]
At [tex]$t=2$[/tex] minutes, the object's temperature is [tex]$150^\circ\text{F}$[/tex], so we set up the equation
[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]
Divide both sides by [tex]$125$[/tex]:
[tex]$$
e^{2k} = \frac{85}{125} = 0.68.
$$[/tex]
Next, take the natural logarithm of both sides to solve for [tex]$k$[/tex]:
[tex]$$
2k = \ln(0.68).
$$[/tex]
So, the constant [tex]$k$[/tex] is
[tex]$$
k = \frac{1}{2} \ln(0.68).
$$[/tex]
Evaluating this (and rounding to the nearest thousandth) gives
[tex]$$
k \approx -0.193.
$$[/tex]
Now, substitute the value of [tex]$k$[/tex] back into the temperature function:
[tex]$$
T(t)=65+125\, e^{-0.193t}.
$$[/tex]
Thus, the equation that models the temperature of the object as a function of time is
[tex]$$
\boxed{T(t)=65+125\, e^{-0.193t}}.
$$[/tex]
