Answer :
To solve this problem, we need to use Newton's Law of Cooling, which helps us understand how the temperature of an object changes over time as it cools down or warms up to match the surrounding environment. The formula is:
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
Where:
- [tex]\( T(t) \)[/tex] is the temperature of the object at time [tex]\( t \)[/tex].
- [tex]\( T_{\text{room}} \)[/tex] is the ambient room temperature.
- [tex]\( T_{\text{initial}} \)[/tex] is the initial temperature of the object.
- [tex]\( k \)[/tex] is the cooling constant.
- [tex]\( t \)[/tex] is the time in minutes.
Given:
- The initial temperature of the soup, [tex]\( T_{\text{initial}} = 100^\circ F \)[/tex].
- The room temperature, [tex]\( T_{\text{room}} = 69^\circ F \)[/tex].
- After 15 minutes, the temperature of the soup [tex]\( T(t) = 95^\circ F \)[/tex].
Let's find the value of the cooling constant [tex]\( k \)[/tex] using the given conditions:
1. Set up the equation with the known values:
[tex]\[ 95 = 69 + (100 - 69) \cdot e^{-15k} \][/tex]
2. Simplify:
[tex]\[ 95 = 69 + 31 \cdot e^{-15k} \][/tex]
3. Subtract 69 from both sides:
[tex]\[ 26 = 31 \cdot e^{-15k} \][/tex]
4. Divide both sides by 31:
[tex]\[ \frac{26}{31} = e^{-15k} \][/tex]
5. Solve for [tex]\( k \)[/tex] by taking the natural logarithm of both sides:
[tex]\[ \ln{\left(\frac{26}{31}\right)} = -15k \][/tex]
6. Divide by -15 to solve for [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{\ln{\left(\frac{26}{31}\right)}}{15} \][/tex]
Using this calculation method, we find the value of [tex]\( k \)[/tex] to be approximately [tex]\( 0.011726 \)[/tex].
Now, we evaluate which of the given options match this constant [tex]\( k \)[/tex]:
- Option 1: [tex]\( S(p) = 31 e^{-0.011726 t} + 69 \)[/tex]
- Option 2: [tex]\( S(p) = 31 e^{-0.05435 t} + 69 \)[/tex]
- Option 3: [tex]\( S(p) = 100 e^{-0.011726 t} + 69 \)[/tex]
- Option 4: [tex]\( S(p) = 31 e^{-0.05435 t} + 100 \)[/tex]
Since the calculated value of [tex]\( k \)[/tex] is approximately [tex]\( 0.011726 \)[/tex], we need to check which equation uses this value of [tex]\( k \)[/tex] and the correct initial displacement from room temperature (i.e., the difference between initial and room temperatures).
The correct cooling equation is:
- Option 1: [tex]\( S(p) = 31 e^{-0.011726 t} + 69 \)[/tex]
This matches the scenario because it uses the correct [tex]\( k \)[/tex] value and correctly calculates the soup's temperature based on the initial temperature difference of 31 degrees.
[tex]\[ T(t) = T_{\text{room}} + (T_{\text{initial}} - T_{\text{room}}) \cdot e^{-kt} \][/tex]
Where:
- [tex]\( T(t) \)[/tex] is the temperature of the object at time [tex]\( t \)[/tex].
- [tex]\( T_{\text{room}} \)[/tex] is the ambient room temperature.
- [tex]\( T_{\text{initial}} \)[/tex] is the initial temperature of the object.
- [tex]\( k \)[/tex] is the cooling constant.
- [tex]\( t \)[/tex] is the time in minutes.
Given:
- The initial temperature of the soup, [tex]\( T_{\text{initial}} = 100^\circ F \)[/tex].
- The room temperature, [tex]\( T_{\text{room}} = 69^\circ F \)[/tex].
- After 15 minutes, the temperature of the soup [tex]\( T(t) = 95^\circ F \)[/tex].
Let's find the value of the cooling constant [tex]\( k \)[/tex] using the given conditions:
1. Set up the equation with the known values:
[tex]\[ 95 = 69 + (100 - 69) \cdot e^{-15k} \][/tex]
2. Simplify:
[tex]\[ 95 = 69 + 31 \cdot e^{-15k} \][/tex]
3. Subtract 69 from both sides:
[tex]\[ 26 = 31 \cdot e^{-15k} \][/tex]
4. Divide both sides by 31:
[tex]\[ \frac{26}{31} = e^{-15k} \][/tex]
5. Solve for [tex]\( k \)[/tex] by taking the natural logarithm of both sides:
[tex]\[ \ln{\left(\frac{26}{31}\right)} = -15k \][/tex]
6. Divide by -15 to solve for [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{\ln{\left(\frac{26}{31}\right)}}{15} \][/tex]
Using this calculation method, we find the value of [tex]\( k \)[/tex] to be approximately [tex]\( 0.011726 \)[/tex].
Now, we evaluate which of the given options match this constant [tex]\( k \)[/tex]:
- Option 1: [tex]\( S(p) = 31 e^{-0.011726 t} + 69 \)[/tex]
- Option 2: [tex]\( S(p) = 31 e^{-0.05435 t} + 69 \)[/tex]
- Option 3: [tex]\( S(p) = 100 e^{-0.011726 t} + 69 \)[/tex]
- Option 4: [tex]\( S(p) = 31 e^{-0.05435 t} + 100 \)[/tex]
Since the calculated value of [tex]\( k \)[/tex] is approximately [tex]\( 0.011726 \)[/tex], we need to check which equation uses this value of [tex]\( k \)[/tex] and the correct initial displacement from room temperature (i.e., the difference between initial and room temperatures).
The correct cooling equation is:
- Option 1: [tex]\( S(p) = 31 e^{-0.011726 t} + 69 \)[/tex]
This matches the scenario because it uses the correct [tex]\( k \)[/tex] value and correctly calculates the soup's temperature based on the initial temperature difference of 31 degrees.
