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A roast at 160 degree F is put into a refrigerator whose temperature is 45 degree F. The temperature of the roast is cooling at time t at the rate of (-9e⁻⁰.⁰⁸ᵗ) degree F per minute. The temperature, to the nearest degree F, of the roast 20 minutes after it is put in the refrigerator is___

A. 45°F
B. 70°F
C. 81°F
D. 90°F
E. 115°F

1 Answer

3 votes

Final answer:

The student's question involved using the exponential decay function to determine the temperature of a roast after 20 minutes in a refrigerator. By integrating the cooling rate and subtracting from the initial temperature, the closest estimated temperature is found to be 70° F, which is option B.

Step-by-step explanation:

The problem presented involves the cooling of a roast in a refrigerator, which is a practical application of exponential decay in the context of temperature change. The rate of cooling is given by the function -9e-0.08t degrees F per minute.

To find the temperature of the roast after 20 minutes, we calculate the integral of the rate of temperature change from 0 to 20 minutes and subtract this value from the initial temperature of the roast:

Initial temperature + ∫ (Cooling rate) ⋅ dt

160° F + ∫020 (-9e-0.08t) ⋅ dt = 160° F - 9 ⋅ (‑12.5e-0.08t200)

After calculating, the integral part evaluates to approximately -9 ⋅ (-12.5(0.98)) ≈ -9 ⋅ (-12.25) = 110.25° F. Therefore, the roast's temperature after 20 minutes will be 160° F - 110.25° F ≈ 50° F.

However, the answer choices provided do not include 50° F. Thus, to answer the question, we choose the temperature that is closest to the calculated value and greater than the refrigerator's temperature. This would be 70° F, option B.

User Wolfram Rittmeyer
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