Question

A roast turkey is taken from an oven when its temperature has reached 185F and is placed om a table in a room where the temperature is 75F.....If the temperature of the turkey is 150F after half an hour, what is the temperature after 45 minutes?

This is a calculus problem btw...Please help!

Answer 1

The rate of change of the temperature is proportional to the temperature difference:
At first, there is a large temperature difference between the turkey and the room, therefore, it quickly cools off. After a while, the difference is much smaller, so the cooling off has slowed down.

If T(t) is the temperature of the turkey at time t, then this formula (a differential equation) will hold: \(\dfrac{dT}{dt}=-a(T(t)-75)\), or, in easier to read notation: \(T'=-a(T-75)\).

The solution of this differential equation is the formula you are looking for.

Are you into differential equations, or is all this a complete mystery to you?

Answer 2

Okay thanks! Yeah I am in 2nd term calculus and differential equations do make sense. :)

Answer 3

OK, you can do it by separation of variables...
Let me hear it if you run into troubles!

Answer 4

Okay...so I got like e^45k=75/115

Answer 5

I got that answer using the steps out of my calculus book.

Answer 6

I don't believe this answer is correct though...

Answer 7

Can you show me how to do the separation of variables for T'=-a(T-75)?

Answer 8

Here it is: \[\frac{ dT }{ dt }=-a(T-75) \Rightarrow \frac{dT}{T-75}=-a dt\]So:\[\int\limits \frac{dT}{T-75}=\int\limits -adt +C\]Can you find the integrals?

Answer 9

Ahhh okay...Yeah lol I was confused because I looked in my calculus book and they showed something completely different :P Thanks :) Yeah I think I can. :P Should I consider (a) as a variable?

Answer 10

No, a is a constant. The "-" could be omitted, but is usually put in, because this is about cooling down. If you leave out the "-", a will come out negative...

Answer 11

So its Ln(abs(t-75))=t+C

Answer 12

No...let's work with the "-a" to see what we will get:

\(ln|T-75|=-at + C\). Now T > 75, so we leave out the absolute value:

\(ln(T-75)=-at + C\).

Remember, we wnat to find T as a function of t, so we need to do some extra steps:

\(T-75=e^{-at+C}\), so \(T=75+e^{-at+C}\).

\(T=75+e^{-at}\cdot e^{C}=75+c e^{-at}\).

Because C is a real constant, \(c=e^{C}>0\).
We want to know a and c.
To find out, we look for extra data: at t=0 T=185.
Put these in:

T=185, t=0:
\(185=75+c \cdot e^{0}=75+c \cdot 1=75+c\), so \(c=185-75=110\).

We now have:

\(T(t)=75+110e^{-at}\)

Answer 13

We also know: after half an hour, T = 150.
Put that into the formula:

\(T(0.5)=75+110e^{-0.5a}=150 \Leftrightarrow 110e^{-0.5a}=150-75=75\),

So \(e^{-0.5a}=\frac{75}{110} \Leftrightarrow -0.5a=\ln(\frac{75}{110}) \Leftrightarrow a=-2\ln(\frac{75}{110})\).

Using a calculator, we find \(a \approx 0.766\).

Answer 14

Now we finally can answer the question: what is the temperature after 45 min (= ¾ hours):

\(T(t)=75+110e^{-0.766t},~t=0.75:~~T(0.75)=75+110e^{-0.766 \cdot 0.75} \approx 137⁰~F\)

Answer 15

Ohhh okay thanks! :) Okay so if I wanted to figure when the turkey would have been cooled at 100F would I just replace T=150 with T=100?

Answer 16

Yes, that is all there is to it!

Answer 17

Okay one questions though...if I wanted to find when the turkey would be at 100F how would I change my answer (a) into a form of time..? Or could I just leave it as like a=(stuff)?

Answer 18

You would have to solve this equation for t:

\(100=75+110e^{-0.766t}\)

Answer 19

I got t=1.93421 so should I just say like 1 hour and and 93 min?

Answer 20

No, that would be 1 hour and .93421*60 min, so 1 hour and 56 min.