Question

please help

Answer 1

i got a negative for ln.....

Answer 2

Is the soup being heated or being cooled?
Unfortunately we have no context.

Answer 3

According to your equation, the soup will get hotter!
(note: e^x is always positive for all real values of x)

Answer 4

@mathmate it says.. "A hot bowl of soup cools according to Newton’s law of cooling."

Answer 5

If we don't have the complete question (preferably an image of the question), we cannot check if your work is correct. There are many constants missing.
For now, all I can say is that the form should be something like
\(\frac{dT}{dt}=-k(T-T_0)\)
where T=temperature at time t, k is a cooling constant, T0 is the initial temperature (at t=0).
Integrate and you should get back something along the lines of your formula used.

Answer 6

alright. here is the problem

Answer 7

Correction: T0 is the ambient temperature, lower than the initial temperature.

Answer 8

Ok, so all the constants are given to you.
the full equation looks like this (if you solve the ODE)
\(T(t)=T0+Ce^{-kt}\)
where
T(t)=temperature at time t minutes
T0=temp. at t=0
C=initial difference in temperature
k=0.04 is a cooling constant (based on the surface area, conductivity of the cup, etc. (forced convection assumed).
So using the formula as is:
\(T(t)=68+144e^{-0.04t}\)
Everything on the right hand side is known (t=15, as per given information)
Take out the calculator and find T(15).

Answer 9

thank you! I'm guessing that you're not supposed to solve it this way? I just wanted to get rid of the e by multiplying ln on both sides but it didn't work

Answer 10

@jh99 not sure why you wrote 15 on the LHS.
The LHS if the temperature you are interested in finding. You only sub in t=15 for the RHS. Otherwise you are not solving for any variable (notice you equation you wrote has no variable, so you aren't solving for anything at all!)