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

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 find the amount of sweat evaporated per minute when an average jogger produces a certain amount of calories, which is removed by the evaporation of sweat.

### Step-by-Step Solution:

1. **Identify the energy produced by the jogger**:
The jogger produces \( 14.5 \times 10^4 \) calories per minute.

2. **Identify the energy required for evaporation**:
It is given that to evaporate 1 kg of sweat, \( 580 \times 10^3 \) calories are required.

3. **Set up the relationship**:
The amount of sweat evaporated (in kg) can be calculated using the formula:
\[ \text{Amount of sweat (kg)} = \frac{\text{Energy produced (cal/min)}}{\text{Energy required for evaporation (cal/kg)}} \]

4. **Substitute the values into the formula**:
\[ \text{Amount of sweat (kg)} = \frac{14.5 \times 10^4 \text{ cal/min}}{580 \times 10^3 \text{ cal/kg}} \]

5. **Perform the calculation**:
- First, simplify the expression:
\[ \text{Amount of sweat (kg)} = \frac{14.5 \times 10^4}{580 \times 10^3} \]
- This can be simplified further:
\[ = \frac{14.5}{580} \times \frac{10^4}{10^3} = \frac{14.5}{580} \times 10^1 \]
- Calculate \( \frac{14.5}{580} \):
\[ \approx 0.025 \]
- Now multiply by \( 10^1 \):
\[ 0.025 \times 10 = 0.25 \text{ kg} \]

6. **Conclude the result**:
Therefore, the amount of sweat evaporated per minute is \( 0.25 \) kg.

### Final Answer:
The amount of sweat evaporated per minute is \( 0.25 \) kg.