Question

You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling:

Answer 1

A. 1 hour

B. 35 minutes

C. 15 minutes

D. 45 minutes

Answer 2

work with the difference in the temperature

Answer 3

idk how to do that... can u pls show me??

Answer 4

the initial difference is \(210-68=142\) after 10 minutes the difference is \(200-68=132\) one way to model this is (probably not what your math teacher had in mind
\[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]

Answer 5

you want to know when it is 180 degrees, or when the temperature difference is \(180-68=112\) so set
\[112=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\] and solve for \(t\)

Answer 6

you can also use
\[T=142e^{kt}\] but that requires solving for \(k\)

Answer 7

kk but idk how to solve for t since its in exponent form :(

Answer 8

i get
\[\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}\]
\[\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}\]
\[t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}\]

Answer 9

requires change of base

Answer 10

oh i didn't open the link, you are supposed to do it the hard way, solve for \(k\) first

Answer 11

so i get 32.50 ?? so my answer is B. 35 minutes
???

Answer 12

since i round ?

Answer 13

\[132=142e^{10k}\]
\[\frac{132}{142}=e^{10k}\]
\[\frac{\ln(\frac{132}{142})}{10}=k\]

Answer 14

so k=-.0073025135?

Answer 15

i got 32.5 as well
remember to work with the temperature differences, not the actual temperature

Answer 16

ok so how do i know what my answer is based off of that??

Answer 17

T(t) = Ta + (To - Ta)e^(-kt)

T(t) = Ta + (210 - Ta)e^(-kt) ... Plug in To = 210 (this is the temp of the object)

T(t) = 68 + (210 - 68)e^(-kt) ... plug in Ta = 68

T(10) = 68 + (210 - 68)e^(-k*10) ... Plug in t = 10

200 = 68 + (210 - 68)e^(-k*10) ... Plug in T(10) = 200 (this is the temp of the object at 10 min)

Now solve for k

200 = 68 + (210 - 68)e^(-k*10)

200 = 68 + (142)e^(-10k)

200-68 = (142)e^(-10k)

132 = (142)e^(-10k)

132/142 = e^(-10k)

0.929577 = e^(-10k)

e^(-10k) = 0.929577

-10k = ln(0.929577)

-10k = -0.0730256

k = -0.0730256/(-10)

k = 0.00730256


So your function goes from

T(t) = 68 + (210 - 68)e^(-kt)

to

T(t) = 68 + (210 - 68)e^(-0.00730256t)


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Now plug in T(t) = 180 and solve for t


T(t) = 68 + (210 - 68)e^(-0.00730256t)


180 = 68 + (210 - 68)e^(-0.00730256t)


180 - 68 = (210 - 68)e^(-0.00730256t)


112 = (210 - 68)e^(-0.00730256t)


112 = (142)e^(-0.00730256t)


112/142 = e^(-0.00730256t)


0.78873239 = e^(-0.00730256t)


ln(0.78873239) = -0.00730256t


-0.23732819 = -0.00730256t


-0.23732819/(-0.00730256) = t


32.4993139 = t


t = 32.4993139


So it will take about 32.4993139 minutes for the cup to cool to 180 degrees F

Answer 18

oh wow thanks!! :) but my answer choices are these:

A. 1 hour

B. 35 minutes

C. 15 minutes

D. 45 minutes

so do i round up and get B. 35 min as my answer??

Answer 19

@jim_thompson5910 @satellite73 is B my answer??

Answer 20

so is 35 minutes my answer?? :O

Answer 21

maybe they're thinking in 5 min intervals

Answer 22

because the answer is 32.4993139 minutes

Answer 23

idk... so what is the answer then??

Answer 24

so probs B. 35 minutes then??

Answer 25

yeah probably

Answer 26

kk thanks!

Answer 27

kk thanks!