By the First Law of Thermodynamics, energy can neither be created nor destroyed (conservation of energy). Thus, energy is transferred from one part of the system to another and the total energy of a system (Q) is unchanged:
ΔQsystem = ΔQbody + ΔQwater = 0
Because Q = (mass)(specific heat)(change in temperature), then:
[(mass)(specific heat)(ΔT)]body + [(mass)(specific heat)(ΔT)]water = 0
Now plugging in the variables:
[(75.0 kg)(3470 J/kg°C)(37°C - 34°C)] + [(kg of water)(4186 J/kg°C)(37°C - 60°C)] = 0
Rearranging:
(kg of water)(4186 J/kg°C)(37°C - 60°C) = -(75.0 kg)(3470 J/kg°C)(37°C - 34°C)
And:
kg of water = -(75.0 kg)(3470 J/kg°C)(3°C) / (4186 J/kg°C)(-23°C) = 8.1 kg of water
Converting to liters of water (density of water = 1.0 kg/liter):
(8.1 kg) ÷ (1 kg/liter) = 8.1 liters of water
In order to raise core temperature to normal, this individual would theoretically have to drink 8.1 liters (2.1 gallons) in a short period of time. This assumes that there is no transfer of heat by radiation, conduction, evaporation, respiration, or convection, and no significant thermogenesis from shivering. One liter of a hot drink would be ambitious for most people, and would raise this individual’s temperature about 0.4°C. The following table shows the amount of 60°C water theoretically needed to normalize the body temperature in individuals of varying weights and degrees of hypothermia.