Milk In The Coffee

  • #1

Main Question or Discussion Point

"This famous problem is always interesting. Suppose you want to make your morning coffee cool off within five minutes to make it a more suitable temperature. Do you pour the cold milk first and then wait five minutes before drinking, or d you wait five minutes before adding the milk?"

I *THINK* that you should wait five minutes then add the milk but I'm not sure...

Any ideas?
 

Answers and Replies

  • #2
SpectraCat
Science Advisor
1,395
1
"This famous problem is always interesting. Suppose you want to make your morning coffee cool off within five minutes to make it a more suitable temperature. Do you pour the cold milk first and then wait five minutes before drinking, or d you wait five minutes before adding the milk?"

I *THINK* that you should wait five minutes then add the milk but I'm not sure...

Any ideas?
I think this is an entropy issue ... the greater the temperature difference, the more irreversible the change upon adding the milk, and thus less energy will be wasted if you allow the coffee to cool first, and then add the milk. So I agree with you ...
 
Last edited:
  • #3
160
0
You don't need entropy to understand this. Just apply Newton's cooling law:

T(t) = Tr + (T0 - Tr)exp(-k*t)

And the well-known formula for the temperature of the mixture of two fluids:

Tmix = (T1*M1 + T2*M2)/(M1 + M2)

Where Tr is the temperature of the room and T0 is the initial temperature of the coffee, and k is a constant that depends upon your particular setup. T(t) is the temperature after t amount of time. In the second formula, T1,M1,T2,M2 are the temperature and mass of the first and second liquids, respectively.
 
  • #4
13
0
"This famous problem is always interesting. Suppose you want to make your morning coffee cool off within five minutes to make it a more suitable temperature. Do you pour the cold milk first and then wait five minutes before drinking, or d you wait five minutes before adding the milk?"

I *THINK* that you should wait five minutes then add the milk but I'm not sure...

Any ideas?
i think, when u add milk first then it follows zeroth law of thermodynamis. then after some time it follows the law.but after waiting of 5 minutes there will be no distribution of heat.
 
  • #5
SpectraCat
Science Advisor
1,395
1
You don't need entropy to understand this. Just apply Newton's cooling law:

T(t) = Tr + (T0 - Tr)exp(-k*t)

And the well-known formula for the temperature of the mixture of two fluids:

Tmix = (T1*M1 + T2*M2)/(M1 + M2)

Where Tr is the temperature of the room and T0 is the initial temperature of the coffee, and k is a constant that depends upon your particular setup. T(t) is the temperature after t amount of time. In the second formula, T1,M1,T2,M2 are the temperature and mass of the first and second liquids, respectively.
*facepalm* Of course, how could I have forgotten Newton's cooling law! That certainly accounts for most or all of the difference, and would dominate any effects from my "entropic cooling efficiency" hypothesis (if it is even correct).

However, it seems clear that the *reason* for Newton's cooling law is simply the 2nd law of thermodynamics, and entropy is the driving force. So, entropy *is* the essential consideration, just not in the way I hypothesized. *wink*
 
  • #6
11
0
this was fun.

let;
Tm = temperature of the milk,
Tr = temperature of the room,
T(t) = temperature of the coffee, w/o the milk
T'(t) = temperature of the coffee, w/ the milk
T0 = initial temp of the coffee (HOT!)
K: a constant between 0 and 1 depending on the ratio of mass of the milk and coffee. Something like Mc / (Mc + Mm)..

For a colder coffee, at the end of 5th minute;
if Tr < Tm < T0; put the milk as soon as you can.
if Tr = Tm < T0; it doesnt matter when you put the milk.
if Tm < Tr < T0; put the milk as late as you can.

I came to the conclusion graphically, using exponential cooling model and mixture of fluids. Its easier to visualise that way. But I'll try to pour it on maths:

Let's just consider the case Tm = Tr. We left the milk on the kitchen table last night, its at room temp in the morning.

normally T(t) = Tr + (T0 - Tr)*exp(-kt)
if you put the milk at t=0;
T'(t) = Tm + K*[T(t) - Tm],

so at 5th minute temperature of the coffee is;
T'(5) = Tm + K*[T(5) - Tm] *** (1)

and if you put the milk at t=5th minute;
T'(t) = T(t) - H(t-5)*(K-1)*[T(t) - Tm]
where H(t) is the Heaviside step function.

just after we pour the milk, at t=5+, the temperature is:
T'(5+) = T(5+) + H(0+)*(K-1)*[T(5+) - Tm]
T'(5+) = T(5) + (K-1)*[T(5) - Tm]
T'(5+) = T(5) + K*[T(5) - Tm] - [T(5) - Tm]
T'(5+) = Tm + K*[T(5) - Tm] *** (2)

aha! (1) = (2). it doesn't matter when you put the milk, if Tm=Tr

The other cases; Tr < Tm and Tm < Tr can be proved graphically much easier.
 
Last edited:
  • #7
160
0
if Tr < Tm < T0; put the milk as soon as you can.
if Tr = Tm < T0; it doesnt matter when you put the milk.
if Tm < Tr < T0; put the milk as late as you can.
Well BL4CKCR4Y0NS has cold milk in mind, so yes - he should pour the milk in as late as possible.

Voila - physics applied to make real life better.
 

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