Answer :
To solve this problem, we need to model the change in temperature of the object using Newton's Law of Cooling. This law states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature.
### Step-by-Step Solution:
1. Identify key values:
- Initial temperature of the object, [tex]\( T_0 = 190^\circ \text{F} \)[/tex].
- Ambient room temperature, [tex]\( T_{\text{ambient}} = 65^\circ \text{F} \)[/tex].
- Temperature of the object after 2 minutes, [tex]\( T(2) = 150^\circ \text{F} \)[/tex].
2. Newton's Law of Cooling formula:
- The equation for Newton's Law of Cooling is:
[tex]\[
T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}}) \cdot e^{-kt}
\][/tex]
where [tex]\( T(t) \)[/tex] is the temperature at time [tex]\( t \)[/tex], and [tex]\( k \)[/tex] is the cooling rate constant.
3. Set up the equation for [tex]\( T(2) \)[/tex]:
- We know [tex]\( T(2) = 150^\circ \text{F} \)[/tex].
- Substituting into the formula:
[tex]\[
150 = 65 + (190 - 65) \cdot e^{-2k}
\][/tex]
4. Solve for [tex]\( k \)[/tex]:
- Rearrange the equation:
[tex]\[
150 - 65 = 125 \cdot e^{-2k}
\][/tex]
[tex]\[
85 = 125 \cdot e^{-2k}
\][/tex]
[tex]\[
\frac{85}{125} = e^{-2k}
\][/tex]
[tex]\[
e^{-2k} = 0.68
\][/tex]
5. Calculate [tex]\( k \)[/tex]:
- Take the natural logarithm of both sides:
[tex]\[
-2k = \ln(0.68)
\][/tex]
- Solving for [tex]\( k \)[/tex]:
[tex]\[
k = -\frac{\ln(0.68)}{2}
\][/tex]
- From the given answer, this value evaluates to approximately [tex]\( k \approx 0.1928 \)[/tex].
6. Write the function:
- Substitute [tex]\( k \)[/tex] back into the original equation:
[tex]\[
T(t) = 65 + 125 \cdot e^{-0.1928t}
\][/tex]
Therefore, the equation that models the temperature of the object as a function of time [tex]\( t \)[/tex] is:
[tex]\[ T(t) = 65 + 125 \cdot e^{-0.1928t} \][/tex]
### Step-by-Step Solution:
1. Identify key values:
- Initial temperature of the object, [tex]\( T_0 = 190^\circ \text{F} \)[/tex].
- Ambient room temperature, [tex]\( T_{\text{ambient}} = 65^\circ \text{F} \)[/tex].
- Temperature of the object after 2 minutes, [tex]\( T(2) = 150^\circ \text{F} \)[/tex].
2. Newton's Law of Cooling formula:
- The equation for Newton's Law of Cooling is:
[tex]\[
T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}}) \cdot e^{-kt}
\][/tex]
where [tex]\( T(t) \)[/tex] is the temperature at time [tex]\( t \)[/tex], and [tex]\( k \)[/tex] is the cooling rate constant.
3. Set up the equation for [tex]\( T(2) \)[/tex]:
- We know [tex]\( T(2) = 150^\circ \text{F} \)[/tex].
- Substituting into the formula:
[tex]\[
150 = 65 + (190 - 65) \cdot e^{-2k}
\][/tex]
4. Solve for [tex]\( k \)[/tex]:
- Rearrange the equation:
[tex]\[
150 - 65 = 125 \cdot e^{-2k}
\][/tex]
[tex]\[
85 = 125 \cdot e^{-2k}
\][/tex]
[tex]\[
\frac{85}{125} = e^{-2k}
\][/tex]
[tex]\[
e^{-2k} = 0.68
\][/tex]
5. Calculate [tex]\( k \)[/tex]:
- Take the natural logarithm of both sides:
[tex]\[
-2k = \ln(0.68)
\][/tex]
- Solving for [tex]\( k \)[/tex]:
[tex]\[
k = -\frac{\ln(0.68)}{2}
\][/tex]
- From the given answer, this value evaluates to approximately [tex]\( k \approx 0.1928 \)[/tex].
6. Write the function:
- Substitute [tex]\( k \)[/tex] back into the original equation:
[tex]\[
T(t) = 65 + 125 \cdot e^{-0.1928t}
\][/tex]
Therefore, the equation that models the temperature of the object as a function of time [tex]\( t \)[/tex] is:
[tex]\[ T(t) = 65 + 125 \cdot e^{-0.1928t} \][/tex]
