as usual, the average value over the 30-minute interval is
1/30 ∫[0,30] 20+75e^(−t/50) dt = 76,40
If a cup of coffee has temperature 95∘C in a room where the temperature is 20∘C, then, according to Newton's Law of Cooling, the temperature of the coffee after t minutes is
T(t)=20+75e^(−t/50).
What is the average temperature (in degrees Celsius) of the coffee during the first half hour?
Average temperature =
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