Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 41

If a leaf is to maintain a temperature of \(40^{\circ} \mathrm{C}\) (reasonable for a leaf), it must lose \(250 \mathrm{W} / \mathrm{m}^{2}\) by transpiration (evaporative heat loss). Note that the leaf also loses heat by radiation, but we will neglect this. How much water is lost after 1 h through transpiration only? The area of the leaf is \(0.005 \mathrm{m}^{2}\).

Short Answer

Expert verified
Answer: 0.002 liters

Step by step solution

01

Identify the given information and the missing variable

We are given: - Heat loss through transpiration: 250 W/m² - Temperature to maintain: 40°C - Area of the leaf: 0.005 m² We need to find the mass of water lost after 1 hour.
02

Calculate the total heat loss through transpiration

To find the total heat loss, we multiply the heat loss per square meter by the area of the leaf: Total heat loss = Heat loss per square meter × Area of the leaf Total heat loss = 250 W/m² × 0.005 m² = 1.25 W
03

Calculate the energy lost after 1 hour of transpiration

To calculate the total energy lost, we multiply the total heat loss by the time in seconds (1 hour = 3600 seconds): Energy lost = Total heat loss × Time Energy lost = 1.25 W × 3600 s = 4500 J
04

Use the latent heat of vaporization to find the mass of water lost

The latent heat of vaporization of water is 2.25 x 10^6 J/kg. To find the mass of water lost, we use the following formula: Mass of water lost = Energy lost / Latent heat of vaporization Mass of water lost = 4500 J / (2.25 × 10^6 J/kg) = 0.002 kg
05

Convert the mass of water into liters

To convert the mass of water lost into liters, we use the density of water (1 kg/L): Water lost (in liters) = Mass of water lost / Density of water Water lost (in liters) = 0.002 kg / 1 kg/L = 0.002 L So, after 1 hour of transpiration, the leaf loses 0.002 liters of water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It is a damp, chilly day in a New England seacoast town suffering from a power failure. To warm up the cold, clammy sheets, Jen decides to fill hot water bottles to tuck between the sheets at the foot of the beds. If she wishes to heat \(2.0 \mathrm{L}\) of water on the wood stove from $20.0^{\circ} \mathrm{C}\( to \)80.0^{\circ} \mathrm{C},$ how much heat must flow into the water?
How much heat is required to change \(1.0 \mathrm{kg}\) of ice, originally at \(-20.0^{\circ} \mathrm{C},\) into steam at \(110.0^{\circ} \mathrm{C} ?\) Assume 1.0 atm of pressure.
At a tea party, a coffeepot and a teapot are placed on the serving table. The coffeepot is a shiny silver-plated pot with emissivity of \(0.12 ;\) the teapot is ceramic and has an emissivity of \(0.65 .\) Both pots hold \(1.00 \mathrm{L}\) of liquid at \(98^{\circ} \mathrm{C}\) when the party begins. If the room temperature is at \(25^{\circ} \mathrm{C},\) what is the rate of radiative heat loss from the two pots? [Hint: To find the surface area, approximate the pots with cubes of similar volume.]
Bare, dark-colored basalt has a thermal conductivity of 3.1 $\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ whereas light-colored sandstone's thermal conductivity is only \(2.4 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) Even though the same amount of radiation is incident on both and their surface temperatures are the same, the temperature gradient within the two materials will differ. For the same patch of area, what is the ratio of the depth in basalt as compared with the depth in sandstone that gives the same temperature difference?
The inner vessel of a calorimeter contains \(2.50 \times 10^{2} \mathrm{g}\) of tetrachloromethane, \(\mathrm{CCl}_{4},\) at \(40.00^{\circ} \mathrm{C} .\) The vessel is surrounded by \(2.00 \mathrm{kg}\) of water at $18.00^{\circ} \mathrm{C} .\( After a time, the \)\mathrm{CCl}_{4}$ and the water reach the equilibrium temperature of \(18.54^{\circ} \mathrm{C} .\) What is the specific heat of \(\mathrm{CCl}_{4} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Medical Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Radiation

Read Explanation

Fluids

Read Explanation

Electrostatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free