Question

If an average person jogs, he produces $14.5\times 10^4\, cal \,min^{-1}$. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1 \,kg$ requires $580 \times 10^3 \,cal$ for evaporation) is

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

Answer 1

Answer: (A)

Given,
Calories produced per minute $=14.5 \times 10^{3} cal / min$
Latent heat $=580 \times 10^{3} cal / kg$
Amount of sweat evaporated per minute
$=\frac{\text { Sweat produces } / \text { minute }}{\text { Number of calories required for evaporation } / kg }=$
$\frac{\text { C alories produced (heat produced) / minute }}{\text { Latent heat(in cal } / kg \text { ) }}$
$=\frac{14.5 \times 10^{3}}{580 \times 10^{3}}=\frac{145}{580}=0.25 \,kg$