Question

If an average person jogs, he produces $14.5\times 103cal/min$. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1kg$ requires $580\times {10}^{3}cal$ for evaporation).

• A. $0.25kg$
• B. $2.25kg$
• C. $0.05kg$
• D. $0.20kg$

Answer 1

Answer: (A)

$0.25kg$

$580\times {10}^3$ calories are needed to convert $1kg{H}_{2}O$ into steam

$1cal$ will produce sweat $=\frac{1}{580\times {10}^3}$

$14.5\times {10}^{3}cal$ will produce sweat $=\frac{14.5\times {10}^3}{580\times {10}^3}$

$=\frac{145}{5800}kgperminute=0.025kgpermin$.