This is copied exactly from the teacher. I know it's probably not realistic, it's just a problem. I need this answer ASAP please

Question

A soup with an internal temperature of 350F was taken off the stove to cool in a 71F room. After 15 minutes, the internal temperature of the soup was 175F.

a) Use Newton's Law of cooling to write a formula that models this situation.

b) How many minutes will it take the soup to cool at 85F?

Answer 1

A) We need to first notice that the difference of temperature between the soup and the room is 350-71=279 degrees. This will be total possible change in temperature of the soup. Then we know this is going to be an exponential equation of the form: f(t) = 279e^-kt + T_Room. To find the k value, we simply solve using the initial condition of 15 minutes.

175 = 279e^-15k + 71

e^-15k = 104/279

-15k = ln(104/279)

k = -ln(104/279) / 15 ≈ 0.065788 **We could rewrite e^-k as approximately 0.93633**

So the general function is f(t) = 279e^-0.0658t + 71 or f(t) = 279(0.93633)^t + 71.

B) Now we can use the function to solve for the time when the temperature is 85F.

85 = 279e^-0.0658t + 71

e^-0.0658t = 14/279

-0.0658t = ln(14/279)

t = - ln(14/279) / 0.0658 ≈ 45.4818

So the temperature will cool to 85F after about 45.5 minutes.

Answer 2

Newton's Law of Cooling is at time t, temperature T(t) = E + (T₀ – E)exp(–rt), where E is the environment temperature. In this case, T₀ = 350ºF and E = 71ºF. To solve the previous equation for the rate r, let τ = 15 minutes and use 175ºF = T(τ) so that r = –ln((T(τ) – E)/(T₀ – E))/τ.