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To model the temperature of the hot chocolate over time given the initial conditions, we can use an exponential cooling model. The form of the function is [tex]\( C(t) = a \cdot b^t + c \)[/tex], where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are constants to be determined.
Given:
- Initial temperature of hot chocolate, [tex]\(C(0) = 100\)[/tex] degrees Celsius.
- Temperature after 5 minutes, [tex]\(C(5) = 80\)[/tex] degrees Celsius.
- Outside (ambient) temperature is 0 degrees Celsius.
### Step-by-Step Solution
1. Determine [tex]\(c\)[/tex]:
The ambient temperature, which the hot chocolate is cooling towards, is 0 degrees Celsius. Therefore, [tex]\( c = 0 \)[/tex].
2. Determine [tex]\(a\)[/tex]:
The constant [tex]\( a \)[/tex] represents the difference between the initial temperature of the hot chocolate and the outside temperature at time [tex]\( t = 0 \)[/tex]:
[tex]\[ a = \text{Initial temperature} - \text{Outside temperature} = 100 - 0 = 100 \][/tex]
3. Use the temperature at [tex]\( t = 5 \)[/tex] minutes to find [tex]\( b \)[/tex]:
Use the given condition that the temperature after 5 minutes is 80 degrees Celsius to find [tex]\( b \)[/tex]:
[tex]\[ C(5) = 100 \cdot b^5 + 0 = 80 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 100 \cdot b^5 = 80 \][/tex]
[tex]\[ b^5 = \frac{80}{100} = 0.8 \][/tex]
[tex]\[ b = \left(0.8\right)^{\frac{1}{5}} \approx 0.95635249979 \][/tex]
### Summary of Constants
- [tex]\( a = 100 \)[/tex]
- [tex]\( b \approx 0.95635249979 \)[/tex]
- [tex]\( c = 0 \)[/tex]
### Formula for [tex]\(C(t)\)[/tex]
Putting these constants into the formula [tex]\( C(t) = a \cdot b^t + c \)[/tex], we get:
[tex]\[ C(t) = 100 \cdot (0.95635249979)^t + 0 \][/tex]
Thus, the function that models the hot chocolate’s temperature at time [tex]\( t \)[/tex] minutes is:
[tex]\[ C(t) = 100 \cdot (0.95635249979)^t \][/tex]
This function expresses the temperature [tex]\( C(t) \)[/tex] of the hot chocolate over time [tex]\( t \)[/tex] in minutes, cooling towards the ambient temperature of 0 degrees Celsius.
Given:
- Initial temperature of hot chocolate, [tex]\(C(0) = 100\)[/tex] degrees Celsius.
- Temperature after 5 minutes, [tex]\(C(5) = 80\)[/tex] degrees Celsius.
- Outside (ambient) temperature is 0 degrees Celsius.
### Step-by-Step Solution
1. Determine [tex]\(c\)[/tex]:
The ambient temperature, which the hot chocolate is cooling towards, is 0 degrees Celsius. Therefore, [tex]\( c = 0 \)[/tex].
2. Determine [tex]\(a\)[/tex]:
The constant [tex]\( a \)[/tex] represents the difference between the initial temperature of the hot chocolate and the outside temperature at time [tex]\( t = 0 \)[/tex]:
[tex]\[ a = \text{Initial temperature} - \text{Outside temperature} = 100 - 0 = 100 \][/tex]
3. Use the temperature at [tex]\( t = 5 \)[/tex] minutes to find [tex]\( b \)[/tex]:
Use the given condition that the temperature after 5 minutes is 80 degrees Celsius to find [tex]\( b \)[/tex]:
[tex]\[ C(5) = 100 \cdot b^5 + 0 = 80 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 100 \cdot b^5 = 80 \][/tex]
[tex]\[ b^5 = \frac{80}{100} = 0.8 \][/tex]
[tex]\[ b = \left(0.8\right)^{\frac{1}{5}} \approx 0.95635249979 \][/tex]
### Summary of Constants
- [tex]\( a = 100 \)[/tex]
- [tex]\( b \approx 0.95635249979 \)[/tex]
- [tex]\( c = 0 \)[/tex]
### Formula for [tex]\(C(t)\)[/tex]
Putting these constants into the formula [tex]\( C(t) = a \cdot b^t + c \)[/tex], we get:
[tex]\[ C(t) = 100 \cdot (0.95635249979)^t + 0 \][/tex]
Thus, the function that models the hot chocolate’s temperature at time [tex]\( t \)[/tex] minutes is:
[tex]\[ C(t) = 100 \cdot (0.95635249979)^t \][/tex]
This function expresses the temperature [tex]\( C(t) \)[/tex] of the hot chocolate over time [tex]\( t \)[/tex] in minutes, cooling towards the ambient temperature of 0 degrees Celsius.
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