Question

If an average person jogs, hse produces 14.5 × 10³ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 10³ cal for evaparation) is ______.

पर्याय

• 0.025 kg

• 2.25 kg

• 0.05 kg

• 0.20 kg

Answer 1

If an average person jogs, hse produces 14.5 × 10³ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 10³ cal for evaparation) is 0.025 kg.

Explanation:

The rate of bum calories is equivalent to sweat produced. Then, the amount of sweat evaporated/minute

= `"Sweat produced/minute"/"Number of calories required for evaporation/kg"`

= `"Calories produced (heat produced) per minute"/"Latent heat (in cal/kg)"`

= `(14.5 xx 10^3)/(580 xx 10^3)`

= `14.5/580`

= 0.025 kg