If an average person jogs, he produces 14.5× 104 cal min-1. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaporation) is

Question

If an average person jogs, he produces 14.5× 104 cal min-1. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 103 cal for evaporation) is (A)  0.25 kg (B)  2.25 kg (C)  0.05 kg (D)  0.20 kg

Tardigrade Admin · Accepted Answer

Given, Calories produced per minute =14.5 × 103 cal / min Latent heat =580 × 103 cal / kg Amount of sweat evaporated per minute =( text Sweat produces / text minute / text Number of calories required for evaporation / kg )= ( text C alories produced (heat produced) / minute / text Latent heat(in cal / kg text ) ) =(14.5 × 103/580 × 103)=(145/580)=0.25 kg

Q. If an average person jogs, he produces $14.5 \times 1 0^{4} c a l mi n^{- 1}$ . This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1 k g$ requires $580 \times 1 0^{3} c a l$ for evaporation) is

$0.25 k g$

$2.25 k g$

$0.05 k g$

$0.20 k g$

Q. If an average person jogs, he produces 14.5×104calmin−1. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1kg requires 580×103cal for evaporation) is

0.25kg

2.25kg

0.05kg

0.20kg

Q. If an average person jogs, he produces $14.5 \times 1 0^{4} c a l mi n^{- 1}$ . This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1 k g$ requires $580 \times 1 0^{3} c a l$ for evaporation) is

$0.25 k g$

$2.25 k g$

$0.05 k g$

$0.20 k g$