Chapter 12, Problem 81A

To determine

The amount of perspiration must a player vaporize per hour.

Explanation of Solution

Given:

Thermal generated per second: $Q\text{'}=30\text{W= 30 J/s}$

Formula used:

Thermal energy for vaporization of perspiration:

  $Q=m{L}_v$

Here, m is the mass and $L_v$ is the latent hear of vaporization.

Calculations:

Latent heat of vaporization of water is $L{v}_{}=2260\text{kJ/kg}$ .

Time, $t=1\text{h}=60\times 60\text{s}$

  $\begin{array}{c}Q\text{'}=\frac{m}{t}{L}_v\\ m=\frac{\text{30}\text{W}}{\text{2260}\text{kJ/kg}}\text{×}\left(\text{60×60}\right)\\ =0.0478\text{kg}\\ =47.8\text{g}\end{array}$

Conclusion:

Thus, the amount of perspiration must a player vaporize per hour is 47.8 g.