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A leaf of area $\SI{40}{\centi\meter\squared}$ and mass $\SI{4.5e4}{kg}$ directly faces the Sun on a clear day. The leaf has an emissivity of 0.85 and a specific heat of $\SI{0.80}{kcal\per\kilo\gram\per\kelvin}$. \begin{abcliste} \abc Estimate the rate of rise of the leaf's temperature. \abc Calculate the temperature the leaf would reach if it lost all its heat by radiation to the surroundings at \SI{20}{\degreeCelsius}. \abc In what other ways can the heat be dissipated by the leaf? \end{abcliste}
\begin{abcliste} \abc The heat (energy) absorbed by the leaf is: \begin{align} P &= \epsilon \cdot \tilde P \cdot A\\ &= 0.85 \cdot \SI{1367}{\watt\per\meter\squared} \cdot \SI{4e3}{\meter\squared}\\ &= \SI{4.648}{W} \end{align} If we use $Q=cm\Delta\theta$ and $P=\frac{\Delta Q}{\Delta t}$, we can calculate the rate, at which the temperature rises: \begin{align} \frac{\Delta\theta}{\Delta t} &= \frac{P}{cm}\\ &= \frac{\SI{4.648}{W}}{\SI{3.346e3}{\joule\per\kilo\gram\per\kelvin}\cdot \SI{4.5e4}{kg}}\\ &= \SI{3.087}{\kelvin\per\second} \end{align} The temperatur of the leave would rise by \SI{3}{\degreeCelsius} each second  if there were no cooling effects. \abc  \abc The leaf can, like the human body, dissipate heat by vapourizing its own water. \end{abcliste}
14:17, 12. June 2017  si  Urs Zellweger (urs)  Current Version 