Initial temperature of the body of the child, T1 = 101°F
Final temperature of the body of the child, T2 = 98°F
Change in temperature, ΔT = [(101 - 98) x 5/9] oC
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 × 103 g
Specific heat of the human body = Specific heat of water = c
= 1000 cal/kg/ °C
Latent heat of evaporation of water, L = 580 cal g–1
The heat lost by the child is given as:
Let m1 be the mass of the water evaporated from the child’* body in 20 min.
Loss of heat through water is given by:
Average rate of extra evaporation caused by the **** = m1/t
= 86.2/200 = 4.3 g/min