Question

PLEASE HELP
A pizza pan is removed at 9:00pm from an oven whose temperature is fixed at 450 F into a room that is constant 72 F. After 5 mins the pizza pan is 300 F
(a) determine the time is the temperature of the pan 125 F?
(b) Determine the time that needs to elapse before the pan is 230 F
(c) what do you notice about the temp. as time passes?

Answer 1

Don't know for sure if the temperature decrease would be linear or exponential decay.
If it's linear then you can calculate the temperature drop per minute.
t0=450F
t5=300F
So 150F in 5 minutes
150/5 is 30F drop per minute
Formula = 450 - 30t
t is the minutes
So to know t for temp = 125F:
450-30t = 125
450 - 125 = 30t --> 325 = 30t
t= 325/30= 10.8 minutes

Same thing for B

C: the temperature goes down

BUT: i don't think that the temperature would go down in a linear way. I think it will go fast in the beginning and slower and slower after that. So exponential decay but im way to lazy to calculate that shyt its 3am >.<

Answer 2

it is not linear

Answer 3

first off work with the temperature differences, that is \(450-72=378\) and \(300-72=228\) so in 5 minutes the temperature difference goes from 378 to 228

Answer 4

your job is to model exponential decay from \((0,378)\) to \((5,228)\)
the ratio is \(\frac{228}{378}=\frac{38}{63}\)
the simple way to do this is to write the equation for the temperature differences is
\[378\times\left( \frac{38}{63}\right)^{\frac{t}{5}}\]
although you can also model it as
\[378\times e^{kt}\] but in that case you have to find \(k\)

Answer 5

if you want \(k\) you can put
\[228=378e^{5k}\] making
\[e^{5k}=\frac{38}{63}\] and so
\[k=\frac{\ln(\frac{38}{63})}{5}\]

Answer 6

so how would it look all together put? would it look like 378 * e^(ln(38/63))/5)

Answer 7

yeah but you sort of want a number in the exponent
that is, compute
\[k=-.10111\] and your equation for the differences is
\[378e^{-.10111t}\]

Answer 8

then it will be 125 degrees when the temperature difference is
\[125-72=53\] solve
\[378e^{-.10111t}=53\] for \(t\)

Answer 9

how will you get get t then? do you have to do the natural log?

Answer 10

yes, divide by \(378\) then take the log, then divide by \(-.10111\)

Answer 11

\[378e^{-.10111t}-53\]
\[e^{-.10111t}=\frac{53}{378}\]
\[-.10111t=\ln(\frac{53}{378})\]
\[t=\ln(\frac{53}{378})\div-.10111\]

Answer 12

I got 19.43

Answer 13

how will this get the time? for part a?

Answer 14

that is part A

Answer 15

yeah I just realized it!

Answer 16

lol

Answer 17

so for part b the starting equation will be 230=72+450e^-0.10111 t and so I will have to find t

Answer 18

no
\[230=72+378\times e^{-0.10111 t}\]

Answer 19

I got 8.627

Answer 20

would the temp of the pan be approaching the room tem. 72 F