# Physics problem #3



## brunson (Jul 17, 2009)

An exercise in thermodynamics.

A student and his physics professor are discussing his grades at the campus coffee shop. They both order coffee, which the waitress pours and delivers, piping hot, to the table. The student immediately picks up the small pitcher of room temperature cream that is on the table and puts a fair amount into his coffee, just enough to bring it to the rim of the cup. The two continue their talk, neither touching their cups for 20 minutes. 

Their conversation wraps up and the professor puts the same amount of cream into his coffee as the student. They both take a sip from their cups at the same time. Which person's coffee is hotter or are they the same temperature?

You can assume that the amount of coffee and cream in each cup is identical and that both cups were served at the same temperature.


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## Bryan (Jul 17, 2009)

Can you describe the coffee cup in more detail? Since they're physicists, is this a theoretical perfectly insulated cup? Also, what is the shape of the cup?


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## anythingtwisty (Jul 17, 2009)

I have no idea, but I'll put my educated guess below.
The professor's.The cream from the student was put in earlier, so the heat spread to it and lessened. I'm probably completely wrong.


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## badmephisto (Jul 17, 2009)

I assume the cups are not insulated to the surroundings because that is the key to answering the question  all that matters is that they are the same type of cup.

its more of a math question really... differential equations  concealed in a physical setting. Curious to see what people say


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## brunson (Jul 17, 2009)

Just as badmephisto says, the cups are normal cups that will conduct heat from the coffee to the surrounding environment. And, yes, the problem originated from my PDE 2 class.


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## ManasijV (Jul 22, 2009)

I'm unable to understand the complexity of the problem. Assuming the cream has a specific heat higher than air(does it?), it absorbs more heat from the student's coffee as heat absorbed is directly proportional to time and the specific heat. Since the question is which coffee is hotter i think it has to be the professor's.


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## Stefan (Jul 22, 2009)

As the professor's was hotter during those 20 minutes, it wasted more heat than the student's. So I say the student's is hotter.


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## nitrocan (Jul 22, 2009)

Student's. (It would take me some time to explain.)


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## Bryan (Jul 22, 2009)

StefanPochmann said:


> As the professor's was hotter during those 20 minutes, it wasted more heat than the student's. So I say the student's is hotter.



Huh? Sure, the rate of heat loss is greater when the temperature difference is greater, but the the total amount of heat would always be greater because as it cools, it's rate would cool to be at most equal to the cooler coffee.

The reason I ask about the shape of the cupe is that if you had a beaker shaped cupe, then adding cream could decrease the amount of surface area that the coffeee was exposed to air, thereby decreasing the rate.


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## Stefan (Jul 22, 2009)

Bryan said:


> but the the total amount of heat would always be greater because as it cools, it's rate would cool to be at most equal to the cooler coffee.


I don't understand.



Bryan said:


> The reason I ask about the shape of the cupe is that if you had a beaker shaped cupe, then adding cream could decrease the amount of surface area that the coffeee was exposed to air, thereby decreasing the rate.


If the shape of the cup played (or was supposed to play) a role, it surely would've been mentioned.


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## Bryan (Jul 22, 2009)

StefanPochmann said:


> Bryan said:
> 
> 
> > but the the total amount of heat would always be greater because as it cools, it's rate would cool to be at most equal to the cooler coffee.
> ...



Doh, that first one was worded wrong. It should be "as it cools, it rate would decreases to be at most..."

Basically, let's say you have a 100 degree item and a 75 degree item in a 70 degree room. Because of the larger difference, the 100 degree item will lose heat at the rate of 5 degrees a minute. The 75 degree item will lost heat at a rate of .5 degrees a minute. The rate at which the 100 degree item will cool will constantly decrease as it approaches the 70 degree mark. At some point, that 100 degree item will be at 75 degrees, so its rate would be .5 degrees/minute. Of course, by the time it reaches this, the 75 degree item has cooled a bit.


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## shelley (Jul 22, 2009)

Remember that the professor also adds cream to his coffee just before he drinks it. The cream absorbs the same amount of heat from both cups.

The student's coffee would be hotter; the professor's coffee would have lost more heat during their conversation, due to the greater temperature difference between the coffee and the surroundings.


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## DavidWoner (Jul 22, 2009)

shelley said:


> Remember that the professor also adds cream to his coffee just before he drinks it. The cream absorbs the same amount of heat from both cups.
> 
> The student's coffee would be hotter; the professor's coffee would have lost more heat during their conversation, due to the greater temperature difference between the coffee and the surroundings.



+1 MOAR CHARACTERS


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## nitrocan (Jul 22, 2009)

Vault312 said:


> shelley said:
> 
> 
> > Remember that the professor also adds cream to his coffee just before he drinks it. The cream absorbs the same amount of heat from both cups.
> ...



That's what I thought.


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## jupp (Jul 22, 2009)

I think that the two coffees have the same temperature. It's only a guess, but I think the temperature falls exponential in both cases. So if you drawthe temperature of the two coffees in a graph it would be two paralell functions. The quotient of these functions always stays equal, so it does not matter when to put in the milk. 
i know it's a bad explanation and not a proof.


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## JBCM627 (Jul 22, 2009)

jupp said:


> I think that the two coffees have the same temperature. It's only a guess, but I think the temperature falls exponential in both cases. So if you drawthe temperature of the two coffees in a graph it would be two paralell functions. The quotient of these functions always stays equal, so it does not matter when to put in the milk.
> i know it's a bad explanation and not a proof.



The quotient will be something similar to:

Prof/Student = (Tc+Ti) / (Tc+Ti*e^kt)

Where Tc is the temperature of the cream, Ti is the initial temperature of the coffee, and k I believe should be >= 1. The ratio of temperatures will change, always in favor of the students being warmer as t increases.


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## Bryan (Jul 22, 2009)

jupp said:


> I think that the two coffees have the same temperature. It's only a guess, but I think the temperature falls exponential in both cases. So if you drawthe temperature of the two coffees in a graph it would be two paralell functions. The quotient of these functions always stays equal, so it does not matter when to put in the milk.
> i know it's a bad explanation and not a proof.



I agree. The notion that the cream removes the same amount of heat is wrong. If you add room temperature cream to room temperature coffee, no heat s going to be removed.



http://answers.google.com/answers/threadview/id/757539.html said:


> T(t) = Ts + (To - Ts)*e^(-k*t) ;
> 
> where:
> t is the time in the preferred units (seconds, minutes, hours, etc.)
> ...



Well, we know that k is equal for both.

We know the temperature of the cream is Ts, so if we add the cream and coffee at a m:c (Milk to Coffee) ratio, the new temperature is

(To*c+Ts*m)/(c+m) This is the temp of the students coffee right after he adds the cream (t = 0)

So the student's coffee after time is 

Ts +((To*c+Ts*m)/(c+m) - Ts)*e^(-k*t)

or

Ts + (To*c+Ts*m - (c+m)Ts)/(c+m)*e^(-k*t)

or

Ts + (To*c+Ts*m - Ts*c- Ts*m)/(c+m)*e^(-k*t)

or

Ts + (To*c - Ts*c)/(c+m)*e^(-k*t)

or

Ts + ((To*c - Ts*c)*e^(-k*t))/(c+m)

or

(Ts(c+m) + ((To*c - Ts*c)*e^(-k*t)))/(c+m)

or

(Tsc+Ts*m+To*c*e^(-k*t)-Ts*c*e^(-k*t))/(c+m)

Or

(Ts(c-c*e^(-k*t) + To*c*e^(-k*t) - Ts*c*e^(-k*t) + Ts*m)/(c+m)



And the professor's coffee right before he adds the cream (just normal cooling for time t) is

Ts + (To-Ts)*e^(-k*t)

So when he adds the cream, it's

((Ts + (To-Ts)*e^(-k*t))*c + Ts*m)/(c+m)

or

((Ts*c + To*c*e^-k*t - Ts*c*e^(-k*t) + Ts*m)/(c+m)

Or

(Ts(c-c*e^(-k*t) + To*C*e^(-k*t) - Ts*c*e^(-k*t) + Ts*m)/(c+m)

Which is the temperature of the student's coffee.


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## Holger (Jul 22, 2009)

I don't really understand the equation, but my guess would be that the students coffee were warmer since it has worse condtion for evaporation, which is requieres a lot of energy.


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## Hash (Jul 22, 2009)

I don't know much about physics, but wouldn't the heat of the prof's cup expand more than the students, therefore creating greater surface area and blah blah blah.. I don't know where I'm going with that..


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## qqwref (Jul 22, 2009)

I think Shelley's right. The coffee and cream portions have a given amount of heat (in the absolute sense), and both people end up with both coffee and cream, so the person with the hotter coffee+cream at the end will be the person who has had less heat radiate from their drink. Heat radiation is proportional to the difference between an object's temperature and the room's temperature, so the student's coffee (which was less hot, and thus closer to room temperature, during the 20 minutes) will have lost less heat and thus will end up hotter once both people have added cream.


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## Weston (Jul 22, 2009)

lol cubers are so smart.
im going to study physics in high school next year,
now i know i have a place to ask questions


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## Stefan (Jul 23, 2009)

The way I thought about it was something like this:

Imagine it's winter and the two came out of the freezing cold and now hold their cups to warm their hands. When the student uses the cream, the professor asks _"Why do you cool down your cup when you want to warm your hands with it?"_ and the student replies _"Conservation of energy - we both use the same energy, I'm just using it differently. You get the warmer hands, I get the warmer stomach"_.


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## JBCM627 (Jul 23, 2009)

Bryan said:


> Long Post



Wow, I didn't expect that; I just worked this out fully too. I don't completely follow your math on here, which is missing some ()'s.

I've attached is a plot of time vs temperature for the 2 cups, with arbitrary unitless factors c=m=k=1, Ts=1, To=2. The change happens at t=2.

And if you like, sure its a lot of cream, but changing the factors will only scale/skew the plot, and won't change what it actually looks like.

Red=Prof, Blue=Student.

Mathematica code:

```
k = 1;
to = 2;
te = 1;
tchange = 2;

Temp[t_, To_, Te_] := Te + (To - Te)*Exp[-k*t]
PTemp[t_] := 
 Piecewise[{{Temp[t, to, te], 
    t < tchange}, {Temp[t - tchange, (Temp[tchange, to, te] + te)/2, 
     te], t > tchange}}]
STemp[t_] := Temp[t, (te + to)/2, te]

plot = Plot[{PTemp[t], STemp[t]}, {t, 1, 10}, Axes -> False, 
  Frame -> True, PlotRange -> {{.8, 4}, {.9, 1.4}}, 
  FrameLabel -> {"time", "Temp"}, 
  PlotStyle -> {Directive[Dashed, Lighter[Red]], 
    Directive[Thin, Darker[Blue]]}]
```


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## Stefan (Jul 23, 2009)

Hmm, if Bryan and Jim are right, my fear last night back at home (where I don't have internet can you believe it no you can't) would be true, that I shouldn't have said "conservation of energy" without properly understanding it.

I thought that if the professor wastes more energy during the 20 minutes (and Jim's graph does seem to agree with that), he shouldn't expect to have the same energy left at the end. Can someone explain what's wrong here?


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## jupp (Jul 23, 2009)

shelley said:


> .....
> The cream absorbs the same amount of heat from both cups.
> ...



It does not! The students coffee is hotter when he puts in the milk. So it cools down the students cup much more. I think the two effects even out in total.



StefanPochmann said:


> Hmm, if Bryan and Jim are right, my fear last night back at home (where I don't have internet can you believe it no you can't) would be true, that I shouldn't have said "conservation of energy" without properly understanding it.
> 
> I thought that if the professor wastes more energy during the 20 minutes (and Jim's graph does seem to agree with that), he shouldn't expect to have more energy left at the end. Can someone explain what's wrong here?



The student wastes more energy when he puts the cream into his coffee


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## Stefan (Jul 23, 2009)

jupp said:


> The student wastes more energy when he puts the cream into his coffee


I don't think adding the cream "wastes" energy at all. It becomes part of the mix, and we're interested in the temperature of the mix. Here's how I think of it:

Just to make it clearer/easier, let's say
- the "cream" actually is coffee 0 degrees cold
- the amount added is equal to the amount already in the cup

So when you add the "cream" to the coffee, the mix's temperature is half of the coffee's before. But the volume has doubled. Now...

energy = (energy/volume) * volume

When the cream is added, energy/volume is reduced by half, but volume is doubled, so the overall energy stays the same, it just gets spread inside the mix. It's not wasted.

And again, Jim's code seems to agree with my _"mix's temperature is middle of component's temperatures"_ understanding. So I continue to be puzzled.


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## Herbert Kociemba (Jul 23, 2009)

StefanPochmann said:


> Hmm, if Bryan and Jim are right, my fear last night back at home (where I don't have internet can you believe it no you can't) would be true, that I shouldn't have said "conservation of energy" without properly understanding it.
> 
> I thought that if the professor wastes more energy during the 20 minutes (and Jim's graph does seem to agree with that), he shouldn't expect to have more energy left at the end. Can someone explain what's wrong here?



1. If the temperatures are the same in both cups in the end, both cups must have lost the same amount of energy (in Joule) to the environment- no way to get around this fact.
2. Because this is independent of the time of mixing, the heat loss/time (in Joule/s) must be the same for both cups for all times t.
3. The heat loss/time of the studends cup must therefore for every moment be the same as the heat loss/time of the professors cup despite the fact, that the temperature difference to the envionment is smaller.
4. The only explanation for this is that the surface of the energy radiating part of the cup of the students cup is larger because there is more fluid inside.
5. This is implicitely included in the Mathematice Code of JBCM627 because he does not change the constant k after adding the cream to the coffee.
k is only constant in a certain situation. In detail it depends for example on the mass m of the fluid and the heat loss/(time*deltaT).


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## JBCM627 (Jul 23, 2009)

Herbert Kociemba said:


> 4. The only explanation for this is that the surface of the energy radiating part of the cup of the students cup is larger because there is more fluid inside.
> 5. This is implicitely included in the Mathematice Code of JBCM627 because he does not change the constant k after adding the cream to the coffee.
> k is only constant in a certain situation. In detail it depends for example on the mass m of the fluid and the heat loss/(time*deltaT).


The exposed surface area actually needs to remain the same for k to remain the same, as k depends on the surface area exposed. So here we assume that the cups are completely insulated, and the top of the cup (the same exposed surface area for both cups) is the only part radiating heat.



StefanPochmann said:


> And again, Jim's code seems to agree with my _"mix's temperature is middle of component's temperatures"_ understanding. So I continue to be puzzled.


This is only because of the particular way I coded it. What should really be done is to take the harmonic mean of the coffee and cream temperature, so it isn't necessarily right in the middle. My code was written using the arithmetic mean (meaning the coffee and cream were put in in equal proportions).


I found it helpful to think of it as a half-life problem; I didn't really understand it intuitively until then. Here is an analogous (albeit more tightly constrained) exponential problem:

Say you have a radioactive material with a half life of k. Sample #1 begins halfway decayed, while sample #2 begins undecayed.

After a time k, half the material in both samples has decayed. If we were to suddenly cause half the sample in #2 to decay at this time, would samples #1 and #2 both contain equal amounts of undecayed material?

Here it is much easier to see. If sample #1 starts at 100% and sample 2 at 50%, after a time k they will have 50% and 25% respectively. If we then half the 50%, they both contain 25% undecayed material.


Changing the half life time (k) will only stretch the graph. Changing the 50% will also only stretch the graph, and won't change the result.


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## Herbert Kociemba (Jul 23, 2009)

JBCM627 said:


> Herbert Kociemba said:
> 
> 
> > 4. The only explanation for this is that the surface of the energy radiating part of the cup of the students cup is larger because there is more fluid inside.
> ...



I totally disagree with you in this point. If the top is the only radiating part, k will decrease when you have more fluid/mass in the cup, because the same loss of energy in Joule leads to a lower decrease of the temperature T.
On the contrary, to keep k constant, the quotient surface/mass has to stay constant. Think for example of a system with two cups with temperature T. The decrease of temperature for this system is the same as for a single cup (if the distance of the two cups is large enough). The system has double mass but also double exposed surface.


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## JBCM627 (Jul 23, 2009)

Herbert Kociemba said:


> I totally disagree with you in this point. If the top is the only radiating part, k will decrease when you have more fluid/mass in the cup, because the same loss of energy in Joule leads to a lower decrease of the temperature T.
> On the contrary, to keep k constant, the quotient surface/mass has to stay constant. Think for example of a system with two cups with temperature T. The decrease of temperature for this system is the same as for a single cup (if the distance of the two cups is large enough). The system has double mass but also double exposed surface.


But k doesn't depend on total mass or amount of substance - only the area of the substance exposed.

From Newton's law of cooling, we have:
dQ/dt = h*A*(T_env-T(t))
Substituting Q(t) with C*T(t), we find that:
T_env-T(t) = Exp(-h*A*t/C + c_i) where c_i is a constant of integration. Or simplifying,
T_env-T(t) = T_0*Exp(-t*Ah/C), where T_0 is the initial temperature.

Here I point out that k = Ah/C, so the rate of decay depends on A, h, and C. h and C don't depend on the mass or amount of the sample, as they are constants representing physical attributes of the substance, and will remain the same no matter how much of it you have (assuming the coffee and cream both have similar h and C values, which is reasonable since both are mostly water). A is the only component of k that can be expected to vary here, and since we want k to be the same, we want A to remain the same.

I can see where you would think the rate of heat transfer depends on the amount of substance, and in the limit that you have a large quantity of substance, it probably does depend on how much you have. But this is only because the aforementioned cooling law no longer applies - you can't assume the substance cools uniformly any more.

If it helps, it is similar in the half life problem - the decay rate doesn't depend how much you have or the mass of it, it will still decay at the same rate.


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## Herbert Kociemba (Jul 23, 2009)

JBCM627 said:


> Herbert Kociemba said:
> 
> 
> > I totally disagree with you in this point. If the top is the only radiating part, k will decrease when you have more fluid/mass in the cup, because the same loss of energy in Joule leads to a lower decrease of the temperature T.
> ...



Sorry, but this is obviously wrong. Of course k is dependent on the mass:

dQ = m*C*dT, 
where m is the mass of the body and C is the heat capacity of the body. On the other hand we have 

dQ = h*A*(T_env -T) dt,
where h is the heat transfer coefficient and A the surface.

Setting this equal we get

m*C*dT = h*A*(T_env -T) dt, or

dT/dt = h*A/(m*C)(T_env -T), 

So writing 

dT/dt = -k(T-T_env) we have 

k = h*A/(m*C) 

Assuming that h and C is constant, we have the result that k only is constant if the *quotient* of surface A and mass m is constant.


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## JBCM627 (Jul 23, 2009)

Herbert Kociemba said:


> k = h*A/(m*C)
> 
> Assuming that h and C is constant, we have the result that k only is constant if the *quotient* of surface A and mass m is constant.


Right, I should've realized that  My C should = c/m, then, and I was mistaken in that it is really c which shouldn't change (C should).

I should have realized this with respect to the half life problem as well... I guess it can be thought of that if you add more substance, the half-life shouldn't change, meaning "A"/m should remain constant, where "A" is sort of a made-up area over which the substance can radiate. Since adding more substance gives a larger area over which to radiate, something else would also need to change to keep the half-life constant.

Meaning...


```
k = 1;
to = 2;
te = 1;
tchange = .5;
pctcoffee = .5;

Temp[t_, To_, Te_, m_] := Te + (To - Te)*Exp[-k/m*t]
PTemp[t_] := 
 Piecewise[{{Temp[t, to, te, pctcoffee], 
    t < tchange}, {Temp[
     t - tchange, (Temp[tchange, to, te, pctcoffee]*pctcoffee + 
       te*(1 - pctcoffee)), te, 1], t > tchange}}]
STemp[t_] := Temp[t, (to*pctcoffee + te*(1 - pctcoffee)), te, 1]

plot = Plot[{PTemp[t], STemp[t]}, {t, 0, 6}, Axes -> False, 
  Frame -> True, FrameLabel -> {"time", "Temp"}, PlotRange -> All, 
  PlotStyle -> {Directive[Dashed, Lighter[Red]], 
    Directive[Thin, Darker[Blue]]}]
```

Btw, where is Brunson with the answer he got in class?


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## mr.onehanded (Jul 23, 2009)

The professor's is cooler because the temperature difference between the coffee and the table and surrounding air is larger in the professor's coffee so more energy is transferred out. In the case of the student he adds his cream first so the temperature of the coffee is lowered and the difference between the coffee and surrounding area is lower.

To paraphrase:
I am almost certain that this is the answer, the equation for specific heat is Q = mcΔT Because c and m are equal in both equations the only variable is the change in temperature. Q or the change in temperature is equal to the difference in temperature; the larger the difference the larger the change in temperature. I hope that was clear, if I've made any mistakes please reply.


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## Stefan (Jul 24, 2009)

JBCM627 said:


> Meaning...


Meaning the complicated error-prone computation now agrees with the simple explanation


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## JBCM627 (Jul 24, 2009)

StefanPochmann said:


> JBCM627 said:
> 
> 
> > Meaning...
> ...


It at least leads to a full explanation... the simple ones aren't entirely correct. As shown by my first plot, if you disregard the amount of coffee/cream cooling, you get a really different answer, despite the warmer cup seemingly releasing more energy. Nobody until Herbert gave an argument that completely included the amount of coffee/cream cooling - which makes a lot of difference.


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## spdqbr (Jul 24, 2009)

Bryan said:


> ...Also, what is the shape of the cup?



As a topologist, I must say the cup is doughnut shaped.


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## qqwref (Jul 24, 2009)

spdqbr said:


> Bryan said:
> 
> 
> > ...Also, what is the shape of the cup?
> ...



You seem to be assuming that the cup has a handle. My experience with coffee shops has been that they tend to use somewhat thick paper cups with plastic tops; thus, I would guess that the macroscopic shape of the surface of the coffee cup (not counting the plastic cover, which is usually either genus 1 or genus 2 but doesn't seem to be standardized) is topologically equivalent to a sphere. Of course, this may very well not be a typical coffee shop, since it oddly enough seems to incorporate waitresses, which many modern coffee-based shops don't do.

I have a further question: what grade do you think the student got?


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## Herbert Kociemba (Jul 24, 2009)

JBCM627 said:


> Meaning...
> 
> 
> ```
> ...



I am not completely satisfied with Exp[-k/m*t] factor, because your new approach now completley ignores the increase of the radiating surface if the mass m is increased. It should be Exp[-const*A(m)/m*t], with A(m) meaning the radiating surface in dependence of the mass m in the cup. But A(m) depends indeed on the geometry of the cup. For a usual cup A(m)/m will also decrease for increasing m (as your 1/m does), but not so fast. So the graph will look similar to your new one. But the temperature difference will be smaller. We roughly can say that A(m) is proportional to m^(2/3) so we could take Exp[-k/(m^(1/3))*t] as a better approximation . Then we get the following graph.


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## JBCM627 (Jul 24, 2009)

Herbert Kociemba said:


> I am not completely satisfied with Exp[-k/m*t] factor, because your new approach now completley ignores the increase of the radiating surface if the mass m is increased. It should be Exp[-const*A(m)/m*t], with A(m) meaning the radiating surface in dependence of the mass m in the cup. But A(m) depends indeed on the geometry of the cup. For a usual cup A(m)/m will also decrease for increasing m (as your 1/m does), but not so fast. So the graph will look similar to your new one. But the temperature difference will be smaller. We roughly can say that A(m) is proportional to m^(2/3) so we could take Exp[-k/(m^(1/3))*t] as a better approximation . Then we get the following graph.



Agreed but only in part. Most coffee mugs are highly insulative, so the amount of heat lost to them is much less than the amount lost to the air, where convective instead of conductive heat transfer occurs, along with evaporation (which I will say doesn't change m). It is therefore probably more accurate to say that the area of radiation remains constant, since only the area exposed to air will radiate most of the energy away.

So if we wanted to be really picky about this, we could model A(m) as follows. If we have a perfectly cylindrical cup, then assuming m!=0, we have:
A(m) = A_a + A_b + A_w*m, where A_a is the area exposed to the air (constant), A_b is exposed to the bottom (also constant), and A_w*m is the amount exposed to the wall (linearly dependent on m).

With more constants (A_a = 2, A_b = .2 (convection should cool much faster than conduction, arbitrarily I'll say 10x as fast), A_w =1), we get the attached graph. For my particular code, A_w/A_b is exactly equal to the ratio of the actual areas of both, since when the cup is full m = 1. So also attached is an image of the cup


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## Herbert Kociemba (Jul 24, 2009)

JBCM627 said:


> Agreed but only in part. Most coffee mugs are highly insulative, so the amount of heat lost to them is much less than the amount lost to the air, where convective instead of conductive heat transfer occurs, along with evaporation (which I will say doesn't change m). It is therefore probably more accurate to say that the area of radiation remains constant, since only the area exposed to air will radiate most of the energy away.
> 
> So if we wanted to be really picky about this, we could model A(m) as follows. If we have a perfectly cylindrical cup, then assuming m!=0, we have:
> A(m) = A_a + A_b + A_w*m, where A_a is the area exposed to the air (constant), A_b is exposed to the bottom (also constant), and A_w*m is the amount exposed to the wall (linearly dependent on m).
> ...



Yes, I fully agree now. Whether the model is really adequate, only an experiment will show. Returning to the initial question I would say from my experience, that after 20 minutes both coffees are cold.


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## deadalnix (Jul 24, 2009)

Here is my gess :

Theyre is a certain amount of energy stored in each cup. This amount is the same.

This energy is leaving the cup to go in the air around making the air hoter and the cup cooler. The transfer depends on the difference of temperature.

I you put the cream in first, the difference of temperature of the mix cofee/cream and the air will be smaller than the one between cofee without cream and air.

So the first will losse more energy during the 20 mins.

Anyway, some conditions can change this result. Depending on the shape of the cup, the ratio surface/calorific capacity can also change. This is not the case for a standard cylindrical cup.


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## miniGOINGS (Jul 25, 2009)

Okey dokey, here goes.

Just for simplicity, let’s say that the temperature of the coffee is 100 degrees, and that the cream is room temperature which is, let’s say 50 degrees, and that the volume of the cream is half the volume of the coffee.

The student’s coffee, which starts at 100 degrees, when mixed with the cream (50 degrees) becomes 83.33… degrees [(100 x 2 + 50 x 1)/3 = 83.33…] (two parts coffee, one part cream, three parts total). Now, as time passes, the 83.33... degrees becomes closer and closer to room temperature (50 degrees). Let’s say that each 5 minutes the coffee gets halfway closer to room temperature:

0 minutes: 83.33… degrees
5 minutes: 66.66… degrees [(83.33... - 50)/2 + 50 = 66.66...]
10 minutes: 58.33… degrees
15 minutes: 54.166… degrees
20 minutes: 52.0833… degrees

The professor’s coffee loses heat at a faster rate because it is at a higher temperature, so:

0 minutes: 100 degrees
5 minutes: 75 degrees [(100 - 50)/2 + 50 = 75]
10 minutes: 62.5 degrees
15 minutes: 56.25 degrees
20 minutes: 53.125 degrees

But then he adds the cream, making it 52.0833... degrees [(53.125 x 2 + 50 x 1)/3 = 52.0833…] which is the same temperature of the student’s.


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## Paul Wagner (Jul 25, 2009)

Don't forget the cream loses heat as well.


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## miniGOINGS (Jul 25, 2009)

Paul Wagner said:


> Don't forget the cream loses heat as well.



The cream was always at room temperature, so it doesn't lose heat.

EDIT: I am positive that changing the numbers of the variables will not change the result.

EDIT2: If it doesn't work, don't get mad at me, im only 15, I havn't taken physics yet, i just made this up thinking it would make sense.


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## badmephisto (Jul 25, 2009)

@JBCM and Kociemba: You guys are insane 
One question though: How robust is that final graph to changes in the assumed constants? Can you flip the answer if you just change the constants a little, or is that what you always get qualitatively?


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## Herbert Kociemba (Jul 25, 2009)

badmephisto said:


> @JBCM and Kociemba: You guys are insane
> One question though: How robust is that final graph to changes in the assumed constants? Can you flip the answer if you just change the constants a little, or is that what you always get qualitatively?



Changing some constants will not flip the answer. The quotient of the heat radiating surface A and the mass of the fluid m is relevant. The answer would flip if

A(coffee with milk)/m(coffee with milk) > A(pure coffee)/m(pure coffee).

For this to happen the shape of the cup will have to be quite unusual.


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## panyan (Jul 25, 2009)

the person who pours it in last will be hotter, there is less time for the energy in the coffee to go into the cream


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## deadalnix (Jul 25, 2009)

miniGOINGS said:


> Just for simplicity, let’s say that the temperature of the coffee is 100 degrees, and that the cream is room temperature which is, let’s say 50 degrees, and that the volume of the cream is half the volume of the coffee.



The volume isn't important (except for a very spécific shape for the cup). The calorific capaciy is important. Anyway, we don't have this data, but the results of your calculation are wrong.



miniGOINGS said:


> But then he adds the cream, making it 52.0833... degrees [(53.125 x 2 + 50 x 1)/3 = 52.0833…] which is the same temperature of the student’s.



So, on of the cup loose more energy than the other, but at the end they are at the same temperature. Tjis is really inconsistante with physics basic like energy concervation.


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## qqwref (Jul 25, 2009)

miniGOINGS said:


> Let’s say that each 5 minutes the coffee gets halfway closer to room temperature:



I think HERE is where we have the problem. If you have two objects of different sizes, they won't necessarily both get halfway closer to room temperature in the same interval of time.

Wikipedia: "The law of Heat Conduction, also known as Fourier's law, states that the time rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area at right angles, to that gradient, through which the heat is flowing." In normal English this means that the rate at which heat is lost (here "heat" is referring to an energy, so that for instance if A and B are the same temperature but A is twice the size of B, A will have twice the heat of B) is some constant times the difference in temperature times the surface area through which heat flows.

So let's suppose that our coffee cups are cylindrical and insulating, so that only the top of the cup is losing energy to the air. This would mean that the surface area through which heat flows is the same no matter how much liquid there is. Thus, the rate at which heat is lost is proportional to the difference in temperature between the liquid and room temperature - or, _the rate at which temperature is lost is proportional to the difference in temperature divided by the volume of the liquid_.

Let's take miniGOINGS's example. Suppose the liquid starts at 90 degrees Celsius and room temperature is 30 degrees Celsius. Suppose each person has 100 ml of coffee and that they add 50 ml of cream.

Now the student adds cream at the beginning and thus ends up with a coffee at (90*2 + 30)/3 = 70 degrees, so the temperature difference is 40 degrees and the volume is 150 ml. On the other hand, for the professor, the temperature difference is 60 degrees but his volume is 100 ml. The professor should be losing heat at a rate of about (60/.1)/(40/.15) = 2.25 times that of the student.

So how do we calculate stuff? Well, since you lose heat at a rate proportional to the temperature, you can model the temperature through time as Tr + Ts * e^(-k t), where Tr is room temperature, Ts is the starting difference in temperature, and k is some constant. The rate of temperature change is the derivative (don't worry about this if you don't know what it means), which is Ts * k * e^(-k t). At t = 0, e^(-k t) is 1, so we just want Ts * k to be equal to our rate of temperature loss. Remember that the rate of temperature loss is proportional to the temperature difference (Ts) times one over the volume of liquid? That means that the value of k will be some number divided by the liquid's volume.

So let's arbitrarily set k = .25 / (volume in liters) and see what happens when t = 1 (we can arbitrarily define this to be 20 minutes). We get:
student's temperature = 30 + 40 * e^(- .25/.15) = 37.5 degrees
professor's temperature = 30 + 60 * e^(- .25/.1) = 33.3 degrees.
professor's final temperature = 30 + 2/3 * (60 * e^(- .25/.1)) = 32.2 degrees.
As you can see, the professor's losing heat much more rapidly because of his smaller volume contacting the same surface, so even though he starts out with more heat he ends up (in this case) colder than the student!


Interestingly, it turns out that ONLY the k constant affects things. To see why, let's write out the equation including the cream. Here 2/3 is the proportion of coffee, but this can be any number from 0 to 1.
student's temperature = Tr + (2/3 * Ts) * e^(- k t)
professor's temperature = Tr + 2/3 * (Ts * e^(- k t))
As you can see they evaluate to the same thing - the only difference is the value of the constant k! And remember that k is proportional to one over the liquid's volume (as well as to the surface area through which heat is lost). So the professor's coffee will end up colder as long as his k is larger. This actually happens whenever the surface area of the cup increases by a smaller factor than the factor by which the liquid volume increases. This always happens with normal coffee cups, so in real life the professor's coffee will end up colder.


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## miniGOINGS (Jul 25, 2009)

qqwref said:


> miniGOINGS said:
> 
> 
> > Let’s say that each 5 minutes the coffee gets halfway closer to room temperature:
> ...



Wow, thank's I never thought of that.


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## brunson (Jul 26, 2009)

ROFL!!! I went on vacation for ten days, so this is the first I've read of the thread since I posted it. 

The question was posed to illustrate the concept that the rate of heat transfer between two contacting masses (the coffee +/- the cream and the surrounding environment) is directly related to the difference in temperatures of the two masses.

I believe to be completely pedantic in the problem description (the coffee cups are cylindrical with negligible heat retention properties, the change in mass of the contents of the cup after the cream is added is negligible in relation to the total mass of the coffe/cup/cream system, etc.) would have detracted from the elegance of it. 

For example, when doing falling mass problems the fact that two objects of the same size but different weights accelerating under the influence of gravity through a fluid produce different amounts of fluid drag is usually overlooked since that part of the equation doesn't affect the outcome enough to change the fact that everyone expects the iron ball and the rubber ball to hit the ground at roughly the same time, except in a fluid dynamics class. 

I clearly underestimated ability of this group to hone in on every possible remote nuance of the problem. I think everyone's assignment at this point it to google "Occam's Razor" and the meaning of the phrase "missing the forest for the trees". Thank you all for a very enjoyable read. Hehe.


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## spdcbr (Jul 26, 2009)

*You can assume that the amount of coffee and cream in each cup is identical and that both cups were served at the same temperature.* 

Wouldn't the student cream already have melted/dissolved in the coffee?

EDIT: Yay! I'm a genius!


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## Bryan (Jul 26, 2009)

brunson said:


> ROFL!!! I went on vacation for ten days, so this is the first I've read of the thread since I posted it.
> 
> The question was posed to illustrate the concept that the rate of heat transfer between two contacting masses (the coffee +/- the cream and the surrounding environment) is directly related to the difference in temperatures of the two masses.
> 
> ...



Blah blah blah. You still didn't answer what the answer was in your class.


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## spdcbr (Jul 26, 2009)

Bryan said:


> brunson said:
> 
> 
> > ROFL!!! I went on vacation for ten days, so this is the first I've read of the thread since I posted it.
> ...



I know! Okay just say: student, professor, or same. I don't want anymore weird physics crap.


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## qqwref (Jul 26, 2009)

brunson said:


> I clearly underestimated ability of this group to hone in on every possible remote nuance of the problem. I think everyone's assignment at this point it to google "Occam's Razor" and the meaning of the phrase "missing the forest for the trees".



I disagree. I think that, if the problem didn't mention a condition that can change the answer, then it's not our fault but yours (for leaving it out). There are theoretically possible cups which actually make the professor's and student's coffee the same temperature at the end. Suppose they use cylindrical cups, but the bottom and top (a cover which is placed right over the surface of the coffee) are insulated, whereas the sides are conductive. Then the surface area of the liquid is proportional to the volume, and since the rate of heat loss is some constant times the surface area times the temperature difference, the addition of cream would not change the rate of heat loss at all (if we use coffee and cream in a 1:1 mix, the surface area doubles, even though the temperature halves).

I think it might even be possible to make a cup where the student's coffee ends up colder (!). Imagine something where the radius of the cup starts out large and decreases as you go up, and where again the bottom and top are insulated. But then the surface area increases faster than the volume, and when you add cream the higher surface area would overcome the lower temperature (and cause the coffee to lose heat faster).

Of course, in physics you might only worry about physically likely conditions (that is, coffee cups which are likely to actually be used by a cafe in real life), but I'm a mathematician and we are only concerned about making sure our conditions are theoretically possible


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## Herbert Kociemba (Jul 26, 2009)

At least for me the physical background of the original problem is sufficiently clarified now and I enjoyed the discussion here. What has become clear now is that the answer depends on the shape of the cup and which parts of the cup are heat conductive. There are some questions beyond that which I think are interesting:

1. Suppose the cup has rotational symmetry and only the top of the cup is heat conducting. Does there exist a shape that no matter how much coffee is inside the cup and how much milk is added the students cup is colder than the professors cup? (I suspect no). (EDIT: variation: has the same temperature)

2. Suppose the cup is cylindrical and again only the top is heat conducting. We know, that the professors cup is colder than the students. For which ratio of coffee and milk the temperature difference has its maximum?


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## JBCM627 (Jul 26, 2009)

qqwref said:


> I disagree. I think that, if the problem didn't mention a condition that can change the answer, then it's not our fault but yours (for leaving it out). There are theoretically possible cups which actually make the professor's and student's coffee the same temperature at the end.
> ...
> I think it might even be possible to make a cup where the student's coffee ends up colder (!).
> ...


So I guess you missed this part:


brunson said:


> the cups are normal cups





Herbert Kociemba said:


> 1. Suppose the cup has rotational symmetry and only the top of the cup is heat conducting. Does there exist a shape that no matter how much coffee is inside the cup and how much milk is added the students cup is colder than the professors cup? (I suspect no).


For this specific case, no - if we add no cream, then they will end up the same temperature no matter what cup shape. If we added no coffee, I'd like to see Brunson's face.

If both cream and coffee must be added, perhaps some sort of shape that flares out logarithmically?

Edit: I think might work, A(m) proportional to A_0 + Ln(m) ?


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## nitrocan (Jul 26, 2009)

What is this effort to turn the question the other way around?

Of course the cups are very intelligently designed robots that sense if the cream is put in by a smart student who knows that it'll be better to put it earlier. Then, the robotic cups, just to mess around with the student, cool down the coffee and the student gets owned.

I guess the student wasn't smart enough to guess that now was he? You don't mess around with the professor.


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## brunson (Jul 27, 2009)

Right... I also didn't mention that the professor was on fire.


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## arud45 (Sep 19, 2009)

I know this is a big bump, but I read something like this in "A brief history of time," and figured I'd comment. According to Hawking all hot objects cool at a rate proportional to the 4th degree of their absolute temperature. Then he uses the case of coffee and milk (not sure if milk = cream, though it sounds like they should), saying that the person that waits till the end to add milk will have a cooler cup than the person that puts it in at the beginning.


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## brunson (Sep 27, 2009)

Experimental evidence:

http://www.thenakedscientists.com/HTML/content/kitchenscience/exp/-0c2ee7e805/


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## guinepigs rock (Jun 11, 2010)

the professors is hotter because the cream in the students cup had time to sit and worm up and the professors never had the cream in it till he took a sip so the professors is hotter.


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## Erdos (Jun 11, 2010)

Newton's law of Cooling -> Student's cup is hotter.

Not exactly sure why we're debating this, since this was figured out in the 18th century although obviously with different situations but similar results.

EDIT:



guinepigs rock said:


> the professors is hotter because the cream in the students cup had time to sit and worm up and the professors never had the cream in it till he took a sip so the professors is hotter.


Yes, that would be true...if we were talking about which cream would be hotter. However, this is about which cup of coffee is hotter. You probably misinterpreted the question.


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## peterbat (Jun 11, 2010)

arud45 said:


> I know this is a big bump, but I read something like this in "A brief history of time," and figured I'd comment. According to Hawking all hot objects cool at a rate proportional to the 4th degree of their absolute temperature. Then he uses the case of coffee and milk (not sure if milk = cream, though it sounds like they should), saying that the person that waits till the end to add milk will have a cooler cup than the person that puts it in at the beginning.



Haha, yay, someone mentioned radiative cooling (there's a reason why the inside of a thermos is reflective -- it's to minimize energy loss due to radiation, i.e. light!) . So, I haven't read Hawking's book (shame on me as a physicist!), but I should mention that the cooling rate is proportional to the difference in the fourth powers of the cup's temperature and the outside environment's temperature.

There is also the heat exchange that occurs during mixing, which, as others have said, produces a weighted average of the two initial temperatures (a linear combination). This, however, does not change the total energy of the mixture.

So I'll argue it like this:

1) The (cream+coffee) system has initially the same total internal energy for both the student's and the professor's case, but the professor's cream isn't mixed in initially. Both coffees are at a temperature T1, let's say.

2) The student immediately adds his/her cream, reducing the temperature of their cup to a value T2. The professor's cup, however, is still at temperature T1, which is hotter than T2.

3) Now both cups start to lose energy, but at different rates:
student rate = (constant) * (Tstudent(t)^4 - Tenv^4)
professor rate = (same constant) * (Tprofessor(t)^4 - Tenv^4),
where Tstudent(t=0) = T2 and Tprofessor(t=0) = T1.

4) If you think about it, you'll realize that the student's rate of heat loss will always be less than the professor's during radiative cooling, so over the course of their conversation, the student's (mixed) coffee+cream will have lost less energy than the professor's (unmixed) coffee+cream.

5) The professor adds his/her cream at the end, which does not change the energy of their (coffee+cream) system. Since the student's coffee+cream lost less total energy, their coffee will be warmer than the professor's.

**So professor's coffee is cooler than the student's.

I should also mention that T^4 power transfer via electromagnetic radiation was not discovered until the 19th century, and not fully understood until the 20th century.


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## Zubon (Jun 11, 2010)

This is a pretty famous thermodynamics question.

To remove ambiguity, you should say that the professor and the student stir their coffees after adding the cream. 

If the coffee was *very *hot, the cream *very *cold and the coffee drunk *immediately *after adding thick cream *without mixing*, the results could change.


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