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The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where is the temperature of the coffee, A is
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The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where 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
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Mathematics
3 years
2021-08-01T21:20:56+00:00
2021-08-01T21:20:56+00:00 1 Answers
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Answers ( )
Answer:
B) 67°C.
Step-by-step explanation:
Newton’s Law of Cooling is given by:
Where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.
We are given that the coffee cools from 100°C to 90°C in one minute at a room temperature A of 25°C.
And we want to find the temperature of the coffee after four minutes.
First, solve the differential equation. Multiply both sides by dt and divide both sides by (T – A). Hence:
Take the integral of both sides:
Integrate:
Raise both sides to e:
The temperature of the coffee T will always be greater than or equal to the room temperature A. Thus, we can remove the absolute value:
We are given that A = 25. Hence:
Since the coffee cools from 100°C to 90°C, the initial temperature of the coffee was 100°C. Thus, when t = 0,T = 100:
Hence:
We are given that the coffee cools from 100°C to 90°C after one minute at a room temperature of 25°C.
So, T = 90 given that t = 1. Substitute:
Solve for k:
Therefore:
Then after four minutes, the temperature of the coffee will be:
Hence, our answer is B.