/*! 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 24 The Kelvin temperature of an obj... [FREE SOLUTION] | 91Ó°ÊÓ

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The Kelvin temperature of an object is \(T_{1},\) and the object radiates a certain amount of energy per second. The Kelvin temperature of the object is then increased to \(T_{2},\) and the object radiates twice the energy per second that it radiated at the lower temperature. What is the ratio \(T_{2} / T_{1} ?\)

Short Answer

Expert verified
The ratio \(T_{2}/T_{1} = 2^{1/4}.\)

Step by step solution

01

Understanding Radiation Power

The power of radiation emitted by an object is proportionate to the fourth power of its temperature in Kelvin, as described by the Stefan-Boltzmann law: \[ P = \sigma A T^4 \]where \(P\) is the power radiated, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area of the radiating body, and \(T\) is the absolute temperature.
02

Setting Up the Equation for Initial Temperature

When the object is at temperature \(T_{1}\), it radiates energy with power \(P_{1} = \sigma A T_{1}^4.\)
03

Finding the Power at New Temperature

When the object's temperature increases to \(T_{2}\), it radiates twice the initial energy. Thus, the power at \(T_{2}\) is given by: \[ P_{2} = 2P_{1} = 2\sigma A T_{1}^4 \] and also \[ P_{2} = \sigma A T_{2}^4. \]
04

Setting the Equations for Power Equal

Equate the expressions for \(P_{2}\) to find the relationship between \(T_{2}\) and \(T_{1}\): \[ \sigma A T_{2}^4 = 2\sigma A T_{1}^4. \]Since \(\sigma\) and \(A\) are constants, they can be cancelled out, leaving:\[ T_{2}^4 = 2T_{1}^4. \]
05

Solving for the Temperature Ratio

Take the fourth root of both sides to find \(T_{2}/T_{1}\): \[ \left(\frac{T_{2}}{T_{1}}\right)^4 = 2 \] Taking the fourth root gives:\[ \frac{T_{2}}{T_{1}} = 2^{1/4}. \]
06

Final Simplification and Solution

The ratio \(T_{2}/T_{1} = 2^{1/4}\), which can be approximated, but exact solutions are sufficient for this problem.

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Key Concepts

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

Thermal Radiation
Thermal radiation is a fascinating phenomenon that involves the emission of energy in the form of waves or particles originating from a body due to its temperature. All objects emit thermal radiation, even if they are not visibly glowing. This emission occurs because the atoms and molecules within the object are in constant thermal motion.
Thermal radiation is part of the electromagnetic spectrum and can include visible light, infrared, and even ultraviolet radiations. As an object heats up, its thermal radiation increases, and the emitted energy spans over different wavelengths. The color of heated objects changing from red to blue is a visible demonstration of this process.
  • It happens at all temperatures above absolute zero.
  • The intensity and color of the radiation depend on temperature.
  • In everyday life, thermal radiation is observed in sunlight or the heat from a glowing metal, like in a forge.
Understanding thermal radiation is essential as it lays the foundation for more advanced topics, such as the Stefan-Boltzmann law.
Kelvin Temperature
The Kelvin temperature scale is crucial in scientific measurements, providing an absolute reference for thermal motion in particles. It begins at absolute zero, the point where all kinetic motion ceases, equating to 0 Kelvin, or -273.15 degrees Celsius. The Kelvin scale is vital for understanding thermal phenomena, as it accurately reflects the proportional relationship between thermal energy and temperature seen in the Stefan-Boltzmann Law.

Using Kelvin helps avoid negative temperature values, simplifying calculations and making thermal relationships clearer. Scientists prefer Kelvin for expressing laws of physics like thermal radiation because it more meaningfully represents the dynamics of energy changes.
  • Kelvin does not use degrees, unlike Celsius or Fahrenheit.
  • Temperature differences are the same in Kelvin and Celsius.
  • An increase in Kelvin always means an increase in thermodynamic activity.
In the context of the Stefan-Boltzmann Law, Kelvin enables precise calculations of radiation power as seen in our problem where we deal with temperature aspects head-on.
Power Emission
Power emission from an object is the amount of energy it radiates per second. The Stefan-Boltzmann Law tightly ties power emission to temperature, showing that emission scales with the fourth power of the object's absolute temperature.
  • The formal expression is given as: \[P = \sigma A T^4\]
  • Here, \(P\) is the power emitted, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area of the emitting object, and \(T\) is the absolute temperature in Kelvin.
The equation indicates that even small increases in temperature can lead to large increases in emitted power. For example, doubling the temperature increases power emission by a factor of 16. This concept is pivotal in understanding radiation and thermal dynamics in various scientific and engineering fields.

In our exercise, we deduce how a change in temperature affects power emission and learn that the nature of this relationship plays a crucial role in objects emitting twice their initial power when heated.
Temperature Ratio
The temperature ratio in the given problem corresponds to the comparative values of two Kelvin temperatures, \(T_2\) after an increase and \(T_1\) before. This is crucial because the Stefan-Boltzmann Law allows us to equate the changes in energy emission with changes in temperature ratios.

Analyzing the problem, initially the object emits radiated power at \(T_1\). When the temperature is raised to \(T_2\), its power doubles. Using the Stefan-Boltzmann equation, we express this as:\[\frac{T_2}{T_1} = 2^{1/4}\].
  • This fourth root relationship arises due to power's dependence on the fourth power of temperature, \(T^4\).
  • The temperature ratio, \(2^{1/4}\), is approximately 1.189, symbolizing an increase in temperature necessary to double the energy output.
Through this, we identify the critical impact small increases in ratio have on emission, illuminating broad applications in understanding thermodynamics across physical sciences.

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Most popular questions from this chapter

A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to \(35^{\circ} \mathrm{C}\) from \(175^{\circ} \mathrm{C},\) its net radiant power decreases to \(12.0 \mathrm{W}\). What was the net radiant power of the baking dish when it was first removed from the oven? Assume that the temperature in the kitchen remains at \(22^{\circ} \mathrm{C}\) as the dish cools down.

In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is \(12.8^{\circ} \mathrm{C} .\) The temperature at the inside surface of the wall is \(20.0^{\circ} \mathrm{C} .\) The wall is \(0.10 \mathrm{m}\) thick and has an area of \(9.0 \mathrm{m}^{2} .\) Assume that one kilowatt \(\cdot\) hour of electrical energy costs \(\$ 0.10 .\) How many hours are required for one dollar's worth of energy to be conducted through the wall?

In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity \(=0.75\) ). It has a temperature of \(62^{\circ} \mathrm{C} .\) The new owner of the house paints the radiator a lighter color (emissivity \(=0.50\) ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?

A person eats a dessert that contains 260 Calories. (This "Calorie" unit, with a capital \(C,\) is the one used by nutritionists; 1 Calorie \(=4186\) J. See Section \(12.7 .\) ) The skin temperature of this individual is \(36^{\circ} \mathrm{C}\) and that of her environment is \(21^{\circ} \mathrm{C}\). The emissivity of her skin is 0.75 and its surface area is \(1.3 \mathrm{m}^{2} .\) How much time would it take for her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of \(1.40 \mathrm{m}^{2},\) which has a temperature of \(34^{\circ} \mathrm{C}\) and an emissivity of \(0.700 .\) Also suppose that metabolic processes are producing energy at a rate of \(115 \mathrm{J} / \mathrm{s} .\) What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature?

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