100 Cold Weather Survival Time

Stages of Hypothermia
Stage Core Body Temperature °C Symptoms
Mild 35°-33° shivering, poor judgment, amnesia and apathy, increased heart and respiratory rate, cold and/or pale skin
Moderate 32.9°-27° progressively decreasing levels of consciousness, stupor, shivering stops, decreased heart and respiratory rate, decreased reflex and voluntary motion, paradoxical undressing.
Severe < 26.9° low blood pressure and bradycardia, no reflex, loss of consciousness, coma, death


Throughout this unit we have analyzed the rate of heat loss during a cold weather survival situation in which a person is wearing a single layer of thin clothing against an 10 mph wind and a -3 °C air temperature. Our analysis shows that collectively the person would experience a 200 W due to and a 1100 W heat loss rate due to . We also found that using a space blanket to reduce the and would leave only 160 W of rate due to across the clothing.  The typical person has a of 100 W, but shivering can increase that to 250 W.[2] It would be interesting to know, given these values for and , how quickly body temperature would actually change. In order to answer that question we need to learn about and .

Human Heat Capacity

In the fight to maintain body the human body gets assistance from the  fact that the body is primarily made of water. The amount of required to change the temperature of the body is relatively high compared to other objects of the same because water has a very high (c). Specific heat is a material property that defines the amount of thermal energy removed from one unit mass of the material when it’s temperature changes by one unit of temperature. For example, water has a specific heat of 4186 J/(kg C°) so  4186 J of thermal energy must be removed from 1 kg of water in order for the temperature to drop by 1 C°. Multiplying the specific heat of a material by the mass of the material gives the  (C) of the object. For example, the heat capacity for 80 kg of water would be:  4186 J/(kg C°) x 80 kg = 334,880 J/(C°), meaning that 334,880 J must be removed to drop the temperature of 80 kg of water by 1 C°.

Reinforcement Exercises

Calculate the heat capacity of  150 kg of water.

Now that we know how to calculate we are ready to calculate the amount of energy required to change the temperature of the body by a dangerous amount. We just multiply the (m) by the (c) to get , which we then we multiply by a dangerous temperature change (Delta T = T_f-T_i) to get the required (Q). This entire process can be summed up by the equation:

(1)   \begin{equation*} Q = mc\Delta T \end{equation*}

Notice that if the temperature is dropping then final temperature is less than the initial temperature so Delta T will be negative, which makes negative, indicating that is leaving the object. The equation works for heating as well as cooling because in that case Delta T and Q will be positive indicating that thermal energy is entering the material.

Reinforcement Exercises

Calculate the required to raise the temperature of 150 kg of water from 14 °C to 40 °C, as is necessary for a typical bath.

Calculate the cost of one bath when using an electric water heater and energy is sold at 12 cents per \bold{kW\cdot hr}. Assume the water heater is 100% at converting electrical energy into and transferring that to the water as .

One bath per day would cost how much each month?

The following chart provides values for various substances. Notice the relatively high specific heat of water.

Table of Specific Heat Values
Substances Specific heat (c)
Solids J/kg⋅ºC kcal/kg⋅ºC
Aluminum 900 0.215
Asbestos 800 0.19
Concrete, granite (average) 840 0.20
Copper 387 0.0924
Glass 840 0.20
Gold 129 0.0308
Human body (average at 37 °C) 3500 0.83
Ice (average, -50°C to 0°C) 2090 0.50
Iron, steel 452 0.108
Lead 128 0.0305
Silver 235 0.0562
Wood 1700 0.4
Benzene 1740 0.415
Ethanol 2450 0.586
Glycerin 2410 0.576
Mercury 139 0.0333
Water (15.0 °C) 4186 1.000
Air (dry) 721 (1015) 0.172 (0.242)
Ammonia 1670 (2190) 0.399 (0.523)
Carbon dioxide 638 (833) 0.152 (0.199)
Nitrogen 739 (1040) 0.177 (0.248)
Oxygen 651 (913) 0.156 (0.218)
Steam (100°C) 1520 (2020) 0.363 (0.482)

Everyday Examples: Cold Weather Survival Time

Applying the previous equation to the human body we can estimate how long it would take body temperature to drop from a normal 37 °C down to the edge of moderate at 33 °C in the example survival situations discuss so far in this unit. Let’s use relatively common human of about 80 kg and the average of human tissue of 3470 J/(kg C°). [4]First we find the loss required for the temperature drop:

    \begin{equation*} Q = mc\Delta T = (80 \,\bold{kg})\left(3470\,\frac{\bold{J}}{\bold{kg\, C^{\circ}}}\right)(33\,\bold{\, C^{\circ}} -37\,\bold{\, C^{\circ}}) \approx 1,000,000 \,\bold{J} \end{equation*}

We know that in our example of a 10 mph wind and a -3 °C air temperature we found that a person with thin clothing and no space blanket experienced 1100 W of convective heat loss and 200 W of radiative heat loss for a total of 1300 W.  If the person was shivering then their would be roughly 250 W. This person would have a 1150 W thermal power deficitm meaning that they lose 1150 of thermal energy each second. Dividing the dangerous heat loss we calculated above, by the thermal power deficit gives us the time required to lose that much heat:

    \begin{equation*} t =  \frac{1,000,000 \,\bold{J}}{1150 \,\bold{J/s}} \approx 900 \,\bold{s} \end{equation*}

Dividing the 900 by 60 (s/min), we see that would be reached in only 14 minutes.   In reality, the depends on the difference so the heat loss rate would not be constant, but instead it would slow a bit as body temperature decreased. In our example case the 40.9 difference between the body temperature and environment temperature changed by only  about 3 (seven percent). Ignoring this effect gives a reasonable for the time until moderate hypothermia. The next two sections will address this approximation and allow us to calculate the time required for more significant temperature changes.


We found previously that a person in wet cotton, but with a space blanket would experience an 1100 W heat loss rate, giving an 850 W deficit. Calculate the time to moderate hypothermia in this situation, in minutes.

Calculate the time in minutes to moderate hypothermia for an 80 kg person wearing wet down in the mountain climbing example at the start of this chapter, for which the rate of heat loss was 5600 W.

Calculate the time in minutes to reach severe hypothermia body temperature (see hypothermia chart) for a person in the same situation as the previous everyday example, but with a mass of only 60 kg.

Rates of Body Temperature Change

The rate at which is transferred out of the body depends on the difference in temperature between the body and the environment. As the body cools closer to the environmental temperature the rate will decrease. In the previous example we ignored this reality and assumed that the cooling rate was constant, which was reasonable because we only examined a very small temperature change. For our example situation we found that mild would be reached in only twenty minutes. The graph below was produced by calculating the body temperature while accounting for a cooling rate that depends on the temperature difference. This was done using a numerical model:

  1. calculating a for the initial body due to both and
  2. using that e to calculate the amount of transferred over a relatively short time interval
  3. using the amount of transferred to calculate a resulting reduction in body
  4. calculating a new body temperature by subtracting the reduction in body temperature
  5. repeat 1-5 until temperature was well beyond survival temperature, keeping track of the temperatures and times to make the graph

We see that the time to mild is actually more like 30 mins. We also see that severe hypothermia could set in after less than an hour and the minimum survivable temperature could be reached in under two hours. For the purpose of these calculations we assumed a of 250 W while shivering, and that shivering stops when the body reaches 30 °C, at which point thermal power returns to 100 W.  We also assumed the that thermal power dropped to zero when body temperature reached 21 °C.

Body temperature in degrees Celsius on the vertical axis and time in hours on the horizontal axis. Starting from normal body temperature, 37 C at time 0, the curve drops to 35 C and moderate hypothermia over about 1/3 hours, reaches 27 C and sever hypothermia after roughly 1 hour and hits the standard minimum survivable body temperature of 20 C at just after 1.5 hours. The record low survived body temperature of just under 15 C is reached in about 2.5 hours.
Predicted body temperature during a simplified example survival situation.

Approximate body temperature vs. time for an 80 kg person in 25 °F air temperature and 10 mph wind  (-3 °C , 4.5 m/s). For the purpose of these calculations we assumed a thermal power of 250 W while shivering, that shivering stops at 30 °C at which point thermal power returns to 100 W, and that thermal power dropped to zero below 21 °C. Minimum survivable body temperature and record low body temperature survived were found at “How Cold Can a Body Get” by Ali VenosaMedical Daily and “Frozen Alive” by Peter StarkOutside Magazine.

Modeling how body temperature changes with time under different insulation and temperature conditions allows forensic investigators to measure a body’s temperature and work backwards to determine the time at which body temperature started to drop from normal. In this way time of death can be determined (unless of course the person was hypothermic or hyperthermic before death).


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