# Noob physics problem question

1. Jul 25, 2009

### cubejunkies

I received this problem recently from a friend, and I think I have the right answer but I'm not sure. I also might not have enough evidence to support my case very well. Here's the problem:

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.

My guess is that because the professor's cup was hotter for 20 minutes, it wasted more heat, and thus I believe that the student's cup is hotter.

This is a ridiculous question, I'm almost sure of it, but please, I really need to know if I'm right.

Thanks
Anthony

ps I bet this thread is in the wrong section too, but I'm still very new to this...

2. Jul 25, 2009

### fleem

Assuming everything else being equal, you're absolutely right. I suppose there might be practical issues having to do with convection, the behavior of the fat in the cream, etc. But I doubt they make much difference in this case.

3. Jul 25, 2009

### cubejunkies

Alright, thank you.

But the question still remains, is there any way to prove this standpoint even more?

4. Jul 25, 2009

### Nabeshin

Initially I am inclined to agree with you. Let's take a look at the math behind this though...

From newton's law of cooling we have:
$$T(t)=T_{env}+\left(T(0)-T_{env}\right)e^{-rt}$$
Where r is some rate constant.
For simplicity's sake, let's say that when the creamer is added, we take the average of the two temperatures and they mix perfectly and instantaneously.

Case I: Creamer added initially.
$$T(t)=T_{env}+\left(\frac{T(0)+T_c}{2}-T_{env}\right)e^{-rt}$$
Case II: Creamer added at time t
$$T(t)=\frac{T_{env}+\left(T(0)-T_{env}\right)e^{-rt}+T_c}{2}$$

If we subtract case I from case II, we get:
$$\frac{T_{env}}{2}+\frac{T_c}{2}\left(e^{-rt}-\frac{1}{2}\right)$$
But since in the stated problem $$T_c=T_{env}$$,
$$\frac{T_{env}}{2}\left(e^{-rt}+\frac{1}{2}\right)$$

Since this value is always positive, it corresponds to case I always being hotter than case II. In doing the calculation, I'm starting to question my assumption about the averaging of the temperatures not effecting the answer. I might get back to you on this one, but that's it for now.

Last edited: Jul 25, 2009
5. Jul 25, 2009

### Nirgal

I think that the temperatures of the two coffee's will be equal.

There are two ways to look a this. The first way is to assume the coffee and cream mix instantaneously and uniformly at the moment the cream is poured in. The other way is to assume it takes time for the cream to mix.

In the first case where the cream mixes instantaneously, the student will have a hotter cup of coffee. In the second case where the cream takes time to mix then the two coffees will be at equal temperatures when the student and professor pick up their cups after a length of time to drink.

My reasoning is as follows:
Assuming the coffee and cream mix instantaneously and uniformly, the temperature of the coffee with cream will be less than the temperature of coffee without cream. This is intuitive and a statement of the second law of thermodynamics.

The rate at which energy is transferred between a body and it's environment is proportional to the differences between the temperatures of the two bodies. The rate at which energy is radiated from the body (to a black body) is proportional to the fourth power of the temperature of the body. All this means is that when you increase the temperature of the body, like a cup of coffee, the rate at which the thermal energy depletes (is transferred) from the cup of coffee becomes greater as you increase the temperature of the coffee.

It's just like a bucket half full of water with a hole near the bottom. The water will pour out much faster if the bucket is full of water, analogous to energy transferring much faster if the temperature is hotter.

So if the cream mixes instantaneously then when the student pours the cream into the coffee, the students coffee will instantaneously become colder then the professors coffee. But that is actually irrelevant initially and only matters at the end when both decide to sip their coffee after of a length of time. It is initially irrelevant because the professor's coffee will come to the same temperature as the student's coffee after a certain amount of time *(assuming this amount of time is less than the length of time they wait to sip), at which time both will loose thermal energy at the same rate. This is analogous to the bucket example where the rate at which the water pours out of the bucket decreases as the bucket becomes less full. So at the end of a length of time, the two coffees are at the same temperature. When the professor pours in their cream, that coffee will become instantaneously colder than the student's coffee. So you are correct in this case but your reasoning is off a little. It is not that the professor's coffee wasted more energy but is because of the assumption that the cream mixes instantaneously.

But in reality the cream does not mix instantaneously. It takes time for energy in the coffee to transfer into the cream and lower the average temperature of the coffee. So in reality the temperatures of the coffee's would be equal because it takes time for the cream in the professors coffee to mix. Assuming this time is longer than the time it takes the professor to pour the cream in and take a sip, then I think both are at equal temperatures.

If you disallow the assumption that I stated above next to the "*", where the time it takes the professor's coffee to come to the same temperature as the student's is less than the time they wait to sip, then at the end of the scenario, when the professor pours in the cream, their coffee will be hotter than the students coffee. This is because the professor's coffee never came to the same temperature as the student's and therefore was always hotter.

I have to examine in all of these assumptions because we're talking about a generic perfect situation. You have to choose your scenario carefully in order to answer this question.

6. Jul 26, 2009

### diazona

Well, that's actually not true according to Newton's law of cooling. The temperatures of the two coffees get closer and closer but they never quite reach the same value. (Of course eventually the difference will become too small to measure)

P.S. If the mixing time is longer than the time it takes for the professor to take a sip, after he pours the cream in, then you can't really talk about the temperatures of the coffees... since until the coffee and creamer mix uniformly, the mixture has no well-defined temperature.