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If an average person jogs, hse produces 14.5 × 10^3 cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute - Physics

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प्रश्न

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 ______.

विकल्प

  • 0.025 kg

  • 2.25 kg

  • 0.05 kg

  • 0.20 kg

MCQ
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उत्तर

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

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अध्याय 12: Thermodynamics - Exercises [पृष्ठ ८३]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 12 Thermodynamics
Exercises | Q 12.2 | पृष्ठ ८३
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