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Coffee and milk (cooling/heating problem) 
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#1
Sep2204, 08:12 PM

P: 98

Ok, this is supposed to be easy, but i'm just not seeing it. I've done some heating/cooling problems w/no issue, so i'm not sure why I can't do this one. My teacher posted this quickly at the end of class, and neglected to go over it next class, so I have no one to help me now since no one in the class seems to have bothered with it.
Two cups of coffee (each 150g) at 90 degrees celsius. Cup A has 50g of 15 degree milk added and then sits for 5 minutes. Cup B sits for 5 minutes, then has 50g of 15 degree milk added. Which cup, A or B, is hotter after 5 minutes? Take coffee and milk to both have specific heat of 1. All I got is that cup A's starting temp T(0)a = [150(90) + 15(50) ]/ 200 = 71.25 degrees celsius. I'm not sure how to set up the equations for this, though. Do I have enough info, or is this guy mad? (j/k, I know it's me.) thx. :) Angela. 


#2
Sep2304, 06:52 AM

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If you've done some "heating/cooling" problems before then you know that the rate of cooling depends on the difference between the temperature of the coffee and the air. Apparently you are not given the temperature of the air. You might try leaving that as "T" and then see if the difference in temperatures is independent of T.



#3
Sep2304, 08:14 AM

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If you have not come across this before, you might want to Google "Newton's Law of Cooling".



#4
Sep2304, 12:24 PM

P: 98

Coffee and milk (cooling/heating problem)
So, you do, or do not know how to do this problem with the given info? I'll see if I can get an answer in the math lab if not, but there isn't anyone in there who has gone thru diff eq . Thanks. A 


#5
Sep2304, 02:24 PM

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If you "can get an answer in the math lab", then there is something seriously wrong with your math lab they shouldn't be in the business of giving out answers. Here's my suggestion go back to my first response and try doing what I suggested there. 


#6
Sep2304, 02:30 PM

P: 98

Wow. You know what, you can feel free to ignore any questions of mine in the future. I don't need the nasty attitude you are passing out. If I had understood your first reply, then I would not have posed the question differently asking again. I'm sorry you're so high and mighty that you can't bring yourself to the level of actually explaining something properly. No, I don't want the "answer". I know the answer, the teacher gave it. I want to know how to arrive at that answer. I thought that was what this forum was for, excuse me if i've bothered you.
Later. A 


#7
Sep2304, 04:02 PM

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Amb, I'm sure HallsofIvy was only trying to get you to think about the problem without actually giving you the answer. More often than not, students will learn better (and feel better) if they solve the problem themselves.
I'm sure you would make some progress if you really think about (or look up) what he suggested  and if there was something particular about his suggestion that you did not understand, surely he'd clarify if you ask him specifically. 


#8
Sep2304, 04:53 PM

P: 98

You ever just get stuck? That's where I am with this problem. I know it should be easy, but I don't see it. That's why I came on here asking about it. I understand that i'm missing the outside temperature. I also don't have other info like what the temp is at time = 2 or time = 3. I am not sure where to go here, and unfortunately, replacing "M" with "T" has not helped me immensely (or at all, actually). You guys see this, you understand it, I do not. But, I would like to, and that is why I asked for help. Maybe this isn't really the forum for people who are beginners. fwiw, the equation i'm using is that Temp(t) = Ce^kt + M where M is the outside temperature. In the problems in the book, we're giving a couple of values like "the liquid at t=2 is 80 degrees" or the like. I am just not sure where to go in this problem without the same type of information.
Thanks anyhow, though. I guess i'll just move on. It's not like it's an actual homework problem or test question,.. just one posed in class that was never properly answered. A 


#9
Sep2304, 06:08 PM

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The equation you've quoted is the right one (it would be instructive to understand how you get it). The constants C and M (as you had suspected) have special meaning.
C is the temperature difference Ti  Ts, where Ti is the initial temperature and Ts is the temperature of the surroundings. M is Ts. So, actually, T(t) = Ts + (Ti  Ts) e^Kt. Now we consider the 2 cases : (i) First case : Initial mixing : If you mix 50g of milk with 150g of coffee, what will the temperature of the, mixture be ? Use energy conservation (heat lost = heat gained) to find the mixed temperature. This is your initial temperature, Ti_mix. Using the cooling rate equation, find T_mix(t=5) in terms of Ts and K. (ii) Second case : Final mixing : Use the rate law separately for the coffee and the milk, to determine their temperatures after t=5 minutes, in terms of K, Ts. Now use energy conservation to determine the temperature after they are mixed. You now have 2 expressions for the final temperature, one for each case. Compare them to determine if one is always greater than the other, irrespective of the values of K and Ts. 


#10
Sep2304, 06:24 PM

P: 98

Very helpful, thanks! I actually only went through the derivation of that equation once, and then did all of the problems with the T(t) = M + Cekt. I actually had forgotten what C actually was. The problems in the book basically give you two values of temperature, so you can solve for c, then solve for k, then solve the equation for T(t). I'm going to do the derivation again and then go thru this without looking at your explanation after i'm done with the newtonian stuff i'm working on now. Thanks for your help! :D
Angela. 


#11
Sep2304, 07:16 PM

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I really wish I hadn't given away so much, and I'll feel better (and so will you) if you tried to solve it by yourself.



#12
Jun910, 07:38 AM

P: 1

Hallsofivy is a very kind person. Very nice and kind, and also clever. Extremely clever. Patient as well.



#13
Jun1010, 08:48 PM

P: 376

Then the total calories = 750 cal + 13500 cal And the mixed temperature would be: 200 g (T)= 14250 cal 


#14
Jun1210, 01:14 AM

P: 83

you have to look at the transient heat transfer (non steady state) of each cup. I believe the general equation is p*c*V*dT/dt where p=density, c=constant, V=volume, T=temperature, t= time
Also, the original question needs more info. What if air temperature is 90 degree C with no convection/radiation. Cup B temperature wont change until cooler milk is added 


#15
Jun1610, 12:03 PM

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P: 1,965

It is a purely qualitative question. You can appeal to Newton's 'law' but whoever's law "the rate of cooling is a monotonic function of the difference of temperatures between the coffee and its surroundings" gives the same qualitative conclusion and is sufficient.
Since the time the question was asked in 2004 the temperature difference in two cups of coffee will theoretically still exist but will have become very small, possibly negligible compared to other factors not stated in the problem. 


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