A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^–1.

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, R_ave = m₁/t

⇒ R_ave = (86.2 g)/(200 min)

⇒ R_ave = 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.