Newton's Law of Cooling?

[Chels_92]

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

 

[chrisr]

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.

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

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

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

$\displaystyle f(T)=275e^{-0.4t}+75\ degrees\ fahrenheit.$

 

[BigGlenntheHeavy]

$\displaystyle Another \ way, \ to \ wit:$

\(\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.\)

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

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

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

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

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

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

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

$\displaystyle Therefore, \ T(5) \ = \ about \ 112^{o} \ F.$

 

[BigGlenntheHeavy]

$\displaystyle chrisr's \ solution \ is \ better \ than \ mine. \ In \ chrisr's \ solution, \ given \ the \ rate \ of \ change, \ the$

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

$\displaystyle \ the \ room) \ to \ find \ the \ temperature \ of \ the \ pizza \ 5 \ minutes \ after \ it \ was \ taken \ out \ of \ the \ oven.$

 

[BigGlenntheHeavy]

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)?

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

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

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

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