Math  /  Calculus

QuestionA bottle of ginger ale initially has a temperature of 72F72^{\circ} \mathrm{F}. It is left to cool in a refrigerator that has a temperature of 34F34^{\circ} \mathrm{F}. After 10 minutes the temperature of the ginger ale is 62F62^{\circ} \mathrm{F}. Complete parts a through c. a. Use Newton's Law of Cooling, T=C+(T0C)ekt\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)e0.0305tT=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? 58F58^{\circ} \mathrm{F} (Round to nearest degree as needed.) c. When will the temperature of the ginger ale be 52F52^{\circ} \mathrm{F} ? \square minute(s) (Round to nearest minute as needed.)

Studdy Solution
To find when the temperature is 52F52^{\circ} \mathrm{F}, set T=52 T = 52 in the model:
52=34+38e0.0305t52 = 34 + 38 e^{-0.0305t}
18=38e0.0305t18 = 38 e^{-0.0305t}
1838=e0.0305t\frac{18}{38} = e^{-0.0305t}
Taking the natural logarithm of both sides:
ln(1838)=0.0305t\ln\left(\frac{18}{38}\right) = -0.0305t
Solving for t t :
t=ln(1838)0.0305t = \frac{\ln\left(\frac{18}{38}\right)}{-0.0305}
Calculating:
tln(0.4737)0.0305t \approx \frac{\ln(0.4737)}{-0.0305}
t0.74720.0305t \approx \frac{-0.7472}{-0.0305}
t24.5t \approx 24.5
Rounded to the nearest minute, t25 t \approx 25 minutes.
The temperature of the ginger ale will be 52F52^{\circ} \mathrm{F} after approximately 2525 minutes.

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