Question

A roast turkey is taken from an oven when its temperature has reached 185F and is placed on a table in a room where the temperature is 75F

a)
If the temperature of the turkey is 150F after half an hour what is the temperature after 45min
b) when will the turkey have cooled to 100F

So I got the following equation using Newtons law of cooling:


110e^((ln(75/100)/30)x) + 75 = T(t)

where T(t) = the temperature of the turkey at time T

For
a) I got 146 Degrees F
b) I got 154 minutes

The text book states that the answers are

a) 137
b) 116min

what am I doing wrong?

Answer 1

Just rough work I set

185 = T(0)
to solve for c
and
150 = T(30)
to solve for k

Answer 2

\(T=A+Be^{kt}\)
\(T=185, A =75, t=0\)
=> \(185=75+B^{0*k}\) =>\( B= 110 \)
Therefore, \(T=75+110e^{kt}\)
Then..
\(T=150, t=30\)
\(150=75+110e^{30k}\) solve for \(k\) => \(k= -0.01\)
Then the equation is \(T= 75+1110e^{-0.1k}\)

Answer 3

typo... \(T=75+110e^{-0.1k}\)

Answer 4

wait..another one lol \(T=75+110e^{-0.1t}\)

Answer 5

For the temperature after \(45min\) is when you sub \(45\) into the equation and you will get the answer just like your textbook!

Answer 6

lol, another typo my bad \(T=75+110e^{-0.01t}\)

Answer 7

Do you understand?

Answer 8

lol

Answer 9

Didnt i get the same answer though, ln(75/100)/30) = roughly -0.001

Answer 10

sorry typo -0.01

Answer 11

i don't know i don't undestand your solution..but you said that you didnt get teh same answer as your textbook

Answer 12

https://www.wolframalpha.com/input/?i=110e^%28%28ln%2875%2F100%29%2F30%29x%29+%2B+75

here it is in latex

Answer 13

\[\large 110e^{\frac{1}{30}\log\left(\frac{75}{100}\right)k} +75\]?

Answer 14

yes except that k should be a t :)

Answer 15

\[\large T= 110e^{\frac{1}{30}\log\left(\frac{75}{100}\right)t} +75\]

Answer 16

looks like the same as mine

Answer 17

yeah can you try plugging in the numbers to see if you get the same answer

Answer 18

I think the text book is wrong but I wanted to make sure because this happens a bunch now

Answer 19

another typo lol \[\large T= 110e^{\frac{1}{30}\log\left(\frac{75}{110}\right)t} +75\]

Answer 20

I got the same as your textbook I assumed they round it off I got \(136.9292118\) and when you round it off its \(137\)

Answer 21

ok weird because when I use wolfram

a)
https://www.wolframalpha.com/input/?i=110e^%28%28ln%2875%2F100%29%2F30%29x%29+%2B+75%2C+x+%3D+45

b)
https://www.wolframalpha.com/input/?i=110e^%28%28ln%2875%2F100%29%2F30%29x%29+%2B+75+%3D+100

Answer 22

you made a mistake its
\[\huge T= 110e^{\frac{1}{30}\log\left(\frac{75}{110}\right)t} +75\]

Answer 23

Its not \[\huge\color{ red} {T= 110e^{\frac{1}{30}\log\left(\frac{75}{100}\right)t} +75}\]

Answer 24

lol figures I made a typo my brain is garbage sometimes

Answer 25

thanks for the help mimi

Answer 26

np