Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - General Problems - Page 411: 66

Answer

$4.3\times 10^{-3}\;\rm kg/h$

Work Step by Step

In general, to maintain a constant temperature in the leaf, the net gained energy must be equal to the net lost energy at the same time. This means that the rate of energy absorbed by the leaf from the sun must be equal to the radiated (transmitted) energy by radiation and evaporation. We need to find the rate of losing water $\dfrac{\Delta m_{water}}{\Delta t}$ to maintain a constant temperature of $35^\circ \rm C$. Thus, $$\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm gained\;from \;sun}=\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm lost \;by\;radiatin}+\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm lost \;by\;evaporation}\tag 1$$ We know, from equation 14-8, that the rate of absorbing energy from the sun by an object that faces it is given by $$\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm gained\;from \;sun}=1000\epsilon A_{\text {faces sun}}\cos \theta$$ We also know that rate of losing energy by radiation is given by $$\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm lost \;by\;radiatin}= \epsilon \sigma A_{\text {radiates}}\left[T_1^4-T_2^4\right]$$ where $T_1$ is the temperature of the leaf and $T_2$ is the temperature of the environment surrounding it. And we know that the energy loss rate by evaporation is given by $$\left[\dfrac{\Delta Q}{\Delta t}\right]_{\rm lost \;by\;evaporation}=\dfrac{m_{water}L_{water}}{\Delta t}$$ where $L_{water}$ is the latent heat of evaporation. Plugging the last 3 formulas into (1) and then solving for $\dfrac{\Delta m_{water}}{\Delta t}$; $$1000\epsilon A_{\text {faces sun}}\cos \theta=\epsilon \sigma A_{\text {radiates}}\left[T_1^4-T_2^4\right]+\dfrac{m_{water}L_{water}}{\Delta t}$$ $$1000\epsilon A_{\text {faces sun}}\cos \theta-\epsilon \sigma A_{\text {radiates}}\left[T_1^4-T_2^4\right]=\dfrac{m_{water}L_{water}}{\Delta t}$$ $$\dfrac{m_{water}}{\Delta t}=\dfrac{1000\epsilon A_{\text {faces sun}} \overbrace{ \cos \theta }^{{\color{red}=\;\cos90=1}}-\epsilon \sigma A_{\text {radiates}}\left[T_1^4-T_2^4\right]}{L_{water}}$$ Noting that the area of radiation is the two sides of the leaf while the area that absorbs heat from the sun is just one side. Thus, $A_{\text {radiates}}=2A_{\text {faces sun}}=2A$ $$\dfrac{m_{water}}{\Delta t}=\dfrac{1000\epsilon A -\epsilon \sigma (2A)\left[T_1^4-T_2^4\right]}{L_{water}}$$ $$\dfrac{m_{water}}{\Delta t}=\epsilon A \;\dfrac{1000 - 2\sigma \left[T_1^4-T_2^4\right]}{L_{water}}$$ Plugging the known; $$\dfrac{m_{water}}{\Delta t}=0.85\times40\times 10^{-4} \dfrac{1000 - \left( 2\times 5.67\times10^{-8}\right)\left[308^4-297^4\right]}{2.45\times 10^6}$$ $$\dfrac{m_{water}}{\Delta t}=1.2\times 10^{-6}\;\rm kg/s=\rm \dfrac{1.2\times 10^{-6}\;kg}{s} \cdot \dfrac{3600\;s}{1\;h}$$ $$\dfrac{m_{water}}{\Delta t}=\boxed{\bf4.3\times 10^{-3}\;\rm kg/h}$$
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