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A bottle of white wine at room temperature (68°F68\degree F) is placed in a refrigerator at 44 p.m. Its temperature after tt hr is changing at the rate of −18−e−0.6t°F/hour-18-e^{-0.6t}\degree F/hour. By how many degrees will the temperature of the wine have dropped by 77 p.m.? What will the temperature of the wine be at 77 p.m.?

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Q. A bottle of white wine at room temperature (68°F68\degree F) is placed in a refrigerator at 44 p.m. Its temperature after tt hr is changing at the rate of −18−e−0.6t°F/hour-18-e^{-0.6t}\degree F/hour. By how many degrees will the temperature of the wine have dropped by 77 p.m.? What will the temperature of the wine be at 77 p.m.?
  1. Calculate Integral: Calculate the integral of the rate of change from t=0t=0 to t=3t=3. ∫03(−18−e−0.6t) dt=[−18t−(e−0.6t)/(−0.6)]\int_{0}^{3} (-18-e^{-0.6t}) \, dt = [-18t - (e^{-0.6t})/(-0.6)] from 00 to 33.
  2. Evaluate Limits: Evaluate the integral at the upper and lower limits and subtract to find the total temperature drop.\newlineTemperature drop = [−18(3)−(e(−0.6⋅3)/(−0.6))]−[−18(0)−(e(−0.6⋅0)/(−0.6))][-18(3) - (e^{(-0.6\cdot 3)}/(-0.6))] - [-18(0) - (e^{(-0.6\cdot 0)}/(-0.6))].
  3. Simplify Expression: Simplify the expression to find the temperature drop.\newlineTemperature drop = [−54−(e−1.8)/(−0.6)]−[0−(1)/(−0.6)][-54 - (e^{-1.8})/(-0.6)] - [0 - (1)/(-0.6)].
  4. Calculate Value: Calculate the numerical value of the temperature drop. Temperature drop = [−54−(e−1.8)/(−0.6)]−[0+1/0.6][-54 - (e^{-1.8})/(-0.6)] - [0 + 1/0.6].
  5. Correct Mistake: There's a mistake in the previous step, we need to correct the calculation.\newlineTemperature drop = [−54−(e−1.8)/(−0.6)]−[0−(1)/(−0.6)][-54 - (e^{-1.8})/(-0.6)] - [0 - (1)/(-0.6)].

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