Answer :

Given,

Mass of the child, m = 30kg


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 = 3°F


ΔT = [3 × (5/9)] °C


ΔT = 1.667 °C


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-1 °C-1


Latent heat of evaporation of water, L = 580 cal g–1


The heat lost by the child is given as


∆θ = mc∆T


∆θ = 30 kg × 1000 cal kg-1 °C-1 × 1.667 °C = 50000 cal


Let m1 be the mass of the water evaporated from the child’s body in 20 min.


Loss of heat through water is given by


∆θ = m1L


m1 = ∆θ/L


m1 = (50000 cal)/(580 cal g–1) = 86.2 g


Hence, average rate of extra evaporation caused by the drug, Rave = m1/t


Rave = (86.2 g)/(200 min)


Rave = 4.3 g/min.


NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.


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