How Long Until Coffee Cools to 70° Using Newton's Law?

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Homework Help Overview

The discussion revolves around a problem involving the cooling of a cup of coffee using Newton's Law of Cooling. The coffee starts at a temperature of 100° and is placed in a room at 68°, with temperature readings taken at one-minute intervals.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of Newton's Law of Cooling through differential equations and integration. There are attempts to derive the cooling constant from temperature data over time. Questions arise regarding the accuracy of the problem statement and the implications of latent heat on the cooling process.

Discussion Status

The discussion is ongoing, with participants providing different interpretations of the problem. Some guidance has been offered regarding the potential impact of latent heat and the validity of the initial temperature readings, but no consensus has been reached on the correct approach or answer.

Contextual Notes

There is a noted discrepancy in the thermal time constants derived from the temperature data, which raises questions about the consistency of the problem setup. Participants are also considering the effects of latent heat, which complicates the calculations.

baileyyc
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a cup of coffee is heated to 100° then left in a room with a constant temperature of 68° after a minute the temperature of the coffee has dropped to 95° after another minute to 92 degrees. how much time must pass before the temperature of the coffee has dropped to 70°?
 
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Hi baileyyc! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
this is how I tried to solve the problem:

y'=k(y-68)
dy/dt=k(y-68)
∫(1/y-68)dy=∫kdt
e^ln|y-68|=e^kt+c1
y-68=e^kt+c1
y=68+ce^kt
100=68+ce^k(0)
100=68+c
c=32

when t=1, y=95 when t=2, y=92
95=68+32e^k(1) 92=68+32e^k(2)
27=32e^1k 24=32e^2k
ln(27/32)=k ln(24/32)=2k
k ~ -.1699 k ~ -.1438

plugging k into the equation plugging k into the equation
y=68+32e^-.1699t y=68+32e^-.1438t

when y=70 when y=70
70=68+32e^-.1699t 70=68+32e^-.1438t
2=32e^-.1699t 2=32e^-.1438t
ln(2/32)=-.1699t ln(2/32)=-.1438t
t~16.32 minutes t~19.28 minutes

I don't know which one is the answer.. can you help me?
 
baileyyc,

Are you sure the problem is stated correctly ? The reason I ask is because with those times and temperatures the thermal time constants differ by 18% and that is why you are getting two different answers.

Thanks
Matt
 
yes I am typing it correctly
 
baileyyc said:
I don't know which one is the answer.. can you help me?

Hi baileyyc! :smile:

I think this must be a trick question :frown:

if the coffee is at 100º, then it must be boiling, so it'll have some latent heat, but you don't know how much.

So I'd ignore the first minute, and calculate on the basis of the second minute only. :wink:
 

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