Newton's Law of Cooling?

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Chels_92

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I think this is a Newton's Law of Cooling question but the formula I have for that isn't working...

A pizza, heated to a temperature of 350 degrees Fahrenheit, is taken out of an oven and placed in a 75 degrees Fahrenheit room at time t=0 minutes. The temperature of the pizza is changing at a rate of -110e^-0.4t degrees Fahrenheit per minute. To the nearest degree, what is the temperature of the pizza at
time t=5 minutes?

The formula I tried to use was T= (To - Ts)e^kt + Ts

The problem I'm having is with the "changing at a rate of..." I don't know what to do with that information. I'd really appreciate the step by step of how to do this particular question.

Thank you!
Chels
 
You are given the rate of change of the pizza temperature after removal from the oven.

A negative rate means the temperature is decreasing as expected.
Differentiating the temperature function gives the rate of change.

\(\displaystyle \frac{d}{dt}f(T)=-110e^{-0.4t}\)

\(\displaystyle Hence,\ f(T)=\frac{-110}{-0.4}e^{-0.4t}+c=275e^{-0.4t}+c\)

\(\displaystyle f(0)=350,\ therefore\ c=350-275=75.\)

\(\displaystyle f(T)=275e^{-0.4t}+75\ degrees\ fahrenheit.\)
 
\(\displaystyle Another \ way, \ to \ wit:\)

\(\displaystyle Newton's \ Law \ of \ Cooling \] states \ that \ the \ rate \ of \ change \ of \ the \ temperature \ of \ an \ object \ is \ proportional \ to \ the \ difference \ between \ the \ temperature \ of \ the \ object \ and \ the \ temperature \ of \ the \ surrounding \ medium.\)

\(\displaystyle Ergo, \ \frac{dT(t)}{dt} \ = \ k[T(t)-L]\)

\(\displaystyle Given, \ \frac{dT(t)}{dt} \ = \ -110e^{-.4t}, \ L \ = \ 75, \ T(0) \ =350, \ find \ T(5).\)

\(\displaystyle k[T(t)-75] \ = \ -110e^{-.4t}\)

\(\displaystyle [T(t)-75] \ = \ \frac{-110e^{-.4t}}{k}\)

\(\displaystyle T(t) \ = \ \frac{-110e^{-.4t}}{k}+75\)

\(\displaystyle T(0) \ = \ 350 \ = \ \frac{-110}{k}+75, \ k \ = \ \frac{-110}{275}\)

\(\displaystyle Hence, \ T(t) \ = \ \frac{-110e^{-.4t}}{\frac{-110}{275}} \ +75 \ = \ 275e^{-.4t}+75\)

\(\displaystyle Therefore, \ T(5) \ = \ about \ 112^{o} \ F.\)
 
\(\displaystyle chrisr's \ solution \ is \ better \ than \ mine. \ In \ chrisr's \ solution, \ given \ the \ rate \ of \ change, \ the\)

\(\displaystyle temperature \ of \ the \ room \ (75^{o} F) \ is \ a \ red \ herring, \ as \ it \ isn't \ necessary \ (the \ temperature \ of\)

\(\displaystyle \ the \ room) \ to \ find \ the \ temperature \ of \ the \ pizza \ 5 \ minutes \ after \ it \ was \ taken \ out \ of \ the \ oven.\)
 
Note: The above problem could have been written this way, to wit:

A pizza, heated to a temperature of 350 degrees Fahrenheit, is taken out of an oven and placed in a room at time t=0 minutes. The temperature of the pizza is changing at a rate of -110e^-0.4t degrees Fahrenheit per minute. To the nearest degree, (a)what is the temperature of the pizza at time t=5 minutes and (b)what is the temperature of the room (assume the room is at a constant temperature)?

\(\displaystyle \frac{dT(t)}{dt} \ = \ -110e^{-.4t} \ \implies \ T(t) \ = \ 275e^{-.4t}+C\)

\(\displaystyle T(0) \ = \ 350 \ = \ 275+C, \ C \ = \ 75\)

\(\displaystyle a) \ Hence, \ T(t) \ = \ 275e^{-.4t}+75, \ T(5) \ = \ 112^{o}(about).\)

\(\displaystyle b) \ \lim_{t\to\infty}[275e^{-.4t}+75] \ = \ 75, \ equals \ the \ temperature \ of \ the \ room, \ Why?\)
 
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