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 minutes, 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.

Question

Explain why :

a body with large reflectivity is a poor emitter
a brass tumbler feels much colder than a wooden tray on a chilly day
an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
the earth without its atmosphere would be inhospitably cold
heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Answer 1

Given Data

Initial temperature	\(101^\circ\text{F}\)
Final temperature	\(98^\circ\text{F}\)
Time interval	\(20\ \text{min}\)
Mass of child, \(m\)	\(30\ \text{kg} = 3.0 \times 10^{4}\ \text{g}\)
Specific heat of body, \(c\)	\(\approx 1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}\)
Latent heat of evaporation, \(L\)	\(580\ \text{cal g}^{-1}\)

Step 1: Temperature Drop in °C

Temperature change in Fahrenheit: \[ \Delta T_F = 101 - 98 = 3^\circ\text{F} \] Convert to Celsius: \[ \Delta T_C = \frac{5}{9}\,\Delta T_F = \frac{5}{9} \times 3 = \frac{5}{3} \approx 1.67^\circ\text{C} \]

Step 2: Heat Lost by the Body

Heat lost: \[ Q = m c \Delta T \] \[ Q = (3.0 \times 10^{4}) \times 1 \times \frac{5}{3} = 5.0 \times 10^{4}\ \text{cal} \]

Step 3: Mass of Sweat Evaporated

This heat is used to evaporate sweat: \[ Q = m_{\text{evap}}\, L \] \[ 5.0 \times 10^{4} = m_{\text{evap}} \times 580 \] \[ m_{\text{evap}} = \frac{5.0 \times 10^{4}}{580} \approx 86.2\ \text{g} \]

Step 4: Average Rate of Extra Evaporation

Time \(t = 20\ \text{min}\) \[ \text{Rate} = \frac{m_{\text{evap}}}{t} = \frac{86.2\ \text{g}}{20\ \text{min}} \approx 4.3\ \text{g min}^{-1} \]

Final Answer

The average rate of extra evaporation of sweat caused by the drug is approximately \[ \boxed{4.3\ \text{g min}^{-1}}. \]

Solution and Explanation