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Math  /  Algebra

QuestionYou are taking a road trip in a car without A/C. The temperture in the car is 105 degrees F. You buy a cold pop at a gas station. Its initial temperature is 45 degrees FF. The pop's temperature reaches 60 degrees FF after 25 minutes. Given that T−AT0−A=e−kt\frac{T-A}{T_{0}-A}=e^{-k t} where T=T= the temperature of the pop at time tt. T0=T_{0}= the initial temperature of the pop. A=A= the temperature in the car. k=k= a constant that corresponds to the warming rate. and t=t= the length of time that the pop has been warming up.
How long will it take the pop to reach a temperature of 75 degrees FF ? It will take □\square minutes.

Studdy Solution

STEP 1

1. The temperature in the car is 105 degrees F.
2. The initial temperature of the pop is 45 degrees F.
3. The pop's temperature reaches 60 degrees F after 25 minutes.
4. We are using the formula T−AT0−A=e−kt\frac{T-A}{T_{0}-A}=e^{-k t} to model the warming process.
5. We need to find the time it takes for the pop to reach 75 degrees F.

STEP 2

1. Identify known values and substitute them into the formula.
2. Solve for the constant k k .
3. Use the value of k k to find the time t t when the pop reaches 75 degrees F.

STEP 3

Identify known values and substitute them into the formula.
- A=105 A = 105 (temperature in the car) - T0=45 T_0 = 45 (initial temperature of the pop) - T=60 T = 60 (temperature of the pop after 25 minutes) - t=25 t = 25 (time in minutes)
Substitute these values into the formula:
60−10545−105=e−k⋅25\frac{60 - 105}{45 - 105} = e^{-k \cdot 25}

STEP 4

Solve for the constant k k .
Calculate the left side of the equation:
−45−60=34\frac{-45}{-60} = \frac{3}{4}
Thus, the equation becomes:
34=e−25k\frac{3}{4} = e^{-25k}
Take the natural logarithm of both sides to solve for k k :
ln⁡(34)=−25k\ln\left(\frac{3}{4}\right) = -25k
Solve for k k :
k=−ln⁡(34)25k = -\frac{\ln\left(\frac{3}{4}\right)}{25}

STEP 5

Use the value of k k to find the time t t when the pop reaches 75 degrees F.
Substitute T=75 T = 75 into the original formula:
75−10545−105=e−kt\frac{75 - 105}{45 - 105} = e^{-kt}
Calculate the left side of the equation:
−30−60=12\frac{-30}{-60} = \frac{1}{2}
Thus, the equation becomes:
12=e−kt\frac{1}{2} = e^{-kt}
Substitute the expression for k k from STEP_2:
12=e(ln⁡(34)25)t\frac{1}{2} = e^{\left(\frac{\ln\left(\frac{3}{4}\right)}{25}\right)t}
Take the natural logarithm of both sides:
ln⁡(12)=(ln⁡(34)25)t\ln\left(\frac{1}{2}\right) = \left(\frac{\ln\left(\frac{3}{4}\right)}{25}\right)t
Solve for t t :
t=ln⁡(12)ln⁡(34)25t = \frac{\ln\left(\frac{1}{2}\right)}{\frac{\ln\left(\frac{3}{4}\right)}{25}}
Calculate t t :
t≈57.78t \approx 57.78
Therefore, it will take approximately 58\boxed{58} minutes for the pop to reach 75 degrees F.

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