Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to 34.0 degree Celsius overnight and rise to 40.0 degree Celsius during the day. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/(kg⋅K). The heat of vaporization of water at 34∘C is 2.42 × 106 J/kg.)

To see how effective this mechanism is for saving water, calculate how many liters of water a 400 kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0 degree Celsius by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0 degree Celsius.

First, the energy involved with the body temperature change of 34.0C to 40.0C is calculated. (Because a change in temperature is present in the equation, it is not necessary to convert °C to K for the units to properly cancel).

Q = mcΔt = (400 kg)(3480Jkg⁻¹K⁻¹)(40.0C - 34.0C) = 8.352 x 10⁶ J

In order to keep the temperature constant and eliminate this amount of heat through sweating, a mass of water M must be drunk and then evaporated by sweating:

Q = ML >> M = Q/L

M = (8.352 x 10⁶ J) / (2.42 x 10⁶ J/kg) = 3.45 kg

The mass of water is then converted to liters using the density of 1.0g/mL or 1.0 kg/L.

(3.4512 kg) / (1.0 kg/L) = 3.45 L