I need help please!

Question

The temperature of a cup of coffee varies according to Newton's Law of Cooling: , where T is the water temperature, A is the room temperature, and k is a positive constant.

If the coffee cools from 180°F to 100°F in 10 minutes at a room temperature of 75°F, how long (to the nearest minute) will it take the water to cool to 80°F?

Options:

1)	13
2)	18
3)	21
4)	27

Answer 1

dT/dt = -k*(T-A)

we can rearrange into

1/(T-A) * dT = -k*dt

We can integrate both sides:

ln(|T-A|) = -kt + c , since integral of 1/U du is ln(|u|), here u = T-A and du = dT

raise e to both sides:

|T-A| = e^-kt+c

Since e^c can just be set as another constant, rewrite as

|T-A| = c*e^-kt

Since we know the A is 75 and all the other temperatures are above 75, we can drop the absolute value and plug in 75 for A:

T - 75 = c*e^-kt

Now we can use the boundary conditions given to find c and k:

condition 1: T=180, t=0

condition 2: T=100, t=10

Having a condition where t=0 will allow us to solve for c:

180 - 75 = c*e^-k*0

c = 105

Now we can use condition 2 to solve for k:

100 - 75 = 105*e^-k*10

k = 0.14351

Now we can solve for t to cool to 80 degrees.

80 - 75 = 105*e^-0.14351t

t = 21.2 minutes