A feverish person is running a temperature of `101^@F` and is given paracetamol to lower his temperature. Due to medicine, there is an increase in rate of evaporation of sweat from his body. The mass of the person is 50 kg and the specific heat of human body is approximately about `1 cal g^(-1) ""^@C^(-1)` . If the fever is brought down to `98^@F` in 25 mins, the average rate of extra evaporation caused by the medicine is R in g `"min"^(-1)` , then the value of 87R/100 is (Latent heat of water at that temperature=580 g/cal)

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Convert the mass of the person to grams
The mass of the person is given as 50 kg. Since we need to work in grams, we convert it:
\[ \text{Mass} = 50 \, \text{kg} = 50 \times 1000 \, \text{g} = 50000 \, \text{g} \]

**Hint:** Remember that 1 kg = 1000 g.

### Step 2: Calculate the change in temperature in Celsius
The initial temperature is \(101^\circ F\) and the final temperature is \(98^\circ F\). We first convert these temperatures to Celsius using the formula:
\[ T(°C) = \frac{5}{9}(T(°F) - 32) \]
Calculating for \(101^\circ F\):
\[ T_1 = \frac{5}{9}(101 - 32) = \frac{5}{9} \times 69 = \frac{345}{9} \approx 38.33^\circ C \]
Calculating for \(98^\circ F\):
\[ T_2 = \frac{5}{9}(98 - 32) = \frac{5}{9} \times 66 = \frac{330}{9} \approx 36.67^\circ C \]
Now, the change in temperature \(\Delta T\) is:
\[ \Delta T = T_1 - T_2 = 38.33 - 36.67 = 1.66^\circ C \]

**Hint:** Use the temperature conversion formula correctly to find the change in temperature.

### Step 3: Calculate the heat lost by the body
Using the formula for heat loss:
\[ Q = mc\Delta T \]
Where:
- \(m = 50000 \, \text{g}\)
- \(c = 1 \, \text{cal/g}^\circ C\)
- \(\Delta T = 1.66^\circ C\)

Substituting the values:
\[ Q = 50000 \times 1 \times 1.66 = 83000 \, \text{cal} \]

**Hint:** Remember that the specific heat capacity is given in cal/g°C, and ensure units are consistent.

### Step 4: Relate heat lost to the mass of sweat evaporated
The heat lost by the body is equal to the heat gained by the sweat evaporating:
\[ Q = mL \]
Where \(L\) is the latent heat of vaporization of water, given as \(580 \, \text{cal/g}\). Let \(m\) be the mass of sweat evaporated:
\[ 83000 = m \times 580 \]
Solving for \(m\):
\[ m = \frac{83000}{580} \approx 143.10 \, \text{g} \]

**Hint:** Use the relationship between heat lost by the body and heat gained by the evaporated sweat.

### Step 5: Calculate the rate of evaporation
The time taken for the evaporation is given as 25 minutes. The rate of evaporation \(R\) in g/min is:
\[ R = \frac{m}{t} = \frac{143.10}{25} \approx 5.724 \, \text{g/min} \]

**Hint:** Rate is calculated as total mass divided by time.

### Step 6: Calculate \( \frac{87R}{100} \)
Now we need to find \( \frac{87R}{100} \):
\[ \frac{87R}{100} = \frac{87 \times 5.724}{100} \approx 5.00 \]

**Hint:** Simply multiply \(R\) by 87 and then divide by 100 to find the final value.

### Final Answer
Thus, the value of \( \frac{87R}{100} \) is approximately \(5.00\).