Question

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

Answer 1

Answer:

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