Question

The differential equation below models the temperature of a 93°C cup of coffee in a 19°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 70°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. dy dt = − 1 51 (y − 19).

Not 10%% sure about this one I got a general solutioin y = c e^(-1.51t) + 19 but I cant see how to find the constant c.

Answer 1

The differential is
\(\frac{dT}{dt} = \frac{-1}{51}(T-19) \) right?

Answer 2

Ill just check that ..

Answer 3

I took it to mean -1.51 but its -1/51.

Answer 4

Yeah so I got

\(y=Ce^{-\frac{1}{51}t} +19\)

Now to find C, you can plug in the initial temperature (93, 0), setting y=93 and t=0.

Answer 5

OK THanks.

Answer 6

np, lmk what you get for C. After that's done to find the temp at 70°, you'll just set y=70 and solve for t

Answer 7

Sorry I got called away.
I get C to be 74 tight?

Answer 8

Yes

Answer 9

19 secionds for time when its 70 degrees?

Answer 10

18.984≈19 seconds, you're correct

Answer 11

Thanks for your help!!

Answer 12

np