Question

A potato is put into an oven that has been heated to 350 degrees Fahrenheit. Its temperature as a function of time is give by T(t)=a(1-e^(-kt))+b. The potato was 50 degrees F when it was first put into the oven. If the potato is 60 degrees F after 2 minutes, what is the value of k? Explain

Answer 1

T(t)=a(1-e^(-kt))+b. The potato was 50 degrees F when it was first put into the oven. If the potato is 60 degrees F after 2 minutes, what is the value of k? Explain

T(0)=50ºF = a(1-e^(-k(0)))+b
T(2)=60ºF = a(1-e^(-k(2)))+b

\(\large \frac{50-b}{a}=1-e^{0}\)

\(\large \frac{60-b}{a}=1-e^{-2k}\)

Can you get it from here, @Tiffanyhavel ?

Answer 2

I don't understand what a and b would be.

Answer 3

You can use that first equation to find b.

Answer 4

Finding 'a' might be a little trickier; you have to know something about thermal equilibrium.

Answer 5

Let me know what you get for b, and I'll give you a hint for finding a.

Answer 6

I figured the problem out

Answer 7

Excellent! What did you do to solve for a? I'm sure there's a different method than what I did.

Answer 8

@cliffsedge I've tried to work out a and b and i'm trying to correct my answer, what would the final answer be?

Answer 9

My hint about the thermal equilibrium means that after enough time, the potato will be the same temperature as the oven and remain so from then on, so imagine setting t=∞ then e^(-kt) would be zero and the second equation would be
(60-b)/a = 1.
From the first equation, it is obvious that b=50, so now finding a is easy.

Answer 10

great!! thank you so much!

Answer 11

So the b=50 and a=10. Then you would plug them in to find k?

Answer 12

I'm SO confused....