Heat Balance in the Human Body

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How much energy does our body use? How do we keep cool when it is really hot out? Can we justify eating more when studying?

Big Ideas: 
  • Steady state: power in = power out
  • The body maintains the power balance with the environment by eating, clothing and sweating.

Introductory video produced by Van Ly and Alex Hass.

Homeotherms (mammals, humans) regulate their internal body temperature within a range of about 1C despite large fluctuations in the temperature of their surroundings. This is achieved by balancing the flow of heat energy. We can divide the main components of this power balance in classic physicist's fashion (conduction, convection, radiation, evaporation) as follows:

$ M + R = C+\lambda+G+S $1

Here M  is the metabolic rate (generated internally by digesting food), R is the power absorbed in the form of radiation, C  is the power lost by convection, λ is the power lost through the evaporation of sweat, G  is the power lost through conduction and S  is the power stored. Depending on circumstances R, C, G and may have negative values. For this article, we will assume S is negligible compared to the other quantities (which is a healthy state for a mammal to be in): this is what we call "steady state".

1. Metabolic rate, M

In the human body, the rate of internal energy production is called the metabolic rate.  As food is broken down, and glucose is oxidized, heat is released.  This oxidation process can be written in chemist's fashion as follows2:

C6H12O6 + 6O2 --> 6H2O + 6CO2 + 686 kcal

Recall that 1 food calorie =  1 kcal  is approximately 4200 J and one mol of glucose is 180 g.  So, for 1 gram of sugar, 16 000 J of energy is produced. 

$ (\dfrac{686 \textnormal{ kcal}}{180 \textnormal{ g}})(\dfrac{4200 \textnormal{ J}}{1 \textnormal{ kcal}}) = 16 000 \textnormal{ J} $

So in physicist's terms, glucose generates 16 MJ/kg (strictly this is called the enthalpy of oxidation).

The Basal Metabolic Rate (BMR) or MB is the energy the body consumes when it is completely at rest (i.e. if you were lying in bed all day).  It is the energy required to perform only critical body functions such as breathing and circulation.

$ {M}_{B} \propto { m}^{n} $,

where m is the mass of the body and the exponent n lies between 2/3 and 3/4. The heat from the  basal metabolism is mainly dissipated through the skin, hence the basal rate is approximately proportional to the surface area of the animal (i.e. for animals of different size but the same shape, n = 2/3). The mean surface area of a human is approximately 1.7 m2 and a reasonable approximation for the BMR of a human is about 100 W (2000 kcal/d).  Hence MB ≈ 60 W/m2 of body surface area. Any exercise on top of the BMR requires more power.

To maintain a steady flow of heat from the human body's core at 37 C to the environment, the skin has to have a temperature of about 33 C. A rise in M means various physical processes have to come into play to maintain the skin temperature, primarily the removal of clothing and sweating. We will not discuss here how the heat flows inside the body, only how it flows to and from the environment.  

A person's weight is a balance between their caloric intake and their energy expenditure. Different foods (carbohydrates, proteins and fats) have different caloric values, or a different amount of energy in each gram1.  If a person consumes less than they expend, they will lose weight, but if they consume more than they expend, they will gain weight.  Any time you consume food you either use it, or your body stores it and you gain weight.  When the body has used the energy from all of the consumed food, it then starts using up your energy stores. You need to balance how much food you eat with how much energy you use, or how active you are, and this article assumes these are balanced.

Movement

Different activities require different amounts of energy.  The power used also depends on the weight of the person (heavier = more power for the same activity) and the intensity of the activity (more intense = more power). How would you rank the following activities: sleeping, swimming, running, walking, water aerobics, biking, and skating?

Check the table below to see if you were right (the ranges in numbers accounts for the intensity at which the exercise is done).  Note: the following numbers were taken for a 65 kg female; a 90 kg male would burn about 1.5 times more calories doing the same exercise at the same intensity3.  

<br />
\begin{tabular}{ | c c |}<br />
\hline<br />
  Activity & kcal/h \\ \hline<br />
  Sleeping & 60 \\<br />
  Walking (5 km/h) & 280 \\<br />
  Water Aerobics & 300 - 500 \\<br />
  Skating & 300 - 600 \\<br />
  Swimming & 300 - 600 \\<br />
  Running & 400 - 700 \\<br />
  Biking & 450 - 700 \\<br />
\hline<br />
\end{tabular}<br />
\

 

Note that 1 kcal/h = (4184 J)/(3600 s) = 1.16 W, so the numbers in watts are only slightly higher than the numbers in kcal/h.

Thinking

What about thinking?  After spending hours studying hard, do you feel especially hungry and reward your efforts with a tasty snack?  What causes this increased hunger?  How much more food can you justify eating?

In a recent study, participants were asked to spend 45 minutes sitting in a chair one day and 45 minutes reading an article and summarizing it another day.  After they did these tasks, participants were given a survey on how hungry they were and then were given the chance to eat.  It was found that when sitting in the chair, participants expended ~60 kcal and when they were reading and writing they expended 85 kcal.  All participants indicated a similar level of hunger, but those who sat ate 900 kcal of food and those who studied at 1150 kcal of food.  The difference in the energy expenditure of sitting vs. studying was only 25 kcal, but the difference in the food consumed afterwards was 250 kcal45!

Activities requiring significant cognitive demand favour an overconsumption of food without increased feelings of hunger.  Why?  Thinking requires glucose.  The body keeps some glucose in the bloodstream, but stores most of it as glycogen, which it has to convert back to glucose to use. While you are thinking, your brain quickly depletes the glucose stores in your blood, and tells you unconsciously to eat so that you can return the glucose levels to normal.  This results in you eating more while you study, even though you don't necessarily feel hungry.

So, next time you are studying, try to resist the urge to constantly eat, or to eat a giant snack when you do decide to visit your kitchen.  If you do give in, try rewarding a long day studying by taking a break and going for a bike ride!

2. Radiation, R

The total radiation power absorbed from the environment is R. We can further decompose this term into longwave radiation from the local environment (thermal infrared from buildings, the ground, flora, and the sky) and short wave radiation from direct sunlight (visible and near infrared):

$ R = {R}_{L}+{R}_{S} $

You can read more on radiation in the article on Thermal Radiation.

The longwave radiation component can be calculated using the following formula:

$ R_L = A \epsilon \sigma (T_{skin}^4 - T_{env}^4) $

where A is the surface area of the body, ε is the emissivity of skin (~0.98), σ is the Stefan-Boltzmann constant, Tskin is the (absolute) skin temperature (i.e. 306 K for bare skin) and Tenv is the (absolute) temperature of the environment. A complication occurs when we consider clothing. Clothing reduces radiation loss by lowering our effective surface temperature. However, when clothed, different parts of our body will have different surface temperatures, and we should calculate RL separately for each.  Note also that in many cases Tenv is not uniform (see annotated photographs at the bottom of the Thermal Radiation article), making the application of this formula yet more complicated.

Figure 1. On a typical day, different amounts of heat are radiated from different objects in the same environment. The above pictures were taken with a normal camera (left) and an infrared camera (right) on a cool day in June. Objects emitting more radiation appear lighter and those emitting less radiation appear dark- the people in this picture are much hotter than the environment and so stand out as being much lighter in colour in IR camera.

When your skin temperature is higher than the temperature of the environment, you radiate heat into the environment, but when it is lower, you absorb heat radiated from the environment. Short wave radiation, RS, coming from the Sun can be up to ~ 1 kW/m2 or more on a clear day on an area facing the Sun directly 6

Example: 54C in Baghdad

The temperature in Baghdad, Iraq, hit 54C in August 2010.  What does this do to the radiation balance of a human body? Let's assume you are inside. For Tskin = 306 K and Tenv = 327 K,  RL ≈ 260 W.

On top of this, if one walks out into bright sunlight 1 kW/m2, and estimating that the area of the body facing the Sun is 0.5 mand that clothing or skin absorbs roughly 50% of the light, one can estimate the power absorbed from the Sun to be:

$ R_S = (0.5)(1000 \textnormal{ W/m}^2)(0.5 \textnormal{ m}^2) = 250 \textnormal{ W} $

This is a big heating load to bear, but it gets worse, as we also need to consider convection, which is not going to cool us in these circumstances.

Figure 2. In the picture above, the surface temperature of the person's clothing is ~29oC but the surface temperature of the concrete is 38oC and so the person appears darker than the surroundings. Notice that the grass looks much cooler than the surroundings - this is due to evaporative cooling; the grass is quite damp, and thus the water molecules with a higher kinetic energy (and temperature) evaporate, leaving behing those with a lower kinetic energy, cooling the grass.

2. Convection, C

Convection is the transfer of heat by gas or liquid in motion (in our case, between our skin or clothing and the surrounding air). Once again, the heat transferred between the two is dependent on the temperature difference between them. There is also another important component to calculating the power transferred: the speed with which the air is moving. To a good approximation, and everything else being equal, convective losses (and gains) are proportional to area and temperature difference:

$ C\approx{K}_{C}A(T_{skin} - T_{env}) $

where A is the surface area of the body,  Tskin is the skin temperature, Tenv is the temperature of the environment and KC is an empirical factor that applies to the geometry of the human body. It varies as the speed of the air varies1:

<br />
\begin{tabular}{ | c c |}<br />
\hline<br />
  Speed of air & {K}_{C} \\ \hline<br />
  0 m/s & 3 W/{m}^{2}/K \\<br />
  2 m/s & 26 W/{m}^{2}/K \\<br />
  10 m/s & 37 W/{m}^{2}/K\\<br />
  20 m/s & 41 W/{m}^{2}/K \\<br />
\hline<br />
\end{tabular}<br />
\

We see that KC is dependent on the speed of moving air, and thus the power transferred is going to change with a change in the air speed. Note that the largest difference is between the still and slow-moving air, and not so much between higher speeds. For a temperature difference of 5 K, the power transferred through convection with still air is only 20 W, whereas if the air is moving at just 2 m/s, the power transferred is 200 W. 

The reason our bodies cool down when wind blows is due to a combined effect of the temperature and wind, known as wind chill - if you have taken a walk on a cold day, you probably felt colder when the wind blew. Our bodies create a form of its own insulation on known as the boundary layer, by warming up a thin layer of air close to the skin. As the wind blows, it takes this protective layer away, and our bodies use energy to warm up a new layer to prevent our skin from being exposed to the outside air. Thus if each layer keeps getting blown away, our skin temperature drops and we feel colder.7 A wind speed of 2 m/s is just enough to blow away the stagnant boundary layer of air around the body and replace with fresh air from the environment. 

A variant of convective loss is the power we lose to our breath. We can estimate this from our breathing rate (typically 6 L/min) and the heat capacity of air (1.3 J/K/L). For a core (breath) temperature of 37 C and an ambient temperature of 20 C, this component of convective loss works out to be a mere 2.2 W. This number rises to about 16 W if we consider the moisture loss in breathing1 (see next section).

Example: 54C in Baghdad

But what about in Baghdad? If the environment is 5 C cooler than our surface temperature, we lose 20 W, but if its 21 C hotter, we will gain 110 W from convection (and ten times that if the air is moving). So we have a sum total of over 600 W from radiation and convection (360 W if one is out of the Sun) to get rid of. This is getting extreme, but there is a way out, just.

4. Evaporative loss, λ

The latent heat of a substance is the amount of energy released or absorbed during a change of state, with no change in temperature. The latent heat of vaporization of water is thus the energy absorbed as water changes from a liquid to a gas. At at 30oC, it is approximately 2.4 MJ/L.

Sweating is a primary mechanism that our body uses to cool us; on a hot day, we sweat and the sweat evaporates off our skin, cooling us. The energy to evaporate our sweat, the latent heat of vaporization, is provided by our bodies. This evaporation in turn cools our bodies. However, being wet in the winter can also chill us because of the same evaporation effect. Do you think we would still keep cool if the sweat dripped off our bodies instead of evaporating?

We can sweat a maximum of 1 or 1.5 L/h. Given that 2.4 MJ is lost in the evaporation of 1 L of water/sweat (for now we approximate them as the same), we can find the power loss due to this evaporation:

$ \lambda\approx (\dfrac{2.4\textnormal{MJ}}{\textnormal{L}})(\dfrac{1.5\textnormal{ L}}{\textnormal{h}})(\dfrac{1\textnormal{ h}}{3600\textnormal { s}}) \approx 1000\textnormal { W} $

The environmental humidity level plays a big role in our ability to sweat. If the air is dry and there is no humidity, there is no barrier to how much we can sweat. We never have zero absolute humidity - the closest it ever gets is 0.03 % in Antartica 8, where the air is too cold to hold any water. Fortunately, our bodies won't usually perspire in Antarctica, otherwise with no barrier to sweating, we would probably get dehydrated fairly quickly! On the other hand, the higher the humidity, the more difficult it is for our sweat to evaporate. Thus on hot humid days, we feel more uncomfortable than on hot dry days9. This also explains why we use a fan, even when it is very hot and the moving air increases our convective heating: the fan blows away the boundary layer of air saturated with water, and allows us to continue evaporating sweat.

Without visible sweating, our total evaporative loss (including breathing) is 25% - 30% of our metabolic rate. This is approximately 30 W or 50 mL/h of water.

Example: 54C in Baghdad

Fortunately when the mercury hit 54 C, the relative humidity was less than 20%, so evaporative cooling of 1 L per hour (plus a lot of acclimatization) could cope.

 

5. Conduction, G

 

Conduction is the transfer of heat from high temperature to low temperature by direct contact between solid materials. As we have been considering heat loss from the exterior of the body and clothing, the only solid contact when standing is through shoes, or if seated, whatever one is sitting on. For shoes, the conduction rate is dependent on the temperature difference between the skin and the floor, and the thermal properties of the sole material10.

Conduction losses are found using the following formula:

$ G = \dfrac{kA({T}_{skin} - {T}_{env})}{d} $

Where G is the conduction loss in W, k is the thermal conductivity (measured in W/(m.K)), d is the thickness of the material (in m), Tskin is the temperature of the skin and Tenv is the temperature of the environment.  Since we are only require the temperature difference, we could use either K (kelvin) or oC (degrees celsius). 

Example: flip-flops

A good example would be to consider conduction losses from our feet: what would be our conduction losses from our feet if we are wearing rubber flip flops?

For a rubber (polyurethane) flip flop:

<br />
\begin{eqnarray}<br />
A &=& (30\textnormal{ cm})(20\textnormal{ cm}) = 0.06 \textnormal{ m}^2\nonumber\\<br />
k &=& 0.02 \textnormal {W/mK}\nonumber\\<br />
d &=& 1 \textnormal{ cm} = 0.01 \textnormal{ m}\nonumber\\<br />
{T}_{skin} &=& 306\textnormal { K}\nonumber\\<br />
{T}_{material} &=& 293\textnormal { K}\nonumber<br />
\end{eqnarray}<br />

Thus, the conduction is found as:

<br />
\begin{eqnarray}<br />
G &=& \dfrac{(0.02)(0.06)(306 - 293)}{0.01}\nonumber\\<br />
&=& 1.56 \textnormal{ W} \nonumber<br />
\end{eqnarray}<br />

 

A total loss of 3.12 W (1.56 W/foot * 2 feet) is small, much smaller than that of all the other processes considered.

Conduction losses, however, aren't always so small: misbehaving prisoners in the famous Alcatraz prison in San Francisco Bay were stripped of their clothes thrown into "The Hole" for punishment. "The Hole" was an enclosed prison cell that had no windows or doors and was made of concrete - the room was never heated. Concrete has a fairly high thermal conductivity (0.29 - 1.73 W/mK)11 and so if the prisoners lay down on the floor, their large sufrace area and large thickness, together with the high KC meant that conduction losses were very high - between 80 W - 400 W during 0oC weather. To prevent these losses (and their bodies consequently freezing), the prisoners would prop themselves off the ground by resting their elbows and knees on toilet paper rolls. The reduced surface area in contact with the material and the low conductivity of paper (0.05 W/mK) lowered their conduction losses by a substantial amount, at the expense of an uncomfortable posture.


 

Resources

Comments

I've tried to calculate

I've tried to calculate cooling by longwave radiation and convection by the formulas given in this article, but I get different values for the Baghdad example - same order of magnitude but different: For longwave radiation I have a heat gain of 269 Watts, and for convection a heat gain of 114 Watts. I can't find an error in my calculations. Were some of the values calculated with more precision than shown in the article?

The error was mine. Thanks!

The error was mine. Thanks!

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