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Sagot :
To model the cooling of a cup of hot chocolate using the formula [tex]\( C(t) = a \cdot b^t + c \)[/tex] and the given conditions, we will derive the constants [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] step-by-step.
### Given Information:
1. Initial Temperature of Hot Chocolate (T_initial): 100 degrees Celsius
2. Temperature after 5 minutes (T_after_5_min): 80 degrees Celsius
3. Outside Temperature (T_outside): 0 degrees Celsius
4. Time Interval (t): 5 minutes
### Step 1: Identify the parameters of Newton’s Law of Cooling
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature. The general form of the temperature model is:
[tex]\[ T(t) = T_{\text{outside}} + (T_{\text{initial}} - T_{\text{outside}}) \cdot \exp(-kt) \][/tex]
where [tex]\( k \)[/tex] is the cooling constant.
Given the temperatures at specific times, the equation can be set up as:
[tex]\[ 80 = 0 + (100 - 0) \cdot \exp(-5k) \][/tex]
### Step 2: Solve for the cooling constant [tex]\( k \)[/tex]
From the equation, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 80 = 100 \cdot \exp(-5k) \][/tex]
[tex]\[ 0.8 = \exp(-5k) \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln(0.8) = -5k \][/tex]
[tex]\[ k = -\frac{\ln(0.8)}{5} \][/tex]
From the calculation:
[tex]\[ k \approx 0.044628710262841945 \][/tex]
### Step 3: Convert the model to the desired form [tex]\( C(t) = a \cdot b^t + c \)[/tex]
We have:
[tex]\[ T(t) = T_{\text{outside}} + (T_{\text{initial}} - T_{\text{outside}}) \cdot \exp(-kt) \][/tex]
### Constants identification:
- [tex]\( T_{\text{outside}} = c = 0 \)[/tex] (since outside temperature is 0)
- [tex]\( T_{\text{initial}} = 100 \)[/tex] (Starting temperature)
Thus,
[tex]\[ T(t) = 0 + 100 \cdot \exp(-kt) \][/tex]
### Reformatting to match [tex]\( C(t) = a \cdot b^t + c \)[/tex]
Here,
- [tex]\( a = 100 \)[/tex]
- [tex]\( b = \exp(-k) = \exp(-0.044628710262841945) \approx 0.956352499790037 \)[/tex]
- [tex]\( c = 0 \)[/tex]
Therefore, the formula for the temperature at time [tex]\( t \)[/tex] is:
[tex]\[ C(t) = 100 \cdot (0.956352499790037)^t + 0 \][/tex]
### Summary:
The final function modeling the cooling of the hot chocolate is:
[tex]\[ C(t) = 100 \cdot 0.956352499790037^t \][/tex]
This function gives the temperature of the hot chocolate at any time [tex]\( t \)[/tex] in minutes.
### Given Information:
1. Initial Temperature of Hot Chocolate (T_initial): 100 degrees Celsius
2. Temperature after 5 minutes (T_after_5_min): 80 degrees Celsius
3. Outside Temperature (T_outside): 0 degrees Celsius
4. Time Interval (t): 5 minutes
### Step 1: Identify the parameters of Newton’s Law of Cooling
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature. The general form of the temperature model is:
[tex]\[ T(t) = T_{\text{outside}} + (T_{\text{initial}} - T_{\text{outside}}) \cdot \exp(-kt) \][/tex]
where [tex]\( k \)[/tex] is the cooling constant.
Given the temperatures at specific times, the equation can be set up as:
[tex]\[ 80 = 0 + (100 - 0) \cdot \exp(-5k) \][/tex]
### Step 2: Solve for the cooling constant [tex]\( k \)[/tex]
From the equation, we can solve for [tex]\( k \)[/tex]:
[tex]\[ 80 = 100 \cdot \exp(-5k) \][/tex]
[tex]\[ 0.8 = \exp(-5k) \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln(0.8) = -5k \][/tex]
[tex]\[ k = -\frac{\ln(0.8)}{5} \][/tex]
From the calculation:
[tex]\[ k \approx 0.044628710262841945 \][/tex]
### Step 3: Convert the model to the desired form [tex]\( C(t) = a \cdot b^t + c \)[/tex]
We have:
[tex]\[ T(t) = T_{\text{outside}} + (T_{\text{initial}} - T_{\text{outside}}) \cdot \exp(-kt) \][/tex]
### Constants identification:
- [tex]\( T_{\text{outside}} = c = 0 \)[/tex] (since outside temperature is 0)
- [tex]\( T_{\text{initial}} = 100 \)[/tex] (Starting temperature)
Thus,
[tex]\[ T(t) = 0 + 100 \cdot \exp(-kt) \][/tex]
### Reformatting to match [tex]\( C(t) = a \cdot b^t + c \)[/tex]
Here,
- [tex]\( a = 100 \)[/tex]
- [tex]\( b = \exp(-k) = \exp(-0.044628710262841945) \approx 0.956352499790037 \)[/tex]
- [tex]\( c = 0 \)[/tex]
Therefore, the formula for the temperature at time [tex]\( t \)[/tex] is:
[tex]\[ C(t) = 100 \cdot (0.956352499790037)^t + 0 \][/tex]
### Summary:
The final function modeling the cooling of the hot chocolate is:
[tex]\[ C(t) = 100 \cdot 0.956352499790037^t \][/tex]
This function gives the temperature of the hot chocolate at any time [tex]\( t \)[/tex] in minutes.
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