Question

The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

74
67
60
42

Answer 1

(b) 67

Explanation:

A solution to the differential equation describing the temperature according to Newton's Law of Cooling could be written as ...

T = (final temp) + (initial difference)×(decay factor)^t

where the decay factor is the fraction of change during 1 unit of time period t.

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Here, the initial difference of 100-25 = 75 degrees decays to 90-25 = 65 degrees in 1 minute. So, the units of t are minutes, the decay factor is 65/75, the initial difference is 75 degrees, and the final temperature is 25 degrees. That lets us write the equation as ...

T = 25 +75(65/75)^t

Then for t=4, the temperature is ...

T = 25 +75(13/15)^4 ≈ 67.3 . . . . degrees

After 4 minutes the temperature of the coffee is about 67 degrees.