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

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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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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: