Let’s say you had two quantities of some mass (soup, for example), one chilled and one warmed. If the frozen soup was at 0°C and the warmed soup was at 40°C, and both were then put into a room with an ambient temperature of 20°C, would one expect them to come to ambient temperature at the exact same time?
The assumption would be that they sat undisturbed, and one could know the average temperature of the mass (the outside of the frozen soup is going to come to ambient temperature faster than the middle, obviously).
Depends on quite a lot of factors. Conductive heat transfer is symmetric and linear; other modes can be more complicated.
If the chilled soup was merely chilled to just above freezing point but not actually frozen, then the hot soup would probably cool to room temperature before the cold soup warmed to room temperature. This is because the hot soup can lose heat by evaporation, and additionally the soup is cooled at its top surface, encouraging convection currents in the soup which mix it so more warm soup is continually brought to the surface to be cooled. Evaporation works against the warming of the cold soup.
If the cold soup is actually frozen solid, it has a large latent heat of fusion to be overcome before it can even begin to rise in temperature, so it’s at even more of a disadvantage in the race.
In general, heat transfer between two masses in contact is linear with respect to the temperature difference. After a degree in chemical engineering I have delta H = cp delta T burned into my brain.
Imagine - two identical liquids (soup, or water, or whatever). One is 20 C, one is 40 C. Identical masses are placed in identical volume containers of the same geometry. Both containers are then placed in a room at 30 C. Both then have the same delta T (10) so the heat transfer will be identical and they will heat and cool at the same rates, coming to the same temperature at the same time.
Now the niggling details:
cp (constant pressure heat capacity) does vary a bit. For a liquid in normal states and at normal temperatures this doesn’t vary much. But it does vary a little. This will throw off the time to ambient temp by a bit.
More significantly - As pericynthion points out, however - if one of the states is in a different phase (frozen soup or water) things will be completely different. There is a significant energy requirement in going from the liquid phase to the gas phase. So if one starts frozen and one starts liquid the temperature rate changes will be completely different - all heat going into the frozen state will be used to change the state to the liquid state (while maintaining the temperature) until all the solid frozen state molecules are in the the liquid state.
Even more nitpicky goodness - if the containers containing the liquid aren’t sealed, both will be losing some molecules (and heat) through vapor pressure evaporation.
So then if both of the masses are solids at room temp (let’s say a block of metal or wood or some such), then one would expect them to come to the same temp at the same time? These wouldn’t be losing anything through vapor pressure evaporation, or what-have-you, I presume?
I’m pretty sure the biggest difference between the two would be that the hot soup would lose lots of heat by evaporation, while the cold soup would only gain a little by condensing atmospheric moisture on it. So, the hot soup would cool faster.
Another factor in the hot soup’s favor is that, if you model radiation transfer with a heat transfer coefficient (which is a little bit funny), it is proportional to the cube of the mean temperature of the soup and the surroundings, so the hot mean is 30 C or 303 K, while the cold mean is 10 C or 283 K. (303^3)/(283^3) is 1.23, so about 23% more radiative transfer for the hot soup.
But there are numerous mechanisms. Conductivity, for example, is often about inversely proportional to absolute temperature, favoring the cold soup. Convective heat transfer isn’t symmetrical if the soup is sitting on a countertop, as the air flows are shaped differently in the two cases. It would be interesting to work this out in detail, but I’m pretty sure the hot cools faster.
Also, this difference will depend on the relative humidity of the room. 40°C soup in a 20°C room with 100 percent relative humidity will cool more slowly than in a 20°C room with 20 percent relative humidity. Similarly, there’ll be more condensation on the cold soup in the higer humidity room. Conceivably (though I wouldn’t bet on it), in a 20°C room with 100 percent humidity, the condensation would be enough to make the cold soup cool quicker.
If you had the block of wood or metal instead, and low enough humidity that there’s no condensation, these differences would go away.
Still might have differences in convection between the two. air coming over the top surface of the hot block is going to rise and get replaced by cool air. Air coming over the top surface of the cold block is going to cool, settle down, and stratify.
You might get closer to equal heat transfer rates by suspending the two blocks in mid-air.