The latent heat of vaporization of H 2 O at bodytemperature (37°C) is 2.42 x 10

Question

The latent heat of vaporization of H2O at bodytemperature (37°C) is 2.42 x 106 J/kg. To cool thebody of a 73.9-kg jogger [average specific heat capacity = 3500J/(kg C°)] by 1.13 C°, how many kilograms of water in theform of sweat have to be evaporated? Please help!! The latent heat of vaporization of H2O at bodytemperature (37°C) is 2.42 x 106 J/kg. To cool thebody of a 73.9-kg jogger [average specific heat capacity = 3500J/(kg C°)] by 1.13 C°, how many kilograms of water in theform of sweat have to be evaporated? Please help!!

Explanation / Answer

The latent heat of vaporization of H2O at bodytemperature (37°C) is L = 2.42 x 106 J/kg
mass of the body m = 73.9-kg average specific heat capacity C = 3500 J/(kg C°) temp difference dt = 1.13 C° Heat lost due to this temp difference Q = m C dt                                                            = 292274.5 J So, mass of water in the form of sweat have to beevaporated is M = Q / L                                                                                                        = 0.1207 kg

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