Statement

On a hot, dry day, a large tree with an adequate water supply may evapotranspire about $500\,\text{L}=5{,}0\times10^{-1}\,\text{m}^3$ of water. Since transpiration occurs mainly in daylight, assume that it is spread over twelve hours. The tree canopy covers a ground area $A_{\text{tree}}=50\,\text{m}^2$ . The density of water and its latent heat of vaporization at atmospheric pressure are

\rho_{\text{water}}=1{,}0\times10^3\,\text{kg}\!\cdot\!\text{m}^{-3}, \qquad L_v=2{,}26\times10^6\,\text{J}\!\cdot\!\text{kg}^{-1}.

Calculate the mass of water evapotranspired during the day and the energy required to vaporize it.
Explain why this produces a cooling effect.
Calculate the tree's average cooling power per square metre during the twelve hours of active transpiration.
For comparison, consider an air conditioner supplied with electrical power $P_{\mathrm{elec}}=7{,}0\times10^2\,\text{W}$ , with cooling coefficient of performance $\mathrm{COP}=3{,}0$ , cooling a room of area $A_{\mathrm{room}}=20\,\text{m}^2$ . By definition, $\mathrm{COP}=P_{\mathrm{cool}}/P_{\mathrm{elec}}$ . Calculate its cooling power and cooling power per square metre, then compare them with the tree's values.
Why can we not conclude that the tree and the air conditioner provide exactly the same perceived cooling, even if their powers per unit area have the same order of magnitude?
Could many houseplants cool a home? (General knowledge: the answer depends on biological processes.)

Hint

Use

m=\rho V

,

Q_{\mathrm{evap}}=mL_v

, and

P=Q/\tau

. For the air conditioner,

P_{\mathrm{cool}}=\mathrm{COP}\,P_{\mathrm{elec}}

.

Detailed solution

Solution

1. The corresponding mass is

m=\rho_{\text{water}}V =1{,}0\times10^3\times5{,}0\times10^{-1} =5{,}0\times10^2\,\text{kg}.

The vaporization energy is

Q_{\mathrm{evap}}=mL_v =5{,}0\times10^2\times2{,}26\times10^6 =1{,}13\times10^9\,\text{J},

whose order of magnitude is one gigajoule.

2. Vaporization is an endothermic phase change: it requires the energy

Q_{\mathrm{evap}}

. This energy comes mainly from the solar radiation absorbed by the foliage: instead of heating the leaf (sensible heat), the absorbed solar energy is converted into latent heat of vaporization. The leaf therefore stays cooler than it would if all the absorbed energy had gone into raising its temperature, which is why evapotranspiration has a cooling effect. 3. Twelve hours correspond to

\tau=12\times3600=4{,}32\times10^4\,\text{s}

. Hence

P_{\text{tree}}=\frac{Q_{\mathrm{evap}}}{\tau} =\frac{1{,}13\times10^9}{4{,}32\times10^4} \approx2{,}62\times10^4\,\text{W},

and

\frac{P_{\text{tree}}}{A_{\text{tree}}} =\frac{2{,}62\times10^4}{50} \approx5{,}2\times10^2\,\text{W}\!\cdot\!\text{m}^{-2}.

Plausibility check: since this energy comes mainly from absorbed solar radiation (question 2), it is reassuring that $5{,}2\times10^2\,\text{W}\!\cdot\!\text{m}^{-2}$ has the same order of magnitude as the solar irradiance received at ground level on a clear day, which reaches on the order of $10^3\,\text{W}\!\cdot\!\text{m}^{-2}$ at its peak and a few hundred $\text{W}\!\cdot\!\text{m}^{-2}$ on daily average. This confirms that a well-watered tree, on a hot sunny day, can genuinely dissipate a substantial fraction of the solar energy it receives through transpiration.

4. The useful cooling power is

P_{\text{cool}}=\mathrm{COP}\,P_{\mathrm{elec}} =3{,}0\times7{,}0\times10^2 =2{,}1\times10^3\,\text{W}.

Per unit floor area,

\frac{P_{\text{cool}}}{A_{\text{room}}} =\frac{2{,}1\times10^3}{20} \approx1{,}1\times10^2\,\text{W}\!\cdot\!\text{m}^{-2}.

In this model, the tree's evapotranspiration power per unit area is about five times that of the air conditioner.

5. This comparison concerns energy fluxes only. An air conditioner cools a controlled enclosed volume, whereas the vapour from a tree disperses into the atmosphere and wind brings in warm air. Perceived cooling depends on ventilation, humidity, ambient temperature, and distance from the tree. Transpiration also varies with sunlight, wind, air humidity, species, and available water, and can decrease sharply during drought. The tree additionally provides shade, which directly reduces the solar radiation received by the ground and by people. The calculation therefore compares orders of magnitude, not equal thermal comfort. 6. In practice, houseplants provide very little cooling. In most plants, light encourages stomata to open, allowing water vapour to escape. Indoor light is generally weak, so stomata open less and transpiration decreases. In a poorly ventilated room, rising humidity also progressively slows evaporation. Many plants may therefore cause slight local evaporative cooling, but cannot replace an air conditioner. Removing heat durably would require moist air to be vented outdoors. Intense artificial lighting might stimulate transpiration, but its electrical energy would itself end up almost entirely as heat in the room. CAM plants adapted to arid environments, including cacti, are an exception: their stomata open mainly at night.