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29 June, 02:03

A 68-kg man whose average body temperature is 39°C drinks 1 L of cold water at 3°C in an effort to cool down. Taking the average specific heat of the human body to be 3.6 kJ/kg°C, determine the drop in the average body temperature of this person under the influence of the cold water

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  1. 29 June, 02:27
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    The man's average body temp. will fall by 0.6°C to 38.4°C

    Explanation:

    The enthalpy (heat content) of the water, using a datum of 0°C, is

    Hw = Mw kg x Cp, w (specific heat capacity kJ/kg °C) x Tw °C

    = 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ

    Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ

    So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.

    But Hman (after drink) has mass 68 + 1 = 69 kg

    Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C

    So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew

    Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C
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