Question

In hot weather, a human body cools by evaporation of sweat. The amount of water that must evaporate to cool the body by \(1~^\circ\mathrm{C}\) is __________________________% of the body mass. (Round off to two decimal places) 
[Given: latent heat of vaporization of water \(L_v=2.25\times10^6~\mathrm{J\,kg^{-1}}\); specific heat capacities of body and water \(c=4.2\times10^3~\mathrm{J\,kg^{-1}\,K^{-1}}\).]

Hint:

A handy ratio: \(\dfrac{c}{L_v}\approx \dfrac{4.2\times10^3}{2.25\times10^6}\approx 1.87\times10^{-3}\) per °C \(\Rightarrow\) \(\approx 0.19%\) of body mass per °C must evaporate.

Answer 1

Correct Answer: 0.18

Step 1 (Energy balance for 1°C cooling).
Let \(M\) be body mass and \(m\) be evaporated sweat mass. The sensible heat removed from the body is \[ Q_{\text{out}} = c\,M\,\Delta T = (4.2\times10^3)\,M\,(1)=4.2\times10^3 M~\mathrm{J}. \]

Step 2 (Latent heat supplied by evaporation).
Evaporation removes \(Q_{\text{evap}}=m L_v\). Set \(Q_{\text{evap}}=Q_{\text{out}}\): \[ m=\frac{4.2\times10^3 M}{2.25\times10^6}=1.867\times10^{-3}\,M. \]

Step 3 (Convert to percent of body mass).
\[ \frac{m}{M}\times 100 = 0.1867%\ \Rightarrow\ \boxed{0.19%}. \]

Step 4 (Interpretation).
For a \(70~\mathrm{kg}\) person: \(0.001867\times 70 \approx 0.131~\mathrm{kg}\approx 131~\mathrm{mL}\) of sweat must evaporate to cool by \(1^\circ\mathrm{C}\). Final Answer:
\[ \boxed{0.19%} \]