# Homework Help: Using newton's law of cooling

1. Mar 11, 2009

### baileyyc

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°?

2. Mar 11, 2009

### tiny-tim

Welcome to PF!

Hi baileyyc! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

3. Mar 12, 2009

### baileyyc

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?

4. Mar 12, 2009

### CFDFEAGURU

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

5. Mar 12, 2009

### baileyyc

yes I am typing it correctly

6. Mar 12, 2009

### tiny-tim

Hi baileyyc!

I think this must be a trick question

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.

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