Answer :
To model the temperature of an object cooling according to Newton's Law of Cooling, we use the formula:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \times e^{kt} \][/tex]
where:
- [tex]\( T(t) \)[/tex] is the temperature of the object at time [tex]\( t \)[/tex].
- [tex]\( T_{\text{room}} \)[/tex] is the constant temperature of the surrounding environment (60 degrees Fahrenheit).
- [tex]\( T_{\text{initial}} \)[/tex] is the initial temperature of the object (150 degrees Fahrenheit).
- [tex]\( k \)[/tex] is a constant that describes the rate of cooling.
- [tex]\( t \)[/tex] is the time in minutes.
We are given that after 2 minutes, the temperature of the object is 130 degrees Fahrenheit. We can use this information to find the value of [tex]\( k \)[/tex].
1. Substitute the values into the formula for [tex]\( T(t) \)[/tex] at [tex]\( t = 2 \)[/tex]:
[tex]\[ 130 = 60 + (150 - 60) \times e^{2k} \][/tex]
2. Simplify the left-hand side:
[tex]\[ 130 = 60 + 90 \times e^{2k} \][/tex]
3. Subtract 60 from both sides to isolate the exponential term:
[tex]\[ 70 = 90 \times e^{2k} \][/tex]
4. Divide both sides by 90:
[tex]\[ \frac{70}{90} = e^{2k} \][/tex]
[tex]\[ \frac{7}{9} = e^{2k} \][/tex]
5. Solve for [tex]\( k \)[/tex] by taking the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{7}{9}\right) = 2k \][/tex]
6. Divide by 2 to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{7}{9}\right)}{2} \][/tex]
From the calculations, the numerical results are:
- The fractional term [tex]\( e^{2k} \approx 0.7778 \)[/tex]
- The value of [tex]\( k \approx -0.1257 \)[/tex]
Thus, the equation of the function that models the temperature of the object as a function of time is:
[tex]\[ T(t) = 60 + 90 \times e^{-0.1257t} \][/tex]
This equation describes how the temperature of the object decreases over time towards the temperature of the room.
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \times e^{kt} \][/tex]
where:
- [tex]\( T(t) \)[/tex] is the temperature of the object at time [tex]\( t \)[/tex].
- [tex]\( T_{\text{room}} \)[/tex] is the constant temperature of the surrounding environment (60 degrees Fahrenheit).
- [tex]\( T_{\text{initial}} \)[/tex] is the initial temperature of the object (150 degrees Fahrenheit).
- [tex]\( k \)[/tex] is a constant that describes the rate of cooling.
- [tex]\( t \)[/tex] is the time in minutes.
We are given that after 2 minutes, the temperature of the object is 130 degrees Fahrenheit. We can use this information to find the value of [tex]\( k \)[/tex].
1. Substitute the values into the formula for [tex]\( T(t) \)[/tex] at [tex]\( t = 2 \)[/tex]:
[tex]\[ 130 = 60 + (150 - 60) \times e^{2k} \][/tex]
2. Simplify the left-hand side:
[tex]\[ 130 = 60 + 90 \times e^{2k} \][/tex]
3. Subtract 60 from both sides to isolate the exponential term:
[tex]\[ 70 = 90 \times e^{2k} \][/tex]
4. Divide both sides by 90:
[tex]\[ \frac{70}{90} = e^{2k} \][/tex]
[tex]\[ \frac{7}{9} = e^{2k} \][/tex]
5. Solve for [tex]\( k \)[/tex] by taking the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{7}{9}\right) = 2k \][/tex]
6. Divide by 2 to solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{7}{9}\right)}{2} \][/tex]
From the calculations, the numerical results are:
- The fractional term [tex]\( e^{2k} \approx 0.7778 \)[/tex]
- The value of [tex]\( k \approx -0.1257 \)[/tex]
Thus, the equation of the function that models the temperature of the object as a function of time is:
[tex]\[ T(t) = 60 + 90 \times e^{-0.1257t} \][/tex]
This equation describes how the temperature of the object decreases over time towards the temperature of the room.
