/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 Neglecting the effects of radiat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Neglecting the effects of radiation absorption, emission, and scattering within their atmospheres, calculate the average temperature of Earth, Venus, and Mars assuming diffuse, gray behavior. The average distance from the sun of each of the three planets, \(L_{s p}\), along with their measured average temperatures, \(\bar{T}_{p}\), are shown in the table below. Based upon a comparison of the calculated and measured average temperatures, which planet is most affected by radiation transfer in its atmosphere? \begin{tabular}{lcc} \hline Planet & \(\boldsymbol{L}_{s-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\\ \hline Venus & \(1.08 \times 10^{11}\) & 735 \\ Earth & \(1.50 \times 10^{11}\) & 287 \\ Mars & \(2.30 \times 10^{11}\) & 227 \\ \hline \end{tabular}

Short Answer

Expert verified
The calculated average temperatures of Venus, Earth, and Mars are found using the Stefan-Boltzmann Law: \(\bar{T}_\text{p} = \left(\frac{Q_\text{p}}{4 \sigma}\right)^{1/4}\). Comparing these with the given measured average temperatures, we find the differences and determine that Venus has the largest temperature difference, indicating it is the most affected by radiation transfer in its atmosphere.

Step by step solution

01

Find the solar constant

To calculate the average temperature of a planet, we first need to find the solar constant, which is the energy received from the sun per unit area per unit time at a distance of 1 AU (astronomical unit). It is approximately 1361 W/m² for the Sun, which is known as G_0.
02

Calculate the energy received by the planet

To find the amount of energy received by each planet, we will divide the solar constant by the square of the distance from the sun (in astronomical units). We get this formula: \(Q_\text{p} = \frac{G_0}{L_\text{sp}^2}\) Where: \(Q_\text{p}\) = Energy received by the planet (W/m²) \(L_\text{sp}\) = Distance from the Sun to the planet (in meters) \(G_0\) = Solar constant (approx. 1361 W/m²) To convert the distances from meters to astronomical units (AU), use the conversion factor: 1 AU = 149,597,870,700 meters.
03

Calculate the average temperature of each planet

Now that we have the energy received by each planet, we can use the Stefan-Boltzmann Law to calculate the average temperature. The formula for the Stefan-Boltzmann Law is: \(\bar{T}_{p} = \left(\frac{Q_\text{p}}{4 \sigma}\right)^{1/4}\) Where: \(\bar{T}_\text{p}\) = Average temperature of the planet (K) \(\sigma\) = Stefan-Boltzmann constant (approx. \(5.67 \times 10^{-8} W m^{-2} K^{-4}\)) Using the formula, calculate the average temperature for Venus, Earth, and Mars.
04

Compare calculated temperatures with measured temperatures

To identify which planet is most affected by radiation transfer in its atmosphere, we need to compare the calculated and measured temperatures. Calculate the difference between the two temperatures for each planet, and then determine which planet has the largest temperature difference. By following these steps, we can calculate the average temperature of each of the three planets and find which planet is most affected by radiation transfer in its atmosphere.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solar Constant
The solar constant is a fundamental concept in planetary climate modeling. It represents the average energy from the Sun that hits a square meter of Earth's atmosphere per second.
It is incredibly important as it helps determine the energy input for a planet's climate system. The solar constant is about 1361 watts per square meter (W/m²). This measurement is taken at one Astronomical Unit (AU), the average distance from the Earth to the Sun.
To understand the importance, imagine standing near a campfire. The closer you are, the hotter it feels. Similarly, planets closer to the Sun receive more solar energy. The solar constant varies slightly due to variations in the Earth's orbit and solar activity, but these are quite small compared to the overall magnitude.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is crucial for understanding how planets radiate energy into space. It establishes a relationship between a body's temperature and the amount of energy it emits.The law is mathematically expressed as:\[E = \sigma T^4\]where- \( E \) is the energy emitted per square meter (W/m²)- \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4}\)- \( T \) is the absolute temperature of the body in Kelvin (K)Using this law, we can estimate the temperature of a planet based on the energy it emits. Planets must balance the energy they absorb from the Sun with the energy they radiate back into space.
This balance is key to understanding the climate and average temperature of celestial bodies.
Radiation Transfer
Radiation transfer explains how energy, in the form of electromagnetic radiation, moves through the atmosphere. It determines how much of the Sun's energy reaches the surface or gets absorbed, reflected, or scattered. This concept helps in understanding the energy budget of a planet: - Energy from the Sun - Energy absorbed by gases and aerosols - Energy reflected by clouds and surfaces - Energy trapped by the greenhouse gases Radiation transfer is what creates different climates and weather patterns across planets. In our exercise, we neglect these effects to simplify the calculation. However, in reality, they are vital in shaping planetary climates and need to be considered for accurate climate modeling.
Astronomical Unit
An Astronomical Unit, or AU, is a standard unit of measurement used in astronomy to describe distances within our solar system. One AU is the average distance from the Earth to the Sun, approximately 149.6 million kilometers or 93 million miles. This unit helps us easily express and compare distances of other planets from the Sun: - Venus: 0.723 AU - Earth: 1 AU - Mars: 1.524 AU Understanding AU is essential for converting distances in planetary calculations, as seen in our exercise.
By expressing distances in AU, we standardize measurements and allow for straightforward comparisons, facilitating our understanding of planetary dynamics.
Diffuse Gray Model
The Diffuse Gray Model simplifies complex radiation calculations by assuming that a planet's atmosphere absorbs the same fraction of radiation at all wavelengths. This means it treats the entire atmosphere as a single layer with uniform properties. In planetary climate studies, employing the Diffuse Gray Model allows for simplified calculations of average temperatures without considering specific gas absorption characteristics. It is especially useful for theoretical modeling, where a more complex treatment could obscure overall energy balance insights. Although a simplification, this model helps highlight the impact of radiation transfer on planetary climates, as seen in our exercise example with Earth, Venus, and Mars. By using this model, we can focus on broader climate trends without getting bogged down in the details of every atmospheric variable.

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

The oxidized-aluminum wing of an aircraft has a chord length of \(L_{c}=4 \mathrm{~m}\) and a spectral, hemispherical emissivity characterized by the following distribution. (a) Consider conditions for which the plane is on the ground where the air temperature is \(27^{\circ} \mathrm{C}\), the solar irradiation is \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the effective sky temperature is \(270 \mathrm{~K}\). If the air is quiescent, what is the temperature of the top surface of the wing? The wing may be approximated as a horizontal, flat plate. (b) When the aircraft is flying at an elevation of approximately \(9000 \mathrm{~m}\) and a speed of \(200 \mathrm{~m} / \mathrm{s}\), the air temperature, solar irradiation, and effective sky temperature are \(-40^{\circ} \mathrm{C}, 1100 \mathrm{~W} / \mathrm{m}^{2}\), and \(235 \mathrm{~K}\), respectively. What is the temperature of the wing's top surface? The properties of the air may be approximated as \(\rho=0.470 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.50 \times\) \(10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}, k=0.021 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=0.72\).

Solar irradiation of \(1100 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a large, flat, horizontal metal roof on a day when the wind blowing over the roof causes a convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outside air temperature is \(27^{\circ} \mathrm{C}\), the metal surface absorptivity for incident solar radiation is \(0.60\), the metal surface emissivity is \(0.20\), and the roof is well insulated from below. (a) Estimate the roof temperature under steady-state conditions. (b) Explore the effect of changes in the absorptivity, emissivity, and convection coefficient on the steady-state temperature. 12.108 Neglecting the effects of radiation absorption, emission, and scattering within their atmospheres, calculate the average temperature of Earth, Venus, and Mars assuming diffuse, gray behavior. The average distance from the sun of each of the three planets, \(L_{s p}\), along with their measured average temperatures, \(\bar{T}_{p}\), are shown in the table below. Based upon a comparison of the calculated and measured average temperatures, which planet is most affected by radiation transfer in its atmosphere? \begin{tabular}{lcc} \hline Planet & \(L_{x-p}(\mathbf{m})\) & \(\bar{T}_{p}(\mathbf{K})\) \\ \hline Venus & \(1.08 \times 10^{11}\) & 735 \\ Earth & \(1.50 \times 10^{11}\) & 287 \\ Mars & \(2.30 \times 10^{11}\) & 227 \\ \hline \end{tabular}

Plant leaves possess small channels that connect the interior moist region of the leaf to the environment. The channels, called stomata, pose the primary resistance to moisture transport through the entire plant, and the diameter of an individual stoma is sensitive to the level of \(\mathrm{CO}_{2}\) in the atmosphere. Consider a leaf of corn (maize) whose top surface is exposed to solar irradiation of \(G_{S}=600 \mathrm{~W} / \mathrm{m}^{2}\) and an effective sky temperature of \(T_{\text {sky }}=0^{\circ} \mathrm{C}\). The bottom side of the leaf is irradiated from the ground which is at a temperature of \(T_{g}=20^{\circ} \mathrm{C}\). Both the top and bottom of the leaf are subjected to convective conditions characterized by \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, T_{\infty}=25^{\circ} \mathrm{C}\) and also experience evaporation through the stomata. Assuming the evaporative flux of water vapor is \(50 \times 10^{-6} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) under rural atmospheric \(\mathrm{CO}_{2}\) concentrations and is reduced to \(5 \times 10^{-6} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) when ambient \(\mathrm{CO}_{2}\) concentrations are doubled near an urban area, calculate the leaf temperature in the rural and urban locations. The heat of vaporization of water is \(h_{f g}=2400 \mathrm{~kJ} / \mathrm{kg}\) and assume \(\alpha=\varepsilon=0.97\) for radiation exchange with the sky and the ground, and \(\alpha_{S}=0.76\) for solar irradiation.

A wet towel hangs on a clothes line under conditions for which one surface receives solar irradiation of \(G_{S}=900 \mathrm{~W} / \mathrm{m}^{2}\) and both surfaces are exposed to atmospheric (sky) and ground radiation of \(G_{\text {tam }}=\) \(200 \mathrm{~W} / \mathrm{m}^{2}\) and \(G_{\mathrm{g}}=250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Under moderately windy conditions, airflow at a temperature of \(27^{\circ} \mathrm{C}\) and a relative humidity of \(60 \%\) maintains a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. The wet towel has an emissivity of \(0.96\) and a solar absorptivity of \(0.65\). As a first approximation the properties of the atmospheric air may be evaluated at a temperature of \(300 \mathrm{~K}\). Determine the temperature \(T_{s}\) of the towel. What is the corresponding evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long?

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

See all solutions

Recommended explanations on Physics Textbooks

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.