Ncert Exercises & Solutions

Question 11.18:
A child running at 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 the human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal gm $^{- 1}$ .

The initial and final temperatures of the boy are given in the question.

Change in temperature, ΔT = °C $[(100 - 98) \times \frac{5}{9}] {}_{o}C$
Specific heat of the human body = Specific heat of water = 1000 cal/kg °C
Latent heat of evaporation of water, L = 580 cal gm $^{- 1}$

The heat lost by the child is given by,

$∆ θ = mc ∆ T$
$= 30 \times 1000 \times (101 - 98) \times \frac{9}{5}$
$= 50000 cal$

Let $m_{1}$ be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water:

$∆ θ = mL$
$∴ m_{1} = \frac{∆ θ}{L} = \frac{50000}{580} = 86.2 g$

$The average rate of extra evaporation caused by the drug = \frac{m_{1}}{t} = \frac{86.2}{200} = 4.3 g / \min$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions