Meta Information  Exercise contained in  Rate this Exercise  



0 
During exercise, a person may give off $\SI{180}{kcal}$ of heat in $\SI{30}{min}$ by evaporation of water from the skin. How much water has been lost?
$\SI{300}{g}$
Since evaporation to the environment occurs slowly, it is enough to consider the heat of evaporation, i.e. \begin{align} Q &= c \cdot m \cdot \Delta\theta + m\cdot L_v\\ m &= \frac{Q}{c \cdot \Delta \theta + L_v}\\ &= \frac{\SI{7.528e5}{J}}{\SI{4182}{\joule\per\kilo\gram\per\kelvin} \cdot \SI{63}{\degreeCelsius} + \SI{22.57e5}{\joule\per\kilo\gram}}\\ &= \SI{0.299}{kg} \end{align} will be evaporated during the $\SI{30}{min}$ exercise. Here we calculated the given unit of the energy $Q$ into SI units: \begin{align} \SI{180}{kcal} &= \SI{1.8e5}{cal}\\ &= \numprint{1.8e5} \cdot \SI{4.182}{J}\\ &= \SI{7.528e5}{J} \end{align}
13:59, 13. June 2018  diff  Urs Zellweger (urs)  Current Version 
07:57, 12. June 2017  lsg  Urs Zellweger (urs)  Compare with Current 
07:55, 12. June 2017  si  Urs Zellweger (urs)  Compare with Current 