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Sagot :
Certainly! To solve this problem, we will use Newton's Law of Cooling:
[tex]\[ T = T_a + (T_0 - T_a) e^{-k t} \][/tex]
where:
- [tex]\( T \)[/tex] is the temperature of the object after time [tex]\( t \)[/tex]
- [tex]\( T_a \)[/tex] is the ambient temperature (temperature of the refrigerator)
- [tex]\( T_0 \)[/tex] is the initial temperature of the object (the can of soda)
- [tex]\( k \)[/tex] is the decay constant
- [tex]\( t \)[/tex] is the time in minutes
We are given the following values:
- [tex]\( T_a = 34 \)[/tex]°F (temperature of the refrigerator)
- [tex]\( T_0 = 73 \)[/tex]°F (initial temperature of the can of soda)
- [tex]\( T_{40} = 57 \)[/tex]°F (temperature of the can of soda after 40 minutes)
- [tex]\( t_{40} = 40 \)[/tex] minutes (time when the temperature was 57°F)
First, we need to determine the decay constant [tex]\( k \)[/tex]. To do this, we can substitute the given values into the formula:
[tex]\[ 57 = 34 + (73 - 34) e^{-k \cdot 40} \][/tex]
Simplify the equation step-by-step:
[tex]\[ 57 - 34 = 39 e^{-40k} \][/tex]
[tex]\[ 23 = 39 e^{-40k} \][/tex]
To isolate [tex]\( e^{-40k} \)[/tex], divide both sides by 39:
[tex]\[ e^{-40k} = \frac{23}{39} \][/tex]
Taking the natural logarithm (ln) of both sides:
[tex]\[ -40k = \ln\left(\frac{23}{39}\right) \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{\ln\left(\frac{23}{39}\right)}{40} \][/tex]
Calculating the value of [tex]\( k \)[/tex]:
[tex]\[ k \approx 0.013 \][/tex]
Now that we have [tex]\( k \)[/tex], we will use it to find the temperature of the can of soda after 60 minutes:
[tex]\[ T_{60} = 34 + (73 - 34) e^{-0.013 \cdot 60} \][/tex]
Simplify:
[tex]\[ T_{60} = 34 + 39 e^{-0.78} \][/tex]
Calculate the value inside the exponent:
[tex]\[ e^{-0.78} \approx 0.459 \][/tex]
Then:
[tex]\[ T_{60} = 34 + 39 \cdot 0.459 \][/tex]
[tex]\[ T_{60} = 34 + 17.901 \][/tex]
[tex]\[ T_{60} \approx 51.901 \][/tex]
Rounding to the nearest degree:
[tex]\[ T_{60} \approx 52 \][/tex]°F
So, the decay constant [tex]\( k \)[/tex] is approximately [tex]\( 0.013 \)[/tex], and the temperature of the can of soda after 60 minutes will be approximately [tex]\( 52 \)[/tex]°F.
[tex]\[ T = T_a + (T_0 - T_a) e^{-k t} \][/tex]
where:
- [tex]\( T \)[/tex] is the temperature of the object after time [tex]\( t \)[/tex]
- [tex]\( T_a \)[/tex] is the ambient temperature (temperature of the refrigerator)
- [tex]\( T_0 \)[/tex] is the initial temperature of the object (the can of soda)
- [tex]\( k \)[/tex] is the decay constant
- [tex]\( t \)[/tex] is the time in minutes
We are given the following values:
- [tex]\( T_a = 34 \)[/tex]°F (temperature of the refrigerator)
- [tex]\( T_0 = 73 \)[/tex]°F (initial temperature of the can of soda)
- [tex]\( T_{40} = 57 \)[/tex]°F (temperature of the can of soda after 40 minutes)
- [tex]\( t_{40} = 40 \)[/tex] minutes (time when the temperature was 57°F)
First, we need to determine the decay constant [tex]\( k \)[/tex]. To do this, we can substitute the given values into the formula:
[tex]\[ 57 = 34 + (73 - 34) e^{-k \cdot 40} \][/tex]
Simplify the equation step-by-step:
[tex]\[ 57 - 34 = 39 e^{-40k} \][/tex]
[tex]\[ 23 = 39 e^{-40k} \][/tex]
To isolate [tex]\( e^{-40k} \)[/tex], divide both sides by 39:
[tex]\[ e^{-40k} = \frac{23}{39} \][/tex]
Taking the natural logarithm (ln) of both sides:
[tex]\[ -40k = \ln\left(\frac{23}{39}\right) \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{\ln\left(\frac{23}{39}\right)}{40} \][/tex]
Calculating the value of [tex]\( k \)[/tex]:
[tex]\[ k \approx 0.013 \][/tex]
Now that we have [tex]\( k \)[/tex], we will use it to find the temperature of the can of soda after 60 minutes:
[tex]\[ T_{60} = 34 + (73 - 34) e^{-0.013 \cdot 60} \][/tex]
Simplify:
[tex]\[ T_{60} = 34 + 39 e^{-0.78} \][/tex]
Calculate the value inside the exponent:
[tex]\[ e^{-0.78} \approx 0.459 \][/tex]
Then:
[tex]\[ T_{60} = 34 + 39 \cdot 0.459 \][/tex]
[tex]\[ T_{60} = 34 + 17.901 \][/tex]
[tex]\[ T_{60} \approx 51.901 \][/tex]
Rounding to the nearest degree:
[tex]\[ T_{60} \approx 52 \][/tex]°F
So, the decay constant [tex]\( k \)[/tex] is approximately [tex]\( 0.013 \)[/tex], and the temperature of the can of soda after 60 minutes will be approximately [tex]\( 52 \)[/tex]°F.
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