For example, if I take an item out of the refrigerator, set it on the counter for a period of time, allow it to warm up a bit (but not so long that it reaches room temperature and stabilizes), and then put it back in the fridge, will it reach its original temperature in the same amount of time that it was on the counter?

For the sake of simplicity, assume that both the fridge and the air temperature in the kitchen are perfectly normalized, i.e., no fluctuations.


4 Answers 4


As everyone else is saying, if you assume Newton's law of cooling:

$$ \dot Q = m c_p \dot T = h A \Delta T $$

The equation for how you heat or cool is an exponential

$$ T(t) = T_\infty + \Delta T e^{ -\frac{hA}{mc_p} t } $$

The rate constant for growth (or dying) of temperature is the same (assuming other details of the material don't change much), so the half-life is the same regardless of the differences in temperatures or goal temperatures, but this does not mean you have to wait the same amount of time. As is the nature of half lives.

I think the best explanation would be a visual one:

Modelled heating and cooling

Here I'm showing how the temperature would evolve with some made up, yet practical chosen values. If you like, ignore the actual numbers on the scales as they are not that important. What's important is you'll notice that for the red curve I "took it out of the fridge" for 1 hour, and then put it back and it isn't until another 6 hours later that is 41 degrees again. But, for the purple curve, which was out of the fridge long enough to nearly come to room temperature (8 hrs), it takes just as much time to cool down as it took to heat up.

  • $\begingroup$ This is a great explanation that best answers the question at hand, I think. $\endgroup$
    – krs013
    Jul 17, 2014 at 23:43
  • 1
    $\begingroup$ This is a good answer, but it would also be worth mentioning convection. A warm object in a cold fridge will heat up the air surrounding it, which will rise up to the top and draw in more cold air from the sides. This increases the rate of cooling. However, a cold object in a warm room will cool the air surrounding it. Cold air sinks rather than rising, so it will mostly just hang around the object and stay cold. So the rate constant will actually be different in the two cases as well. $\endgroup$
    – N. Virgo
    Jul 18, 2014 at 2:17
  • $\begingroup$ @Nathaniel I don't think the effect will be too large. The convection can't take place until the heat conducts across the stale boundary layer at the surface of the object, this is why newton's law of cooling looks just like conduction, with a different constant out front, the heat transfer coefficient. People normally assume the heat transfer coefficient is independent of temperature, so I doubt it makes qualitative changes to the analysis. $\endgroup$
    – alemi
    Jul 18, 2014 at 2:56
  • $\begingroup$ @alemi that's very dependent on the situation, e.g. the geometry of the object and its surroundings. Sometimes it does make a big difference. In engineering situations where people really care about heat flow rates, people go to a lot of effort to calculate the convective heat flux. Usually you can only do it by numerically simulating the air flow, and software packages to do this are a profitable business. I've used them in the past. In this case I would expect the sign of the temperature difference to have a measurable effect. $\endgroup$
    – N. Virgo
    Jul 18, 2014 at 3:16
  • $\begingroup$ @alemi , your plot is an EXCELLENT representation of what is happening. That is one of the very best plots that I have ever seen regarding the representation of a moderately complicated process (e.g., 1st order exponential heating and cooling). $\endgroup$ Jan 20, 2016 at 1:51

I would use a very simple model, and assume that the item has only one temperature. Also i don't think that this will change the result, the real thing is more complicated.

As far as i know, the thermal energy flux is dependent on the thermal energy difference. So the time-dependent solution is something like an exponential function. That means that the temperature of the item converges exponentially to the temperature of the surrounding. This means you have to wait very long, if the difference is very small.

In your example, the item gains quite fast energy if it is taken out of the fridge, as the thermal difference between item and air is large. If you put it back, the difference is small, and it needs long to equilibrate.

So the energy it will give to the fridge is the same as it got from the room, but it needs longer to give it than to get it.

If you would put it short in the fridge, and than back in the room, it would loose its energy in the fridge faster, than it would get it back in the room.

Additional there are some higher order effects, as the air in the fridge might have, due to an different density and different amount of water in it, another heat capacity, but this shouldn't be to important for this specific question (fridge + room temparture, both air)

  • $\begingroup$ This seems hand-wavey and suspicious. Can you back this up with the relevant equations? $\endgroup$
    – phs
    Jul 17, 2014 at 21:53

Not always. The rate of heat transfer from one body to another depends on the difference in temperature between the two bodies and many other factors. Higher the temperature difference, faster is the rate of heat flow. So, when the object is brought out of the refrigerator, it will depend on the temperature difference between the object and air and when kept back in the refrigerator it will depend on the temperature difference between the object and refrigerator. They may not be the same.

  • $\begingroup$ Also worth mentioning that there are other variables at play beyond simply air temperature that can affect the rate of heat transfer; humidity of air inside vs outside the fridge for instance $\endgroup$
    – John
    Jul 17, 2014 at 20:24

A simple approach is to use Newton's Law of heat transfer: $Q = k\Delta T$ for the rate of heat transfer. Assuming the conduction, convection, and radiation aren't all that much different (same $k$) between sitting in the fridge and sitting on the counter, in one minute, the food will absorb or release an amount of heat proportional to the difference in temperature between it and its environment. Neglect internal conduction: the food is at a uniform temperature at any point, and the heat capacity is constant. If the food comes out of the fridge at 0 degrees C and the room is 30 degrees, the food will absorb 30 units of heat, and warm up by some amount (to still less than 30 degrees). Over the next minute, it will absorb slightly less heat, since $\Delta T$ is a bit smaller. Say the food is up to 20 degrees when you put it back in the fridge. Over the first minute, it will release 20 units of heat, and cool down by some amount (to still above 0 degrees). Assuming the heat capacity is constant, the 30 units of heat being absorbed will raise the temperature more than the drop due to 20 units of heat being lost to the fridge. So, it should take longer to cool back down to 0, as you're starting with a smaller $\Delta T$. If you had let the food warm up all the way to 30 degrees, it should take the same amount of time to cool down as it did to heat up, all other things being equal.

This is grossly simplified, and as others have mentioned, there are many complications in how well heat is transferred between the ambient air and the food, internal conduction, convective effects, radiative effects, the heat addition and removal capacities of the open air and fridge (assuming infinite reservoirs of heat or cold), the food's heat capacity, etc., etc. It would take an infinite amount of time to actually get to 30 (or 0), so you'd have to make 29 or 1 your target for the total time.


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