convection\nthe average person produces heat at the rate of about p = 120 w when at rest.\nat what rate must…

convection\nthe average person produces heat at the rate of about p = 120 w when at rest.\nat what rate must water evaporate from the body to get rid of all this energy?\n(for simplicity we assume this evaporation occurs when a person is sitting in\nthe shade and surrounding temperature are the same as skin temperature,\neliminating heat transfer by other methods)\nlatent heat of vaporization, lv = 334*10³ j/kg\np = φ/t = 120 w = 120 j/s\nφ = mlv\nm·lv = 120 w\nm/t = 120 w/lv = 120 j/s / 334×10³ j/kg\nm/t = ( ? ) kg/s

convection\nthe average person produces heat at the rate of about p = 120 w when at rest.\nat what rate must water evaporate from the body to get rid of all this energy?\n(for simplicity we assume this evaporation occurs when a person is sitting in\nthe shade and surrounding temperature are the same as skin temperature,\neliminating heat transfer by other methods)\nlatent heat of vaporization, lv = 334*10³ j/kg\np = φ/t = 120 w = 120 j/s\nφ = mlv\nm·lv = 120 w\nm/t = 120 w/lv = 120 j/s / 334×10³ j/kg\nm/t = ( ? ) kg/s

Answer

Explanation:

Step1: Recall power - heat relationship

Power $P=\frac{Q}{t}$, and heat for vaporization $Q = mL_v$. So $P=\frac{mL_v}{t}$.

Step2: Solve for mass - flow rate

We want to find $\frac{m}{t}$. Rearranging the formula $\frac{m}{t}=\frac{P}{L_v}$. Given $P = 120\ W=120\ J/s$ and $L_v=334\times10^{3}\ J/kg$. Then $\frac{m}{t}=\frac{120\ J/s}{334\times10^{3}\ J/kg}$.

Step3: Calculate the result

$\frac{m}{t}=\frac{120}{334\times10^{3}}\ kg/s\approx 3.59\times 10^{-4}\ kg/s$.

Answer:

$3.59\times 10^{-4}\ kg/s$