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.
The average rate of extra evaporation caused by the drug is approximately 4.31 g/min.
Explanation
This problem applies the principle of heat transfer during evaporation. The heat lost by the child's body due to temperature decrease equals the heat required to evaporate the sweat. We equate the heat lost (using Q = msΔT) to the heat absorbed during evaporation (using Q = mL) to find the mass of sweat evaporated, then divide by time to get the rate.
Solution Steps
Step 1: Calculate the temperature change in Celsius Temperature change = 101°F - 98°F = 3°F Converting to Celsius: ΔT = 3 × (5/9) = 5/3 °C 1.67°C
Step 2: Calculate the heat lost by the body Mass of child, m = 30 kg = 30,000 g Specific heat of human body specific heat of water = 1 cal g⁻¹ °C⁻¹ Heat lost, Q = m × s × ΔT Q = 30, × (5/3) = 50,000 cal
Step 3: Calculate the mass of sweat evaporated Latent heat of evaporation, L = 580 cal g⁻¹ Using Q = m_evap × L 50,000 = m_evap × 580 m_evap = 50,000/580 = 86.21 g
Step 4: Calculate the average rate of evaporation Time = 20 minutes Rate of evaporation = m_evap / time Rate = 86.21/20 = 4.31 g/min