Question

During a brisk run, an adult human generates heat at a rate of about 1270 W. To remove this heat by evaporative cooling, what mass of water per second m/t must be evaporated from the body as sweat? The latent heat of vaporization of water at 37 °C (typical body temperature) is 24.2 × 105 J/kg.

Answer 1

Answer on Question #37951 – Physics – Thermodynamics

Question:

During a brisk run, an adult human generates heat at a rate of about 1270 W. To remove this heat by evaporative cooling, what mass of water per second m/t must be evaporated from the body as sweat? The latent heat of vaporization of water at $37{}^{\circ}\mathrm{C}$ (typical body temperature) is $24.2 \times 10^{5} \mathrm{~J} / \mathrm{kg}$.

Answer:

Amount of heat equals:

$$Q = L m$$

where $L$ is latent heat of vaporization of water, $m$ – mass of the water.

Dividing by time:

$$\frac {Q}{t} = P = L \left(\frac {m}{t}\right)$$

Therefore mass of water per second $\left(\frac{m}{t}\right)$ equals:

$$\frac {m}{t} = \frac {P}{L} = \frac {1270 \, W}{24.2 \times 10^{5} \, \frac {J}{kg}} = 5.25 \times 10^{-4} \, \frac {kg}{s}$$

Answer: $5.25 \times 10^{-4} \, \frac{kg}{s}$