Ncert Exercises & Solutions

If an average person jogs, he produces $14.5 \times10^3$ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1$ kg requires $580 \times10^3$ cal for evaporation) is:

1.	$0.25$ kg	2.	$0.50$ kg
3.	$0.025$ kg	4.	$0.20$ kg

1. Hint: Use the law of conservation of energy.

Step 1: Find the amount of sweat evaporated.

Amount of sweat evaporated / minute = \frac{Sweat produced / minute}{Number of calories required for evaporation / kg}

$Amount of sweat evaporated / minute = \frac{Sweat produced / minute}{Number of calories required for evaporation / kg}$

= \frac{Amount of heat produced per minute in jogging}{Latent heat (in cal / kg)}

$= \frac{Amount of heat produced per minute in jogging}{Latent heat (in cal / kg)}$

= \frac{14.5 \times 10^{3}}{580 \times 10^{3}} = \frac{14.5}{580} = 0.025 kg

$= \frac{14.5 \times 10^{3}}{580 \times 10^{3}} = \frac{14.5}{580} = 0.025 kg$

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