Math  /  Calculus

QuestionUse Newton's Law of Cooling, T=C+(T0C)ekt\mathrm{T}=\mathrm{C}+\left(\mathrm{T}_{0}-\mathrm{C}\right) \mathrm{e}^{\mathrm{kt}}, to solve. A frozen steak initially has a temperature of 26F26^{\circ} \mathrm{F}. It is left to thaw in a room that has a temperature of 75F75^{\circ} \mathrm{F}. After 9 minutes, the temperature of the steak has risen to 36F36^{\circ} \mathrm{F}. After how many minutes will the temperature of the steak be 50F50^{\circ} \mathrm{F} ?
The temperature will be 50F50^{\circ} \mathrm{F} after \square minutes. (Round to nearest minute as needed.)

Studdy Solution
Use the calculated k k to find the time t t when the temperature is 50F 50^\circ \mathrm{F} :
Substitute T=50 T = 50 into the equation:
50=75+(2675)ekt 50 = 75 + (26 - 75) e^{kt}
Simplify:
50=7549ekt 50 = 75 - 49 e^{kt}
49ekt=7550 49 e^{kt} = 75 - 50
49ekt=25 49 e^{kt} = 25
ekt=2549 e^{kt} = \frac{25}{49}
Take the natural logarithm on both sides:
kt=ln(2549) kt = \ln\left(\frac{25}{49}\right)
Substitute the value of k k :
(19ln(3949))t=ln(2549) \left(\frac{1}{9} \ln\left(\frac{39}{49}\right)\right) t = \ln\left(\frac{25}{49}\right)
Solve for t t :
t=9ln(2549)ln(3949) t = 9 \frac{\ln\left(\frac{25}{49}\right)}{\ln\left(\frac{39}{49}\right)}
Calculate t t using a calculator:
t22.5 t \approx 22.5
Round to the nearest minute:
t23 t \approx 23
The temperature will be 50F 50^\circ \mathrm{F} after 23\boxed{23} minutes.

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