Math / Calculus

QuestionA bottle of ginger ale initially has a temperature of $72^{\circ} \mathrm{F}$ . It is left to cool in a refrigerator that has a temperature of $34^{\circ} \mathrm{F}$ . After 10 minutes the temperature of the ginger ale is $62^{\circ} \mathrm{F}$ . Complete parts a through c. a. Use Newton's Law of Cooling, $\mathrm{T}=\mathrm{C}+\left(\mathrm{T}_{0}-\mathrm{C}\right) e^{\mathrm{kt}}$ , to find a model for the temperature of the ginger ale, T , after t minutes. $T=34+(38) e^{-0.0305 t}$ (Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.) b. What is the temperature of the ginger ale after 15 minutes? $58^{\circ} \mathrm{F}$ (Round to nearest degree as needed.) c. When will the temperature of the ginger ale be $52^{\circ} \mathrm{F}$ ? $\square$ minute(s) (Round to nearest minute as needed.)

Studdy Solution

STEP 1

1. The initial temperature of the ginger ale is $72^{\circ} \mathrm{F}$ .
2. The refrigerator temperature is $34^{\circ} \mathrm{F}$ .
3. After 10 minutes, the temperature of the ginger ale is $62^{\circ} \mathrm{F}$ .
4. We will use Newton's Law of Cooling to model the temperature change.

STEP 2

1. Derive the model for the temperature using Newton's Law of Cooling.
2. Calculate the temperature after 15 minutes using the model.
3. Determine the time when the temperature reaches $52^{\circ} \mathrm{F}$ .

STEP 3

Newton's Law of Cooling is given by:
$T = C + (T_0 - C) e^{kt}$
where: - $T$ is the temperature at time $t$ , - $C$ is the ambient temperature ( $34^{\circ} \mathrm{F}$ ), - $T_0$ is the initial temperature ( $72^{\circ} \mathrm{F}$ ), - $k$ is the cooling constant, - $t$ is the time in minutes.

STEP 4

Substituting the known values into the equation:
$62 = 34 + (72 - 34) e^{10k}$
Simplifying:
$62 = 34 + 38 e^{10k}$
$28 = 38 e^{10k}$
$\frac{28}{38} = e^{10k}$
Taking the natural logarithm of both sides:
$\ln\left(\frac{28}{38}\right) = 10k$
Solving for $k$ :
$k = \frac{\ln\left(\frac{28}{38}\right)}{10} \approx -0.0305$

STEP 5

Using the model $T = 34 + 38 e^{-0.0305t}$ , calculate the temperature after 15 minutes:
$T = 34 + 38 e^{-0.0305 \times 15}$
Calculating:
$T \approx 34 + 38 \times e^{-0.4575}$
$T \approx 34 + 38 \times 0.6321$
$T \approx 34 + 24.0208$
$T \approx 58.0208$
Rounded to the nearest degree, the temperature is $58^{\circ} \mathrm{F}$ .

STEP 6

To find when the temperature is $52^{\circ} \mathrm{F}$ , set $T = 52$ in the model:
$52 = 34 + 38 e^{-0.0305t}$
$18 = 38 e^{-0.0305t}$
$\frac{18}{38} = e^{-0.0305t}$
Taking the natural logarithm of both sides:
$\ln\left(\frac{18}{38}\right) = -0.0305t$
Solving for $t$ :
$t = \frac{\ln\left(\frac{18}{38}\right)}{-0.0305}$
Calculating:
$t \approx \frac{\ln(0.4737)}{-0.0305}$
$t \approx \frac{-0.7472}{-0.0305}$
$t \approx 24.5$
Rounded to the nearest minute, $t \approx 25$ minutes.
The temperature of the ginger ale will be $52^{\circ} \mathrm{F}$ after approximately $25$ minutes.

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