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If an average person jogs, he produces $14.5 \times 10^3 \mathrm{cal} / \mathrm{min}$. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1 \mathrm{~kg}$ requires $580 \times 10^3$ cal for evaporation) is
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The correct answer is:
$0.025 \mathrm{~kg}$
$0.025 \mathrm{~kg}$
As we know that amount of sweat evaporated/minute
$$
\begin{aligned}
&=\frac{\text { Sweat produced } / \text { minute }}{\text { Number of calories required for evaporation } / \mathrm{kg}} \\
&=\frac{\text { Amount of heat produced per minute in jogging }}{\text { Latent heat }(\text { in cal } / \mathrm{kg} \text { ) }} \\
&\begin{array}{l}
580 \times 10^3 \text { calories are needed to convert } \\
1 \mathrm{~kg} \mathrm{H} \mathrm{H}_2 \mathrm{O} \text { into stream. } \\
1 \text { cal. will produce } \text { sweat }=1 \mathrm{~kg} / 580 \times 10^3 \\
14.5 \times 10^3 \text { cal will produce }(\mathrm{sweat})
\end{array} \\
&=\frac{14.5 \times 10^5}{580 \times 10^3} \mathrm{~kg}=\frac{145}{580} \mathrm{~kg} / \mathrm{m} \\
&=0.025 \mathrm{~kg} / \mathrm{min}
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{\text { Sweat produced } / \text { minute }}{\text { Number of calories required for evaporation } / \mathrm{kg}} \\
&=\frac{\text { Amount of heat produced per minute in jogging }}{\text { Latent heat }(\text { in cal } / \mathrm{kg} \text { ) }} \\
&\begin{array}{l}
580 \times 10^3 \text { calories are needed to convert } \\
1 \mathrm{~kg} \mathrm{H} \mathrm{H}_2 \mathrm{O} \text { into stream. } \\
1 \text { cal. will produce } \text { sweat }=1 \mathrm{~kg} / 580 \times 10^3 \\
14.5 \times 10^3 \text { cal will produce }(\mathrm{sweat})
\end{array} \\
&=\frac{14.5 \times 10^5}{580 \times 10^3} \mathrm{~kg}=\frac{145}{580} \mathrm{~kg} / \mathrm{m} \\
&=0.025 \mathrm{~kg} / \mathrm{min}
\end{aligned}
$$
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