Heat Balance in the Human Body

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?

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 conventional fashion (conduction, convection, radiation, evaporation) as follows:

$M + R = C+\lambda+G+S\tag{1}$[note]Monteith, J., & Unsworth, H. M. (2008). Principles of Environmental Physics. Amsterdam: Academic Press.[/note]

Here M  is the metabolic rate (generated internally by digesting food), R is the power absorbed in the form of radiation (minus that lost to radiation from the body), C  is the power lost (usually – if the body is warmer than its surroundings) by convection, λ is the power lost through the evaporation of sweat, G  is the power lost through conduction with the floor (gained if the floor is warmer) and S is the power stored. 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.” Depending on the environment, therefore, R, C, G and may have negative values.

1. Metabolic rate, M

In the human body, the rate of internal energy production is called the metabolic rate.  As the food is broken down, and glucose is oxidized, heat is released.  This oxidation process can be written in chemist’s fashion as follows[note]John R Cameron, James G Skofronick, Roderick M Grant, Physics of the body, Madison, WI: Medical Physics Publishing, 1992. 73-89.[/note]:

C6H12O6 + 6O2 –> 6H2O + 6CO2 + 686 kcal $\tag{2}$

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.

$$(\dfrac{686 \textnormal{ kcal}}{180 \textnormal{ g}})(\dfrac{4200 \textnormal{ J}}{1 \textnormal{ kcal}}) = 16 000 \textnormal{ J}\tag{3}$$

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.

$${M}_{B} \propto { m}^{n}\tag{4}$$

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 (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.

Different foods (carbohydrates, proteins and fats) have different caloric values, i.e. different amounts of energy in each gram[note]Monteith, J., & Unsworth, H. M. (2008). Principles of Environmental Physics. Amsterdam: Academic Press.[/note].  If we consume food energy less than we expend, we burn our own fat and lose weight. If we consume more energy than we expend, we gain weight. We need to balance how much food we eat with how much energy we use, or how active we are, and this article assumes these rates are balanced.


Different activities require different amounts of energy.  The power used also depends on the mass of the person (heavier = more power for the same activity) and the intensity of the activity (more intense = more power). In the table below, we rank the following activities: sleeping, swimming, running, walking, water aerobics, biking, and skating. The numbers in Table 1 were taken for a 65 kg female; so, for example, a 90 kg male would burn about 1.5 times more calories doing the same exercise at the same intensity[note]Health Status: Internet Assessment. Calories Burned Estimator (online) http://www.healthstatus.com/calculate/cbc [18 August 2010]. [/note].

Table 1. Power required for various activities.

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.


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 ate 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 kcal[note]JP Chaput, A Temblay. “Acute Effects of Knowledge-Based Work on Feeding Behavior and Energy Intake.” Physiology & Behavior 90 (2006): 66-72.[/note][note]JP Chaput, V. Drapeau, P. Poirier, N. Teasdale, A. Tremblay. “Glycemic Instability and Spontaneous Energy Intake: Association with Knowledge-Based Work.” Psychosomatic Medicine 70 (2008): 797-804.[/note]!

Activities requiring significant cognitive demand favour 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):

$$R = {R}_{L}+{R}_{S}\tag{5}$$

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

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

$$R_L = A \epsilon \sigma (T_{skin}^4 – T_{env}^4)\tag{6}$$

Where $A$ is the surface area of the body, $ε$ is the emissivity of skin (~0.98), $σ$ is the Stefan-Boltzmann constant, $T_{skin}$ is the (absolute) skin temperature (i.e. 306 K for bare skin) and $T_{env}$ 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 $R_L$ separately for each.  Note also that in many cases $T_{env}$ is not uniform (see annotated photographs at the bottom of the Thermal Radiation article), making the application of this formula yet more complicated.

Fig.1. The same scene viewed in visible light (left) and longwave (thermal) IR (right).

On a typical day, different amounts of heat are radiated from different objects in the same environment. The above pictures in Fig.1 were taken with a normal camera (left) and an infrared camera (right) on a cool day in June. The people in this picture are warmer than the environment and appear much lighter in colour in IR camera image.

Inside a building, when your skin temperature is higher than the ambient temperature, you radiate more heat into the environment than you gain from it, but when it is lower, the reverse is true. Outside on a clear day, short wave radiation, $R_S$, coming from the Sun can be as high as 1 kW/m2 for an area facing the Sun directly[note]/article/useful-numbers[/note].

Example: Surviving 54C

Australian map with new colours

Fig.2. Australia’s 2013 heatwave saw temperatures soar into the low 50s, prompting the country’s meteorologists to extend their temperature colour index[note]McGrath, Matt. (2013, January 9). Retrieved January 10, 2013, from BBC News: http://www.bbc.co.uk/news/science-environment-20958060[/note].

In the last decade temperatures of 54 C have been experienced in Iraq, Australia, Pakistan.  What does this do to the radiation balance of a human body? Let’s assume you are inside. For $T_{skin}$ = 306 K and $T_{env}$ = 327 K,  $R_L$ ≈ 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:

$$R_S = (0.5)(1000 \textnormal{ W/m}^2)(0.5 \textnormal{ m}^2) = 250 \textnormal{ W}\tag{7}$$

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.

Fig.3. Visible light and IR images of a person standing in bright sunlight on hot concrete.

In Fig.3, the surface temperature of the person’s clothing is ~29 C, but the surface temperature of the concrete is 38 C, 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 higher kinetic energy (and temperature) evaporate, leaving behind those with lower kinetic energy, cooling the grass

3. 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:

$$C\approx{K}_{C}A(T_{skin} – T_{env})\tag{8}$$

Where $A$ is the surface area of the body,  $T_{skin}$ is the skin temperature, $T_{env}$ is the temperature of the environment, and $K_C$ is an empirical coefficient that depends on the geometry of the human body. It varies as the speed of the air varies[note]Monteith, J., & Unsworth, H. M. (2008). Principles of Environmental Physics. Amsterdam: Academic Press.[/note] as listed in Table 2.

Table 2. Convection coefficient for the human body in different airspeeds.

We see that $K_C$ is dependent on the speed of moving air, and thus the power transferred is going to change with a change in the airspeed. 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 the 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 are insulated by the boundary layer, a thin layer of stationary (relative to the body) air surrounding us. As the wind blows, it takes this protective layer away, and we lose heat faster. [note]Environment Canada. (2010, September 09). Canada’s Wind Chill Index. Retrieved September 21, 2010, from Environment Canada: http://www.ec.gc.ca/meteo-weather/default.asp?lang=n&n=5FBF816A-1[/note] Note how quickly $K_C$ rises with airspeed; 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 breathing[note]Monteith, J., & Unsworth, H. M. (2008). Principles of Environmental Physics. Amsterdam: Academic Press.[/note](see next section).

Example: 54C

But what if the ambient temperature is 54 C?. 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 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:

$$\lambda\approx (\dfrac{2.4\textnormal{MJ}}{\textnormal{L}})(\dfrac{1.5\textnormal{ L}}{\textnormal{h}})(\dfrac{1\textnormal{ h}}{3600\textnormal { s}}) \approx 1000\textnormal { W}\tag{9}$$

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 [note]Antarctic Connection. (n.d.). Antarctic Weather. Retrieved 2012.09.27, from Antarctic Connection: http://www.antarcticconnection.com/shopcontent.asp?type=weather-main[/note], where the air is too cold to hold any water. However we tend not to perspire much in Antarctica, otherwise, with no barrier to sweating, we would dehydrate very 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 days with the same ambient temperature[note]Heckert, P. A. (2008, June 10). Suite101. Retrieved September 21, 2010, from Physics of Sweating: http://www.suite101.com/content/physics-of-sweating-a56641[/note]. 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.

Fifty years ago, it was common to talk of the Wet-Bulb Temperature[note]https://en.wikipedia.org/wiki/Wet-bulb_temperature.[/note], which could be measured with a wet “sock” placed on the bulb of a thermometer in a strong airflow. The wet-bulb temperature was a measure of human comfort in a given temperature and humidity. In recent years this measure has got lost in a rash of more sophisticated indices like humidex, heat index, and “feels like” values. However as concern grows over the impact of global heating on human health[note]https://en.wikipedia.org/wiki/Wet-bulb_temperature#Wet-bulb_temperature_and_health.[/note], the wet-bulb temperature is making a comeback, as it is a simple measure of survivability, with 35 C (wet) being a hard limit.

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 material[note]Answers.com. (n.d.). Conduction. Retrieved September 21, 2010, from Answers.com: http://www.answers.com/topic/heat-conduction[/note].

Conduction losses are found using the following formula:

$$G = \dfrac{kA({T}_{skin} – {T}_{env})}{d}\tag{10}$$

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), $T_{skin}$ is the temperature of the skin and $T_{env}$ is the temperature of the environment.  Since we only require the temperature difference, we can use either K (kelvin) or C (celsius).

Example: flip-flops

For a rubber (polyurethane) flip flop:

A &=& (30\textnormal{ cm})(20\textnormal{ cm}) = 0.06 \textnormal{ m}^2\nonumber\tag{11}\\
k &=& 0.02 \textnormal {W/mK}\nonumber\tag{12}\\
d &=& 1 \textnormal{ cm} = 0.01 \textnormal{ m}\nonumber\tag{13}\\
{T}_{skin} &=& 306\textnormal { K}\nonumber\tag{14}\\
{T}_{material} &=& 293\textnormal { K}\nonumber\tag{15}

Thus, the conduction is found as:

G &=& \dfrac{(0.02)(0.06)(306 – 293)}{0.01}\nonumber &=& 1.56 \textnormal{ W} \nonumber \tag{16}

A total loss of 3.12 W (1.56 W/foot times two feet) is small, much smaller than that of all the other processes considered. (The feet here are the biological kind, not the measure of length).

Conduction losses, however, aren’t always so small: misbehaving prisoners in the famous Alcatraz prison [note]https://en.wikipedia.org/wiki/Alcatraz_Island[/note] 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)[note]Engineering Toolbox. (n.d.). Thermal Conductivity of Some Common Materials. Retrieved September 21, 2010, from Engineering Toolbox: http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html[/note] and so if the prisoners lay down on the floor, their large surface area and large thickness, together with the high $K_C$ meant that conduction losses were very high – between 80 W – 400 W during 0 C 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[note]Personal communication with U.S. National Parks Service guide during a visit in 2008.[/note].


Updated 2019-09-09