Hot Coffee
John makes a cup of coffee, which is initially at a temperature of 9 0 ∘ C . After 5 minutes, he notices that its temperature has dropped to 8 0 ∘ C . How much more time will it take for the temperature of the coffee to further drop to 7 0 ∘ C ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
Moderator note:
According to Newton's law of cooling, would the coffee ever actually reach room temperature?
If the rate of cooling slows down as the difference in the room temperature and the coffee temperature decreases, then does it mean that the coffee will never reach the exact same temperature as the room? Will it always be slightly warmer? This is not what we usually observe in real life.
Log in to reply
This is true, there will always remain a very tiny little difference in the temperature of the coffee and the surrounding. This difference is so small that we can not really sense it.
feels like the problem of the arrow reaching half of the way to its target, then half again, then half again... carrying on like this it should never reach its target.. yet in real life it does.
Not sure the laws at play are similar, but coffee and ambiant air in the room will reach to an equilibrium and at some point the room will be very very slightly warmer, and coffee a lot cooler (but supposedly slightly superior to initial room's temperature)
How do we know the coffee is in a room? Does it influence the result?
Log in to reply
I don't think this will affect the result until there is a sudden change in the weather conditions.
As temperature does have a fundamental unit, a quanta, wouldn't the cup actually reach room temperature once it approaches this limit to the minuteness in temperature difference?
Log in to reply
Theoretically, it will never reach the room temperature.
Rohit Gupta says... "When the coffee is left in a room, it radiates as well as receives energy to/from the surroundings."... (?)
My understanding is that heat (thermal energy) "always" goes(transfers) from "Hot to Cold"(higher potential thermal energy to lower potential thermal energy) Never from "Cold to Hot".?
Log in to reply
The net flow/transfer of heat is from the hot coffee to the cold surroundings. Although, the room is not at 0 K , therefore, it will also radiate. A part of those radiations will fall on the coffee and the coffee will absorb some of those radiations according to its absorptance.
However, the temperature of the coffee is greater than the room, therefore, it will emit more than it absorbs and hence the net heat flows out of it and it cools down.
Log in to reply
Rohit Gupta -- Your statement that the colder object will radiate heat that will be absorbed by the warmer object seems to conflict with what I have learned, that Heat "always" flows from hot to cold. Never (spontaneously) from cold to hot.
The second law of thermodynamics which is concerned with the direction of natural processes asserts that a natural process runs only in one sense, and is not reversible. For example, heat always flows spontaneously from hotter to colder bodies, and never the reverse, unless "external work" is performed on the system. Thus, an ice cube "must" melt on a hot day, rather than becoming colder.
An explanation for this form of the 2nd law can be obtained from Newton's laws and a microscopic description of the nature of temperature. We know that the flow of heat through conduction occurs when fast (hot) atoms collide with slow (cool) atoms, transferring some of their kinetic energy in the process.
One might wonder why the fast atoms don't collide with the cool ones and subsequently speed up, thereby gaining kinetic energy as the cool ones lose kinetic energy - this would involve the "spontaneous" transfer of heat from a cool object to a hot one, in violation of the 2nd law of thermodynamics.
The answer lies in energy and momentum conservation in a collision - one can show, using these two principles, that in a collision between two objects which conserves energy (called an elastic collision) the faster object slows down and the slower object speeds up.
It is important to emphasize that this statement of the 2nd law of thermodynamics applies to the "spontaneous" flow of heat from hot to cold.
It is possible, of course, to make a cool object in a warm place cooler - this is what a refrigerator does - but this involves the input of some "external" energy. As such, the flow of heat is NOT "spontaneous" in this case.
Log in to reply
@Allan Israel – I agree with you that net heat flows from higher temperature to lower temperature spontaneously. However, this does not imply that an object won't absorb the heat radiations coming from an object with even higher temperature. How would an object know that which radiations are coming from which source and what it is the temperature of that source?
We get infrared from the sun as well as from a human being. Suppose a cup of tea is placed in open with a person standing nearby, the tea will receive infrared from the person as well as from the sun. The tea does not have a filter to know which radiations are from the person at low temperature and which are from the hot Sun. Does it? If no, then the tea will absorb both the radiations without any bias.
When we place the tea in the room, the radiations emitted by the room will fall on the tea and the tea will absorb it. However, if its temperature is greater than the room then it will emit more than it has absorbed and therefore the net heat will flow out of the tea to the room. Hence, the second law of thermodynamics will remain intact.
Log in to reply
@Rohit Gupta – My original understanding was that the coffee was hottest object in the room (surrounding environmental system) and therefore the heat would flow only from the hotter coffee to cooler objects. Any cooler object that was "radiating" heat energy at a less temperature than that of the coffee would not be absorbed (would be reflected) by the hotter coffee because of the 2nd law of thermodynamics that states that the "spontaneous" flow of heat is from hot to cold (higher to lower temperature).
If there were an object radiating hotter heat than the coffee, then the coffee would absorb the hotter radiant heat since the coffee would be relatively "cooler".
I think we are in agreement. Thanks!
What you think is the main and mathematical difference in the heating/cooling between the coffee in a pot that was in the fire and the coffee when it is poured in a cup?
I need to know the thermodynamical difference besides the one in the speed. I think it has something to do in the second derivative , thanks!
Log in to reply
When you pour the coffee in a pot that was in a fire then the transfer of heat takes place mostly by conduction through the surface of the pot. Once the energy is received by the coffee near the surface of the pot it will be distributed throughout through conduction.
When you pour the coffee in a pot and left it in a room then the energy is lost mainly through radiations. Although, the surface of the pot will also be losing the heat through conduction. However this could be neglected as the coffee mugs are generally made poor conductor of heat.
Log in to reply
Hi , thanks for the answer. To be clear, i just want to know a clear difference when the coffe has the conduction of the pot and when it is poured in the cup. I mean in both cases it's getting cooler but obviously its getting cooler faster in the cup since, in the second its poured, the walls of the container help the cooling and it's not like the pot that is still preventing the coffe of getting cooler faster. My sense in comparing the second derivative is, like two functions can both be positive and crecent, but the concavity (how fast it grows) can be different depending on the sign of the second derivative. That's calculus 101 but I think in thermodynamics it has some similar effect. But I'm a mathematician, not a physicist. Thanks!
Log in to reply
@Aharon Ta – @Josh Silverman need some help with this comment from a month ago, it would be very helpful for me, thanks in advance.
@Aharon Ta – Hi @Aharon Ta , I don't see a crisp question here. Can you list some concrete points you want me to weigh in on?
Log in to reply
@Josh Silverman – yes, thanks. @Josh Silverman Whats the main thermodynamic difference in the cooling between a pot that was in the fire and a cup that wasn´t? (I want a numeric or algebraic answer, the logic is already there
since the pot still have the heat and mantains the temperature of the liquid inside even after its taken off the fire and the cup starts cooling the liquid in the instant it's poured inside because the walls of the container cool the liquid.
Thanks in advance!
Log in to reply
@Aharon Ta – Is there a reason you're distinguishing between a pot in a fire and a cup not in a fire? Would it change your question if it was a pot that had been in a fire and a pot that had not been in a fire?
Log in to reply
@Josh Silverman – No, it wouldn´t change the question if it was a pot that had been in a fire and a pot that had not been in a fire. The question still remains the same. And yes there is a reason, I need to know the effect and difference if, for example, I put a piece of raw meat inside the pot that has been in the fire and what changes if I put another piece of raw meat in a pot that was poured on from a pot that was on the fire. I know for sure and experiments that there is a difference, I just want to give it a scientific explanation.
Log in to reply
@Aharon Ta – Ok so would this be the situation you're talking about?
Case 1 - Water at temp T that is the result of heating the water in a pot which is now also at temp T . Case 2 - Water at temp T that was poured into a room temperature pot.
Log in to reply
@Josh Silverman – that's exactly the situation!
Log in to reply
@Aharon Ta – OK well, in both cases you're going to have to model 1. heat lost through the liquid surface 2. heat exchange between the water and the pot. 3. the heat exchange between pot and the room.
The water in the hot pot should take longer to cool down, but that seems obvious so I'm not sure I'm helping you. In the context of this example (the one with case 1 and case 2) can you ask me your question?
Log in to reply
@Josh Silverman – I think we are getting there, in both cases, 1. the heat lost through the liquid surface must be similar. but the heat exchange between the water and the pot is going to be completely opposite, since in case 1 the pot is neutral (or sometimes positive depending if the material is thermic), so it is "helping the water keep its heat agaist the room" and in case 2 , the pot is "helping the room" take heat from the water.
I need to explain this model mathematically to notice the clear difference between both cases. As you mentioned before, it oviously gets cooler faster in case 2, just need to clarify why. Also, there is a surprisingly big difference in the result of trying to cook something in such cases, I need to figure the scientific explanation.
thanks!
Log in to reply
@Aharon Ta – Well to your first question. The hot pot is at the same temperature as the water, so there isn't going to be much heat flow between them, unless one or the other starts to lose heat to the room at a fast rate. In the pre-heated pot case, the water will mainly lose heat to the air.
In the case of pouring hot water into the room temperature pot, a lot of heat from the water is going to transfer to the metal (because it's colder, Newton's law of heat transfer is roughly H ˙ ∝ Δ T ). That means the water will lose a lot of heat quickly by flowing away to the pot plus the heat it would otherwise lose to the air.
Log in to reply
@Josh Silverman – thanks, so in the equation you wrote about Newton's law of heat transfer, can you tell a main difference between cases? (for example, one variable becoming zero in a case or variables cancelling each other).
Log in to reply
@Aharon Ta – I have a feeling that obviously the graph for T vs time in both cases will be decreasing, but the convexity (or second derivative) may change depending on the pot.
Log in to reply
@Aharon Ta – The only difference in the setup would be in the initial conditions. Both systems have the same interactions and, so, the same governing equations.
Can you say more about your convexity hypothesis?
Log in to reply
@Josh Silverman – Of course, if we consider both cases as Functions, let them be case 1 and 2 f(t) and g(t) respectively, with t=time and f(t) being the temperature in the water.
In both of them the first derivative should be negative, since the temperature is decreasing with time, however, maybe the second derivative can have a difference since in f(t) the hot metal pot is fighting against the cooling and the decrement of the function is taking longer time, as opposed to g(t) when the cold metal pot is actually helping the cooling and therefore the decrement happens faster.
To put it in another way, the second derivative (convexity) tells us if the "growth is growing" or the "growth is decreasing", also viceversa.
Just as we see in the classic car moving example, where the speed is the first derivative of distance, and acceleration is the second derivative.
Am I being clear? Or maybe there is no difference in the convexity, but there should be a clear one if we have opposite effects from the pot in the relation of the heat exchange between the liquid and the room.
Thanks, once again
Log in to reply
@Aharon Ta – @Josh Silverman what do you think about the convexity hypothesis?
Hi @Aharon Ta , I second what @Rohit Gupta wrote about conduction in the pot, and infrared radiation during cooling. I don't understand your question about the second derivative—can you say a little more about what you're asking?
I thought similar to you, but not quite
The rate at which the coffee cools is proportional to the difference in temperature of the coffee and the environment. Once it has cooled a bit, this difference is smaller so the coffee cools at a slower rate.
Yes, this is called the Newton's law of cooling.
I used thermodynamics for the solution. A body of material loses heat quicker relative to the temperature of it's surroundings. As the water cools, it takes more time to cool by a given amount because it is approaching equilibrium with the surrounding temperature. As the temperature ratio tends to 1:1 the exchange rate of energy decreases, increasing the time needed.
Yes, the rate of cooling is directly proportional to the temperature difference.
But is thermodynamics really needed to reach the answer? For one, i just thought "what if the room is at 75 degrees?" xD
Well, this answer should actually be, "We don't know," because any number of variables could change. For example, putting the coffee in the refrigerator will cool it faster, as would putting ice in it. However, assuming all things being equal, the coffee would take longer to cool if left alone because, as the coffee cools down, its surroundings heat up, even if just a little bit, and the coffee's rate of cooling is dependent on its surroundings.
In questions like this, it is implied that "all things being equal", otherwise it would be impossible to have one correct answer.
Even if the temperature of the surroundings didn't increase one bit, the cup of coffee will still take more time to go from 80 C to 70 C as compared to from 90 C to 80 C. This is because the rate of cooling depends on the temperature difference between the room and the surroundings, and this difference decreases as the cup cools down.
In the first 5 mins the coffee loses heat to the room, and the cup, in the next period if time it is only really losing heat to the room! And even then at a slower rate than before!!
Can you elaborate on what you are trying to say? It will help others understand and follow your solution more clearly.
Log in to reply
I think what Jon means is that the coffee cup is not initially at 9 0 ∘ . Some heat is used to heat the cup to the same temperature as the coffee.
Right, even if we don't take into consideration the heat transfer from the coffee to the cup, it will take more time for coffee to cool from 8 0 ∘ to 7 0 ∘ as compared to from 9 0 ∘ to 8 0 ∘ .
When the coffee is left in a room, it radiates as well as receives energy to/from the surroundings. The energy it radiates depends on its temperature and the energy it gains depends on the room's temperature. As the temperature of the coffee is more than that of the room, the net heat flows out of the coffee and it cools down.
However, as the temperature of the coffee decreases, the rate at which it radiates energy to the room also decreases. This results in a reduction in the speed of cooling. Hence, the coffee will take more time to cool from 8 0 ∘ C to 7 0 ∘ C than from 9 0 ∘ C to 8 0 ∘ C . So, our answer is more than 5 minutes .
We can explain the same using the Newton's law of cooling , according to which the rate of cooling is directly related to the temperature difference between the surroundings and the object. As the object cools down, this difference decreases and hence the rate of cooling also decreases. This results in a slower cooling from 8 0 ∘ C to 7 0 ∘ C then from 9 0 ∘ C to 8 0 ∘ C .