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

AI Thread Summary
The discussion revolves around calculating the time it takes for a cup of coffee, initially at 100°, to cool to 70° in a room at 68°. The user presents their calculations using Newton's Law of Cooling, noting temperature drops to 95° after one minute and 92° after two minutes, leading to two different decay constants. A participant questions the accuracy of the initial conditions, suggesting the problem may be misleading due to the coffee being at boiling point, which involves latent heat. Ultimately, the suggestion is to disregard the first minute and base calculations solely on the second minute's data for a more accurate estimate.
baileyyc
Messages
4
Reaction score
0
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°?
 
Physics news on Phys.org
Welcome to PF!

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:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top