Question

A child running a temperature of 101° F is given an antipyrine, 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 the temperature is about 580 cal g⁻¹.

Answer 1

Answer:

4.31 g/min

Step by step explanation:

The givens are:

• The change in temperature $\Delta T=101-98=3^0F=3*\frac{5}{9}=\frac{5}{3}^0C$
• the mass 'm' = $30\:kg=30*10^3\:g$
• Specific heat of human body 'c' = $1\:cal\:g^{-1}\:^0C^{-1}$

Heat lost by child in 20 min = $mc\Delta T$$=30*10^3*1*\frac{5}{3}=50000\:cal$

Therefore, heat lost in 1 min = $\frac{50000}{20}=2500\:cal/min$

Therefore, average rate of extra evaporation = $\frac{2500}{580}=4.31\:g/min$

Answer 2

Answer:

See the images attached...and get the solution