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...

Question

The temperature of a cup of coffee varies according to Newton's Law of Cooling: , 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 80°C in 1 minute at a room temperature of 22°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

A	20
B	24
C	46
D	54

Answer 1

You can avoid solving for k

100-20 = 80

22 + (100-22)[(80-22)/(100-22)]^4 = 22 + (80-22)^4/(100-22)^3 = 45.84669 or

45 50225/59319

Answer 2

The solution of the differential equation is T=e^-k(T-A)t + C, assuming that A is a constant.

You need to use the information given to evaluate the C and the k.

The C is determined by the initial condition, i.e. you know T(0) which gives you C.

Then the k will be determined by the initial rate of cooling: 20 degrees in 1 minute.

The use those values to determine T(4).