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11.11: Cold Weather Survival Time

  • Page ID
    17812
  • Stages of Hypothermia
    Stage Core Body Temperature °C Symptoms
    Mild Hypothermia 35°-33° shivering, poor judgment, amnesia and apathy, increased heart and respiratory rate, cold and/or pale skin
    Moderate Hypothermia 32.9°-27° progressively decreasing levels of consciousness, stupor, shivering stops, decreased heart and respiratory rate, decreased reflex and voluntary motion, paradoxical undressing.
    Severe Hypothermia < 26.9° low blood pressure and bradycardia, no reflex, loss of consciousness, coma, death

    [1]

    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 heat loss rate due to thermal radiation and a 1100 W heat loss rate due to forced convection. We also found that using a space blanket to reduce the wind chill and thermal radiation would leave only 160 W of heat loss rate rate due to conduction across the clothing. The typical person has a thermal power of 100 W, but shivering can increase that to 250 W.[2] It would be interesting to know, given these values for thermal power and heat loss rate, how quickly body temperature would actually change. In order to answer that question we need to learn about specific heat and heat capacity.

    Human Heat Capacity

    In the fight to maintain body temperature the human body gets assistance from the fact that the body is primarily made of water. The amount of thermal energy required to change the temperature of the body is relatively high compared to other objects of the same mass because water has a very high specific heat (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 heat capacity (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 heat capacity 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 mass (m) by the specific heat (c) to get heat capacity, which we then we multiply by a dangerous temperature change (quicklatex.com-758dacfb9b0dd9b2c168815956661a6b_l3.png) to get the heat required (Q). This entire process can be summed up by the equation:

    (1) quicklatex.com-1e6d611957572c27ee1fca69555d05fb_l3.png

    Notice that if the temperature is dropping then final temperature is less than the initial temperature so quicklatex.com-4ec078a94a147800b6390c92e7d49dd5_l3.png will be negative, which makes heat negative, indicating that thermal energy is leaving the object. The equation works for heating as well as cooling because in that case quicklatex.com-4ec078a94a147800b6390c92e7d49dd5_l3.png and quicklatex.com-56bc4aea76e43e57eecebbce9b11204b_l3.png will be positive indicating that thermal energy is entering the material.

    Reinforcement Exercises

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

    quicklatex.com-5dc9c906246f8983f77030f9c0b27ff0_l3.png. Assume the water heater is 100% efficient at converting electrical energy into thermal energy and transferring that to the water as heat.
    Figure: One bath per day would cost how much each month?

    The following chart provides specific heat 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
    Liquids
    Benzene 1740 0.415
    Ethanol 2450 0.586
    Glycerin 2410 0.576
    Mercury 139 0.0333
    Water (15.0 °C) 4186 1.000
    Gases
    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)
    [3]

    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 hypothermia at 33 °C in the example survival situations discuss so far in this unit. Let’s use relatively common human mass of about 80 kg and the average specific heat of human tissue of 3470 J/(kg C°). [4]First we find the heat loss required for the temperature drop:

    quicklatex.com-5c0e4b9ba0bb84bc918cc9500c1b042b_l3.png

    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 thermal power would be roughly 250 W. This person would have a 1150 W thermal power deficitm meaning that they lose 1150 Joules 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:

    s}} \approx 900 \,\bold{s} \end{equation*}

    Dividing the 900 s by 60 (s/min), we see that hypothermia would be reached in only 14 minutes. In reality, the rate of heat loss depends on the temperature 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 approximation 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.

    Exercises

    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 thermal energy 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 hypothermia 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 heat transfer rate for the initial body temperature due to both thermal radiation and forced convection
    2. using that heat transfer rate to calculate the amount of heat transferred over a relatively short time interval
    3. using the amount of heat transferred to calculate a resulting reduction in body temperature
    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 hypothermia 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 thermal power 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.

    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 Venosa, Medical Daily and “Frozen Alive” by Peter Stark, Outside 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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