### Brain Teasers

# Cup of Coffee

You are served a hot cup of coffee and room-temperature cream at a restaurant. You want to wait a few minutes before you drink the coffee, and you want it to be as hot as possible when you drink it. Should you pour the cream in the coffee:

a) Immediately

b) Just before you drink it

c) It doesn't matter

( "don't add any cream" is not an option)

a) Immediately

b) Just before you drink it

c) It doesn't matter

( "don't add any cream" is not an option)

### Answer

The driving force for heat transfer is temperature difference. The coffee by itself is very hot and will therefore cool down at a fast rate. Then once the cream is added the temperature will drop even more. If the cream is added immediately, then the temperature will drop initially but will then drop at a slower rate since the coffee with cream is cooler than the coffee alone and therefore the driving force for heat transfer is less.As an added bonus, adding the cream will increase the mass of the contents of the cup. A larger mass takes longer to cool down than a smaller one at the same temperature.

Therefore, the cream should be added immediately.

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## Comments

I'm not sure about where you say "If the cream is added immediately, then the temperature will drop initially but will then drop at a slower rate since the coffee with cream is cooler than the coffee alone and therefore the driving force for heat transfer is less."

There might be less driving force, but the coffee and cream is already at a much lower temperature. Would it not be true that when the coffe on it's only gets to the same temperature as the coffe and cream drops to initially that its rate of change would be less too, because it is at a lower temp, but during this time the coffee and cream has dropped even further. I'm wonderinging what would happen depending on the volumes involved, for example 1 litre coffee, 1ml of cream, or how hot the coffee is, although the size of a cup of coffee and cream added are obviously more specific!

I agree with your answer, just not fully with your reasoning! I think its a brilliant teaser by the way! Good work!

There might be less driving force, but the coffee and cream is already at a much lower temperature. Would it not be true that when the coffe on it's only gets to the same temperature as the coffe and cream drops to initially that its rate of change would be less too, because it is at a lower temp, but during this time the coffee and cream has dropped even further. I'm wonderinging what would happen depending on the volumes involved, for example 1 litre coffee, 1ml of cream, or how hot the coffee is, although the size of a cup of coffee and cream added are obviously more specific!

I agree with your answer, just not fully with your reasoning! I think its a brilliant teaser by the way! Good work!

I tried to come up with a better explanation for this, but I'm not having much luck. I'm afraid that any non-believers will just have to go on faith (unless someone else can explain it better). By the way Cathal, thanks for the compliment but this teaser is far from being an original. You'll have to thank the good people at - oh, wait - I get what's going on here! You're trying to get me to reveal my sources!! Well you can just wait until my pet pig "Muffin" flies to China under his own power!! HA HA HA...I'll NEVER TELL! NEVER!!!!!!!

I saw this problem on a kids science show about 20 years ago. From memory, I think the presenter said that there would be no difference, but I could be wrong. However, he did say that if the cream was high in fat it would leave a thin film of insulating oil in the top of the coffee. Hence the coffee would be slightly warmer if you added the cream first.

Heat loss is an exponential decay graph. The hotter the object is, the faster it cools. As it cools more and more, it starts to cool less rapidly. When the cream is added immediately, it brings the temperature down to a lower level, meaning the heat will be lost less rapidly. If you wait for the coffee to cool first, and then add the cream, the temperature change of the coffee would be less because the two substances are already closer together in temperature scale. Therefore, it shouldn't really matter, because one way you lose heat in the mixture, one way you lose heat into the air. The ammount of heat lost in each option should be equal, if you wait the same ammount of time with each trial.

who would really do all that thinking about when to put in their cream? it doesn't seem like much of a diffrence, but I may be wrong...

It was the great thinkers of ancient times who developed simple games for the most complicated problems they came across as a method to solve them. John Nashes @Game Theory@ is a modern day example (film: A beautiful mind based on him) Its little trivial things like this that can solve much more complicated issues, and its a lot easier to experiment on a cup of coffe for temperature than say a Power Plant, or whatever

Actually, I think the "driving force" aspect of the answer

is wrong. Suppose that the coffee loses half the temperature difference

between itself and the room every minute (a nice exponential decay curve),

and that it starts out with a 128-degree difference (coffee at 198 degrees, room at 70).

The temperature of the cream isn't going to be changing, so suppose that it reduces the

coffee temperature difference by 1/4th (there is 3 times as much coffee as cream).

Then if you let the coffee cool first, after 4 minutes the difference is 1/16th what it

started with, so it's an 8 degree difference (78 degrees). Adding the cream now reduces

the difference by 1/4, to 6 degrees. But if you added the cream first, you would have dropped

the temperature difference to 96 degrees difference. Letting this cool for 4 minutes likewise

drops the difference to 1/16th - to the same six degree difference. The temperature loss by cooling

is less, but the final temperature is the same.

However, the "added bonus" part probably still applies - the surface area through which the heat is

lost (primarily the top) remains the same, but the volumn holding the heat has increased. So you

probably do get some effect due to this.

is wrong. Suppose that the coffee loses half the temperature difference

between itself and the room every minute (a nice exponential decay curve),

and that it starts out with a 128-degree difference (coffee at 198 degrees, room at 70).

The temperature of the cream isn't going to be changing, so suppose that it reduces the

coffee temperature difference by 1/4th (there is 3 times as much coffee as cream).

Then if you let the coffee cool first, after 4 minutes the difference is 1/16th what it

started with, so it's an 8 degree difference (78 degrees). Adding the cream now reduces

the difference by 1/4, to 6 degrees. But if you added the cream first, you would have dropped

the temperature difference to 96 degrees difference. Letting this cool for 4 minutes likewise

drops the difference to 1/16th - to the same six degree difference. The temperature loss by cooling

is less, but the final temperature is the same.

However, the "added bonus" part probably still applies - the surface area through which the heat is

lost (primarily the top) remains the same, but the volumn holding the heat has increased. So you

probably do get some effect due to this.

I got it right, but I didn't even know why that was the answer. You all are geniuses.

The answer supplied is quite correct. The puzzle as I heard originally came from a series called "Why is it so?" by Professor Julius Sumner Miller. (I think he is a physicist). It can be modelled with differential equations but it is probably not necessary if you consider the answer given that heat loss is greatest when temperature difference is greatest. Therefore the sooner the coffee is brought closer to room temperature, the slower will be the heat loss.

Dewtell, your math is totally wrong. I will show you the correct math. Start off with an overall energy balance and simplify.

In - Out + Source -Sink = Accumulation.

=> -Out = Accumulation

In this case, I will assume that the Biot number is less than .1 and thus I won't have to worry about any temperature gradients in the coffee. To also simplify the problem, I will assume that heat will only be convected away from the cup at the liquid surface and I will assume that the surface area doesn't change when the cream is added.

Going back to the equation, after adding in Newton's Law of Cooling and the accumulation term, it becomes:

-A*h*(T-Tc) = V*ro*Cp*dT/dt

A = Area of heat transfer

h= heat transfer coefficient

T= Temperature of coffee

Tc = Air temperature

V= Volume of Coffee

ro= Density

Cp= Heat capacity

dT/dt= Rate of change of Temperature with respect to time

This differential equation can be separated and integrated resulting in an equation that gives the temperature versus time:

T = Tc + (Th-Tc)*exp((-A*h*t)/(V*ro*Cp))

Th= Initial temp. of coffee

I will assume that all the physical properties of coffee are the same as water and I will make a logical guess as to the size of the cup, the initial temperature, and the magnitude of the heat transfer coefficient.

I will also assume ideal mixing so I can calculate the temperature of the coffee after mixing.

No matter what time I plug in I get that the coffee that had the cream added immediately was slightly warmer than the coffee that had the cream added right before drinking.

In - Out + Source -Sink = Accumulation.

=> -Out = Accumulation

In this case, I will assume that the Biot number is less than .1 and thus I won't have to worry about any temperature gradients in the coffee. To also simplify the problem, I will assume that heat will only be convected away from the cup at the liquid surface and I will assume that the surface area doesn't change when the cream is added.

Going back to the equation, after adding in Newton's Law of Cooling and the accumulation term, it becomes:

-A*h*(T-Tc) = V*ro*Cp*dT/dt

A = Area of heat transfer

h= heat transfer coefficient

T= Temperature of coffee

Tc = Air temperature

V= Volume of Coffee

ro= Density

Cp= Heat capacity

dT/dt= Rate of change of Temperature with respect to time

This differential equation can be separated and integrated resulting in an equation that gives the temperature versus time:

T = Tc + (Th-Tc)*exp((-A*h*t)/(V*ro*Cp))

Th= Initial temp. of coffee

I will assume that all the physical properties of coffee are the same as water and I will make a logical guess as to the size of the cup, the initial temperature, and the magnitude of the heat transfer coefficient.

I will also assume ideal mixing so I can calculate the temperature of the coffee after mixing.

No matter what time I plug in I get that the coffee that had the cream added immediately was slightly warmer than the coffee that had the cream added right before drinking.

Hey am I the only one who knew the answer just from years of actually drinking coffee with cream?

While no one may ever read this as I'm responding two years later than the last response, there is one fundamental issue that many of you are missing.

Stirring a liquid is one of the fastest ways to cool it down. By stirring at the start you lose a lot of heat.

Add the cream, but don't stir.

Stirring a liquid is one of the fastest ways to cool it down. By stirring at the start you lose a lot of heat.

Add the cream, but don't stir.

Bassically, since the coffee will need to use its energy on the cream anyway, add the cream immediatelly...

Here's an explanation that's not as precise as dewtell's, but it might be helpful.

- In terms of mass and chemical composition, the coffee + cream mixture is identical regardless of how it's done.

- The total energy of the system starts out the same in either scenario (obviously).

- The cream, at room temperature, neither gains nor loses thermal energy.

Therefore: We need only consider the net energy loss from whatever is in the coffee cup.

The hotter coffee (waiting to add the cream) loses energy at a faster rate, because it has a greater temperature difference compared to the surrounding air.

The imprecision in this scenario comes from ignoring the increased convection area of the coffee + cream system. I.e., increasing the volume raises the fluid level, thus increasing the surface area of the cylinder formed by the liquid.

This is likely to be neglible because of the insulating effects of coffee containers -- most all of the energy is being lost from the disc-shaped top of the cylinder, and that disc does not increase in size.

- In terms of mass and chemical composition, the coffee + cream mixture is identical regardless of how it's done.

- The total energy of the system starts out the same in either scenario (obviously).

- The cream, at room temperature, neither gains nor loses thermal energy.

Therefore: We need only consider the net energy loss from whatever is in the coffee cup.

The hotter coffee (waiting to add the cream) loses energy at a faster rate, because it has a greater temperature difference compared to the surrounding air.

The imprecision in this scenario comes from ignoring the increased convection area of the coffee + cream system. I.e., increasing the volume raises the fluid level, thus increasing the surface area of the cylinder formed by the liquid.

This is likely to be neglible because of the insulating effects of coffee containers -- most all of the energy is being lost from the disc-shaped top of the cylinder, and that disc does not increase in size.

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