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Thus, energy is transferred from one part of the system to another and the total energy of a system (Q) is unchanged: Qsystem = Qbody + Qwater = 0 Because Q = (mass)(specific heat)(change in temperature), then: [(mass)(specific heat)(T)]body + [(mass)(specific heat)(T)]water = 0 Now plugging in the variables: [(75.0 kg)(3470 J/kgC)(37C - 34C)] + [(kg of water)(4186 J/kgC)(37C - 60C)] = 0 Rearranging : (kg of water)(4186 J/kgC)(37C - 60C) = -(75.0 kg)(3470 J/kgC)(37C - 34C) And: kg of water = -(75.0 kg)(3470 J/kgC)(3C) / (4186 J/kgC)(-23C) = 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

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[DOI] [PubMed] [Google Scholar] [359].Vacca V, Marinelli S, Pieroni L, Urbani A, Luvisetto S, Pavone F, Higher pain perception and lack of recovery from neuropathic pain in females: A behavioural, immunohistochemical, and proteomic investigation on sex-related differences in mice, Pain 155(2) (2014) 388402