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16 January, 12:32

Estimate the mass of water that must evaporate from the skin to cool the body by 0.45 ∘C. Assume a body mass of 95 kg and assume that the specific heat capacity of the body is 4.0 J/g⋅∘C.

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  1. 16 January, 13:40
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    0.0757 kg

    Explanation:

    m = mass of the body = 95 kg = 95000 g

    ΔT = change in the temperature of the body = 0.45 °C

    c = specific heat capacity of the body = 4.0 J / (g °C)

    M = mass of water evaporated

    L = latent heat of vaporization of water = 2260 J/g

    Using conservation of heat

    Heat gained by water = heat lost by body

    ML = m c ΔT

    M (2260) = (95000) (4.0) (0.45)

    M = 75.7 g

    M = 0.0757 kg
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