In biology, it is stated that more surface area and less volume helps for keeping bodies cool. How can one explain this phenomena in terms of physics?

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    $\begingroup$ More surface area means more space for heat to leave the body. I'm guessing the "less volume" part means you want more surface area while keeping the volume the same (eg breaking an ice cube to create more surface area but keeping the same exact volume). $\endgroup$ – markovchain Feb 22 '14 at 7:25
  • $\begingroup$ I think you mean keeping them cool. $\endgroup$ – Nathaniel Feb 22 '14 at 7:51
  • $\begingroup$ I am sorry. I meant cool. $\endgroup$ – user41114 Feb 22 '14 at 7:56
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    $\begingroup$ @Nathaniel that completely depends on the surrounding temperature, doesn't it? If $T_{surr} > T_{body}$, the body will tend to heat up and more surface area will allow more absorption of heat. If $T_{surr} < T_{body}$, the body will cool down. $\endgroup$ – mikhailcazi Feb 22 '14 at 9:06
  • $\begingroup$ @mikhailcazi sure, but the stated context is biology. Typically, organisms generate heat in their tissues and have to dissipate it into the environment. $\endgroup$ – Nathaniel Feb 22 '14 at 11:38

Like all good physicists we'll start by assuming the animal is spherical. Actually the calculation I'm going to describe is basically the same whatever shape the animal is, but choosing a sphere means I can write down some simple formulae.

If the radius of the spherical animal is $r$, then the total area of its skin is the surface area of the sphere:

$$ A = 4 \pi r^2 $$

and the volume and mass are:

$$ V = \frac{4}{3}\pi r^3 $$

$$ M = \rho V = \rho\frac{4}{3}\pi r^3 $$

where $\rho$ is the density of the animal.

Under most circumstances the dominant mode of cooling will be convection, and the rate of heat loss per unit area of skin is given by Newton's law of cooling:

$$ \frac{dQ}{dt} = k A \Delta T \tag{1}$$

where $\Delta T$ is the difference between the animal's temperature and the temperature of the air around it, and $k$ is a constant (technically known as a fudge factor) that depends on the details of how the wind is blowing around the animal.

Now, the animal will have an (average) specific heat, $C$, and this tells us how much the temperature of the animal falls when it loses a quantity of heat $\delta Q$:

$$ \delta T = \frac{\delta Q}{C M} $$

The rate of cooling of the animal is $dT/dt$, and we get this by dividing both sides of the above equation by $\delta t$ and letting the changes $\delta$ become small so they turn into differentials:

$$ \frac{dT}{dt} = \frac{1}{CM} \frac{dQ}{dt} $$

and we already have an expression for $dQ/dt$ from our equation (1) above. If we substitute this we get an expression for the rate of cooling:

$$ \frac{dT}{dt} = \frac{1}{CM} k A \Delta T $$

The last step is to replace the mass $M$ and the skin area $A$ with the equations from above, and do some rearranging to tidy up the expression, and we get:

$$ \frac{dT}{dt} = \frac{3k\Delta T}{C} \frac{1}{r} $$

and since $k$, $\Delta T$ and $C$ are all constants we get:

$$ \frac{dT}{dt} \propto \frac{1}{r} $$

So the rate of cooling is inversely proportional to the radius of the animal. Big animals cool slowly and small animals cool fast.

Although I started by assuming the animal is a sphere, the result really only depends on the ratio of surface area to volume, and for all shapes this scales inversely with size. So the result applies to any shape of animal though the constants in the equation will change with shape. Also note that I've simplified the calculation by assuming the thermal conductivity of the animal is high enough that its internal temperature is everywhere constant. In practice the animal's skin will get cold while its core is still warm. Still, the calculation does give you a basic idea of why larger animals cool more slowly than smaller animals.

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  • $\begingroup$ Sidenote: For animals hopefully $dT/dt=$ because of internal heat production :) $\endgroup$ – Bernhard Feb 22 '14 at 8:58
  • $\begingroup$ You can, by the way, incorporate all effects of conduction and radiation inside the $k$ of equation (1) $\endgroup$ – Bernhard Feb 22 '14 at 9:17

According to Stefan-Boltzmann law ,the thermal power radiated per unit surface area from any body is same for that object at a given temperature, regardless of its shape

So when surface area increase thermal radiation increases, and the body cools down faster

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    $\begingroup$ Only radiation is important? What do we define windchill temperature you think? $\endgroup$ – Bernhard Feb 22 '14 at 8:59

To start off I am just an amateur physicist, and do not use this information as credible facts. I composed this out of memory and not factual information.

The higher surface area the higher amount of heat transfer. Higher surface are contributes more space and places for heat to leech out or leech in depending on the surrounding circumstances. Also in the volume case the higher volume the more heat that can be held in.

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To understand this, look at heat as the energy possessed by vibrating atoms. Also remember that heat flows from hotter to colder objects.

When a greater surface area of a particular body is in contact with a cooler body, the vibrating atoms of the hot body can strike a larger amount of the atoms from the cold body, thus effectively transferring heat energy in the form of elastic collisions to the atoms from the cooler body (thus cooling down themselves).

For a particular body if the volume increases the number of atoms at the surface decreases and the atoms in the bulk cannot transfer heat to the colder object.

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    $\begingroup$ Please take care about formatting. You do use a lot of bold-face, but no capitals, this makes your answer very hard to read. $\endgroup$ – Bernhard Feb 22 '14 at 9:00

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