Respuesta :
The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C
From Newton's law of cooling, we have that
[tex]T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}[/tex]
Where
[tex](t) = \ time[/tex]
[tex]T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)[/tex]
[tex]T_{s} = Surrounding \ temperature[/tex]
[tex]T_{0} = Initial \ temperature \ of \ the \ body[/tex]
[tex]k = constant[/tex]
From the question,
[tex]T_{0} = 86 ^{o}C[/tex]
[tex]T_{s} = -20 ^{o}C[/tex]
∴ [tex]T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C[/tex]
[tex]T_{0} - T_{s} = 106^{o} C[/tex]
Therefore, the equation [tex]T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}[/tex] becomes
[tex]T_{(t)}=-20+106 e^{kt}[/tex]
Also, from the question
After 1 hour, the temperature of the ice-cream base has decreased to 58°C.
That is,
At time [tex]t = 1 \ hour[/tex], [tex]T_{(t)} = 58^{o}C[/tex]
Then, we can write that
[tex]T_{(1)}=58 = -20+106 e^{k(1)}[/tex]
Then, we get
[tex]58 = -20+106 e^{k(1)}[/tex]
Now, solve for [tex]k[/tex]
First collect like terms
[tex]58 +20 = 106 e^{k}[/tex]
[tex]78 =106 e^{k}[/tex]
Then,
[tex]e^{k} = \frac{78}{106}[/tex]
[tex]e^{k} = 0.735849[/tex]
Now, take the natural log of both sides
[tex]ln(e^{k}) =ln( 0.735849)[/tex]
[tex]k = -0.30673[/tex]
This is the value of the constant [tex]k[/tex]
Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at [tex]t = 2 \ hours[/tex]
From
[tex]T_{(t)}=-20+106 e^{kt}[/tex]
Then
[tex]T_{(2)}=-20+106 e^{(-0.30673 \times 2)}[/tex]
[tex]T_{(2)}=-20+106 e^{-0.61346}[/tex]
[tex]T_{(2)}=-20+106\times 0.5414741237[/tex]
[tex]T_{(2)}=-20+57.396257[/tex]
[tex]T_{(2)}=37.396257 \ ^{o}C[/tex]
[tex]T_{(2)} \approxeq 37.40 \ ^{o}C[/tex]
Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C
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