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. $\mathrm{T}=\square+\square \mathrm{e}^{\square}$ (Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places 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 find a model for the temperature of the ginger ale over time.

STEP 2

1. Identify known values and set up the equation using Newton's Law of Cooling.
2. Solve for the constant $k$ using the given conditions.
3. Write the final model for the temperature $T$ as a function of time $t$ .

STEP 3

Identify the known values: - $C = 34^{\circ} \mathrm{F}$ (temperature of the refrigerator) - $T_0 = 72^{\circ} \mathrm{F}$ (initial temperature of the ginger ale) - At $t = 10$ minutes, $T = 62^{\circ} \mathrm{F}$ .
Set up the equation using Newton's Law of Cooling: $T = C + (T_0 - C) e^{kt}$ $T = 34 + (72 - 34) e^{kt}$ $T = 34 + 38 e^{kt}$

STEP 4

Use the condition at $t = 10$ minutes, $T = 62$ : $62 = 34 + 38 e^{10k}$
Solve for $e^{10k}$ : $62 - 34 = 38 e^{10k}$ $28 = 38 e^{10k}$ $e^{10k} = \frac{28}{38}$ $e^{10k} = \frac{14}{19}$
Take the natural logarithm of both sides to solve for $k$ : $10k = \ln\left(\frac{14}{19}\right)$ $k = \frac{1}{10} \ln\left(\frac{14}{19}\right)$
Calculate $k$ using a calculator: $k \approx \frac{1}{10} \times (-0.3365)$ $k \approx -0.03365$

STEP 5

Substitute $k$ back into the model: $T = 34 + 38 e^{-0.03365t}$
The model for the temperature of the ginger ale is: $T = 34 + 38 e^{-0.0337t}$
The model for the temperature of the ginger ale is:
$T = 34 + 38 e^{-0.0337t}$

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