A simple question. In several places, including this very recent and already famous paper


it is stated that bodies lose heat in water very fast due to the big heat capacity of water. I don't understand why heat capacity is relevant for the rate of heat loss. Why does it matter is the temperature of the air around me increases faster than that of water? Unless I am using clothes, that hot air is going away due to convection anyway and is being constantly replaced by new, cold air.

My naive explanation of this would be that we lose heat faster in water because we are essentially made of water and, when submerged, heat is transferred simply via conduction. We lose heat fast because the thermal conductivity of water is large.

So, does the heat capacity of a fluid influence significantly the heat loss rate of a body submerged in it?


The heat flow (per unit area) through some thin layer, e.g. a boundary layer of water, is given by:

$$ \frac{dQ}{dt} = \frac{K\Delta T}{d} $$

where $K$ is the thermal conductivity, $d$ is the thickness of the layer and $\Delta T$ is the temperature difference between the two sides of the layer.

So a high thermal conductivity does indeed mean a high heat flow rate. But as your body loses heat, that heat goes into heating up the water. If the water had a low specific heat then it would heat up fast and you'd quickly be surrounded by a layer of water at your body temperature. This layer of water would then act as an insulator.

In the context of the equation above, a low specific heat means $\Delta T$ quickly reduces with time and that reduces the heat flow. Conversely, a high specfic heat means it takes a lot of the heat from your body to heat the water, and that tends to maintain $\Delta T$ at a high value.

So while you are quite correct that high thermal conductivity is a major factor in the heat loss to water, so is the high specific heat of the water.

  • $\begingroup$ But aren't there also convective currents that ensure I have a relatively cold layer of water (or air) all the time around me? If the air around me reaches my temperature, won't it simply ascend and give its place to cold air? $\endgroup$
    – user5800
    May 19 '15 at 16:59
  • 3
    $\begingroup$ @whistles: yes, but how much heat those currents can carry away depends in part on the specific heat. If the specific heat is very low the current warms rapidly to your body temperature without removing much heat. $\endgroup$ May 19 '15 at 17:03
  • $\begingroup$ So, you mean that a warm reservoir at temperature $T$, placed in say cold air at temperature $T_A$, doesn't really create a strong enough convective current for relatively large values of $\Delta T$. Hence, even though the air around the reservoir is slowly being replaced, it's average temperature is stabilized at value significantly higher than $T_A$. On the other hand, in water it takes much longer to warm the water around the reservoir and hence even fairly slow convection is enough to keep it quite cool. Is that it? $\endgroup$
    – user5800
    May 19 '15 at 21:48
  • $\begingroup$ @whistles: Yes, I think that's a reasonable way to put it. $\endgroup$ May 20 '15 at 5:20

So would this be an accurate explanation: The process of heat transfer to air and water are both by conduction and occur continuously, assuming a thermal equilibrium will not be reached, due to their convective properties (as the air or water is heated, it is replaced by colder air or water). Furthermore, while both are poor conductors, water is worse than air. Based upon these facts, it would seem the body should lose heat faster to the air. Water, however, has a higher heat capacity than air which, in essence, means it "absorbs" heat better than air. Though heat is lost faster to water because, even though the rate of heat transfer to air is faster than to water, the "amount" of heat transferred to water is greater than air. Ultimately, with regard to it's affect on the body, the greater rate of heat transfer to air is outweighed by greater amount of heat transferred to water.

  • $\begingroup$ that answer isnt very clear $\endgroup$
    – Dan Z
    May 8 '20 at 7:56