A child running a temperature of `101^(@)F` is given an antipyrine (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 water at that temperature is about `580 cal//g`.

Options

A. `4.3 g//min`
B. `6.2 g//min`
C. `3.6 g//min`
D. `2.4 g//min`

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step-by-Step Solution:

1. **Determine the Temperature Change:**
- The initial temperature of the child is \( 101^\circ F \) and the final temperature is \( 98^\circ F \).
- The change in temperature \( \Delta T \) is:
\[ \Delta T = 101^\circ F - 98^\circ F = 3^\circ F \]

2. **Convert Fahrenheit to Celsius:**
- To convert the temperature change from Fahrenheit to Celsius, we use the conversion formula:
\[ \Delta T_{C} = \frac{5}{9} \Delta T_{F} \]
- Therefore,
\[ \Delta T_{C} = \frac{5}{9} \times 3 = \frac{15}{9} = \frac{5}{3} \text{ degrees Celsius} \]

3. **Calculate the Heat Loss:**
- The mass of the child is \( 30 \, \text{kg} \), which is \( 30,000 \, \text{g} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \)).
- The specific heat capacity of the human body (approximated as water) is \( 1 \, \text{cal/g}^\circ C \).
- The heat loss \( Q \) can be calculated using the formula:
\[ Q = m \cdot C \cdot \Delta T \]
- Substituting the values:
\[ Q = 30,000 \, \text{g} \times 1 \, \text{cal/g}^\circ C \times \frac{5}{3} \, \text{degrees Celsius} \]
- Therefore,
\[ Q = 30,000 \times \frac{5}{3} = 50,000 \, \text{cal} \]

4. **Relate Heat Loss to Evaporation:**
- The heat lost by the body is gained by the sweat evaporating. The heat gained by evaporation can be expressed as:
\[ Q = m \cdot L \]
- Where \( L \) is the latent heat of vaporization of water, which is \( 580 \, \text{cal/g} \).
- Setting the heat loss equal to the heat gained:
\[ 50,000 \, \text{cal} = m \cdot 580 \, \text{cal/g} \]
- Solving for \( m \):
\[ m = \frac{50,000}{580} \approx 86.2 \, \text{g} \]

5. **Calculate the Average Rate of Evaporation:**
- The time taken for this evaporation is \( 20 \, \text{minutes} \).
- The average rate of evaporation \( R \) is given by:
\[ R = \frac{m}{\text{time}} = \frac{86.2 \, \text{g}}{20 \, \text{min}} \approx 4.31 \, \text{g/min} \]

### Final Answer:
The average rate of extra evaporation caused by the drug is approximately \( 4.3 \, \text{g/min} \).

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