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After sitting on a shelf for a while, a can of soda at a room temperature of [tex]73^{\circ} F[/tex] is placed inside a refrigerator and slowly cools. The temperature of the refrigerator is [tex]34^{\circ} F[/tex]. Newton's Law of Cooling explains that the temperature of the can of soda will decrease proportionally to the difference between the temperature of the can of soda and the temperature of the refrigerator, as given by the formula below:

[tex]\[ T = T_a + (T_0 - T_a) e^{-kt} \][/tex]

Where:
- [tex]T_a[/tex] is the temperature surrounding the object,
- [tex]T_0[/tex] is the initial temperature of the object,
- [tex]t[/tex] is the time in minutes,
- [tex]T[/tex] is the temperature of the object after [tex]t[/tex] minutes,
- [tex]k[/tex] is the decay constant.

The can of soda reaches the temperature of [tex]57^{\circ} F[/tex] after 40 minutes. Using this information, find the value of [tex]k[/tex], to the nearest thousandth. Use the resulting equation to determine the Fahrenheit temperature of the can of soda, to the nearest degree, after 60 minutes.



Answer :

GinnyAnswer
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.

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