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A study published in July 2004 indicated that temperature increases in urban areas in the eastern United States are causing plants to bud up to 7 days early compared with plants in rural areas just a few miles away, thereby disrupting biological cycles. Average temperatures in the urban areas were up to \(3.5 \mathrm{C}^{\circ}\) higher than in the rural areas. By what percent will the radiated heat per square meter increase due to such a temperature difference if the rural temperature is \(0^{\circ} \mathrm{C}\) on the average?

Short Answer

Expert verified
The radiated heat will increase by approximately 4.61%.

Step by step solution

01

Identify the Concept

To find out by what percentage the radiated heat increases due to a temperature difference, we need to consider how emitted heat is related to temperature. This involves the Stefan-Boltzmann law, which states that the power radiated by a surface is proportional to the fourth power of its temperature in Kelvin.
02

Express the Law Mathematically

The Stefan-Boltzmann law is given by: \[ E = \sigma \cdot T^4 \] where \(E\) is the energy radiated per unit area, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature in Kelvin.
03

Convert Celsius to Kelvin

To apply the Stefan-Boltzmann law, convert the given temperatures from Celsius to Kelvin. For rural areas: \(T_\text{rural} = 0^{\circ} \mathrm{C} = 273.15 \, \mathrm{K}\) For urban areas: \(T_\text{urban} = 3.5^{\circ} \mathrm{C} = 273.15 + 3.5 = 276.65 \, \mathrm{K}\)
04

Calculate Radiated Energy for Each Temperature

Calculate the radiated energy for rural and urban temperatures: \[E_\text{rural} = \sigma \cdot (273.15)^4\] \[E_\text{urban} = \sigma \cdot (276.65)^4\] Note: the Stefan-Boltzmann constant \(\sigma\) will cancel out in the percentage calculation.
05

Determine the Increase in Radiated Energy

Compute the percentage increase in energy:1. Find the difference: \[\Delta E = E_\text{urban} - E_\text{rural} = \sigma(276.65^4 - 273.15^4)\]2. Find the percentage increase: \[\% \text{ Increase} = \left(\frac{\Delta E}{E_\text{rural}}\right) \times 100\] Thus, \[\% \text{ Increase} = \left(\frac{(276.65^4 - 273.15^4)}{273.15^4}\right) \times 100\]
06

Perform Calculation

Calculate each value: 1. \(273.15^4 \approx 5.5663 \times 10^9\)2. \(276.65^4 \approx 5.8227 \times 10^9\)3. \(\Delta E = 5.8227 \times 10^9 - 5.5663 \times 10^9\)4. \(\Delta E \approx 2.564 \times 10^8\)5. \% \text{ Increase} = \left(\frac{2.564 \times 10^8}{5.5663 \times 10^9}\right) \times 100 \approx 4.61\%\.

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Key Concepts

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

Urban Heat Island Effect
In urban environments, a phenomenon known as the urban heat island effect leads to higher temperatures compared to rural areas. This happens because cities tend to have materials like concrete and asphalt that absorb and retain heat. Also, human activities and lack of vegetation contribute to this warming.
The increased temperatures in urban areas can lead to several issues, including:
  • Earlier budding of plants, potentially by up to 7 days compared to rural environment.
  • Changes in local weather patterns.
  • Increased energy usage for cooling purposes.
Understanding the urban heat island effect is essential in urban planning and environmental conservation efforts, as it impacts biodiversity and quality of life.
Temperature Change Impact
Changes in temperature can have significant effects on both natural systems and human activities. A temperature increase of just a few degrees, such as the 3.5°C seen in urban areas compared to rural ones, can lead to noticeable impacts such as disruptions in biological cycles and increased strain on energy systems.
Some of the broader impacts of temperature changes include:
  • Alteration of growing seasons and ecosystems.
  • Impact on biodiversity and species migration patterns.
  • Increased incidence of heat-related illnesses.
Effective management of temperature changes requires understanding their sources and effects, particularly in the context of urbanization and global warming.
Thermal Radiation Increase
The increase in thermal radiation associated with higher temperatures in urban areas is explained by the Stefan-Boltzmann law. This scientific principle tells us that the energy emitted from a surface is proportional to the fourth power of its temperature.
Due to the temperature difference between urban and rural areas:
  • Urban areas emit a higher amount of thermal radiation.
  • This increased thermal radiation contributes to further warming of the local environment.
In the given study, a 3.5°C rise resulted in approximately a 4.61% increase in the radiant energy per square meter, showcasing how even small temperature changes can significantly affect thermal energy dynamics.
Energy Transfer in Physics
Energy transfer processes are foundational concepts in physics that describe how energy moves from one system to another. In the context of thermal radiation, energy is transferred from hot surfaces to cooler surroundings.
Key principles include:
  • Heat transfer mechanisms including conduction, convection, and radiation.
  • Importance of thermal equilibrium in understanding energy distribution.
By studying how energy transfer works, we can better understand phenomena such as the urban heat island effect, predict how systems will respond to temperature changes, and develop strategies to mitigate negative impacts.

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Most popular questions from this chapter

You are making pesto for your pasta and have a cylindrical measuring cup \(10.0 \mathrm{~cm}\) high made of ordinary glass \(\left(\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) that is filled with olive oil \(\left(\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) to a height of \(1.00 \mathrm{~mm}\) below the top of the cup. Initially, the cup and oil are at a kitchen temperature of \(22.0^{\circ} \mathrm{C}\). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees and in kelvins? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January 23,1916 , and the next day it plummeted to \(-56.0^{\circ} \mathrm{F}\). What was the temperature change in Celsius degrees and in kelvins?

The blood plays an important role in removing heat from the body by bringing this heat directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. We shall assume that the blood is brought to the bottom layer of skin at a temperature of \(37^{\circ} \mathrm{C}\) and that the outer surface of the skin is at \(30.0^{\circ} \mathrm{C}\). Skin varies in thickness from \(0.50 \mathrm{~mm}\) to a few millimeters on the palms and soles, so we shall assume an average thickness of \(0.75 \mathrm{~mm}\). A \(165 \mathrm{lb}, 6 \mathrm{ft}\) person has a surface area of about \(2.0 \mathrm{~m}^{2}\) and loses heat at a net rate of \(75 \mathrm{~W}\) while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, which is then transferred to the liquid for \(120 \mathrm{~s}\) at a constant rate of \(65.0 \mathrm{~W}\). The mass of the liquid is \(0.780 \mathrm{~kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\). (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or its surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

(a) How much heat is required to raise the temperature of \(0.250 \mathrm{~kg}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C} ?\) (b) If this amount of heat is added to an equal mass of mercury that is initially at \(20.0^{\circ} \mathrm{C},\) what is its final temperature?

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