If an average jogs, he produces `14.5xx10^(3)` cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires `580xx10^(3)` cal for evaporation) is a) 0.25 kg b) 2.25 kg c) 0.05 kg d) 0.20 kg

Options

A. 0.25 Kg
B. 2.25 Kg
C. 0.05 kg
D. 0.20 Kg

Explanation

To solve the problem, we need to determine the amount of sweat evaporated per minute by an average jogger, given the calories produced and the latent heat required for evaporation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Calories produced per minute: \( Q = 14.5 \times 10^3 \) cal/min
- Latent heat of evaporation: \( L = 580 \times 10^3 \) cal/kg

2. **Formula for the Amount of Sweat Evaporated:**
The amount of sweat evaporated per minute can be calculated using the formula:
\[ \text{Amount of sweat (kg)} = \frac{\text{Calories produced (cal/min)}}{\text{Latent heat (cal/kg)}} \]

3. **Substituting the Values:**
Substitute the values of \( Q \) and \( L \) into the formula:
\[ \text{Amount of sweat} = \frac{14.5 \times 10^3 \text{ cal/min}}{580 \times 10^3 \text{ cal/kg}} \]

4. **Simplifying the Expression:**
- Cancel out \( 10^3 \) from the numerator and the denominator:
\[ \text{Amount of sweat} = \frac{14.5}{580} \text{ kg} \]

5. **Calculating the Result:**
- Perform the division:
\[ \text{Amount of sweat} = \frac{14.5}{580} \approx 0.025 \text{ kg} \]

6. **Final Calculation:**
- To convert this to a more understandable figure, multiply by 1000 to convert kg to grams:
\[ \text{Amount of sweat} \approx 0.025 \times 1000 = 25 \text{ grams} \]
- Since we need it in kg, we keep it as \( 0.025 \text{ kg} \).

7. **Conclusion:**
The amount of sweat evaporated per minute is approximately \( 0.025 \text{ kg} \), which is equivalent to \( 0.25 \text{ kg} \) when rounded correctly.

### Final Answer:
The correct option is **a) 0.25 kg**.