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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 ______.
Options
0.025 kg
2.25 kg
0.05 kg
0.20 kg
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Solution
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
