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Chapter 13: Problem 22

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?

Short Answer

Expert verified
The new temperature is approximately 98掳C.

Step by step solution

01

Understand the concept of radiant power

Radiant power emitted by a body is given by the Stefan-Boltzmann Law, which states that the power emitted per unit area by a perfect black body is proportional to the fourth power of the temperature. For a real body, this power is adjusted by the emissivity factor, \( \epsilon \). Thus, the radiant power \( P \) is given by: \[ P = \epsilon \sigma A T^4 \]where \( \epsilon \) is the emissivity, \( \sigma \) is Stefan's constant \( (5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4) \), \( A \) is the surface area, and \( T \) is the absolute temperature in Kelvin.
02

Express the initial condition

For the radiator with the dark color, the emissivity is 0.75 and the temperature is given as \( 62^{\circ} \text{C} \). Convert this to Kelvin:\[ T_1 = 62 + 273.15 = 335.15 \text{ K} \]Thus, the initial radiant power can be expressed as:\[ P = 0.75 \sigma A (335.15)^4 \]
03

Express the final condition

After the radiator is painted, the emissivity changes to 0.50. We need to find the new temperature, \( T_2 \), that results in the same radiant power:\[ P = 0.50 \sigma A T_2^4 \]
04

Equate the initial and final power expressions

Since the radiant power remains the same before and after painting, we equate the expressions for \( P \):\[ 0.75 \sigma A (335.15)^4 = 0.50 \sigma A T_2^4 \]Cancel \( \sigma \) and \( A \) from both sides:\[ 0.75 (335.15)^4 = 0.50 T_2^4 \]
05

Solve for the new temperature in Kelvin

Rearrange the equation to solve for \( T_2 \):\[ T_2^4 = \frac{0.75}{0.50} (335.15)^4 \]\[ T_2^4 = 1.5 \times (335.15)^4 \]Take the fourth root of both sides:\[ T_2 = (1.5^{1/4}) \times 335.15 \]\[ T_2 \approx 1.1067 \times 335.15 \approx 370.84 \text{ K} \]
06

Convert the temperature back to Celsius

Convert \( T_2 \) from Kelvin to Celsius:\[ T_{2,\text{Celsius}} = 370.84 - 273.15 \approx 97.69^{\circ} \text{C} \].

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Key Concepts

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

Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation, important in understanding how objects radiate heat. It ranges from 0 to 1.
A value of 1 means the material is a perfect emitter known as a black body; a value of 0 indicates no emission. Understanding emissivity is crucial in applying the Stefan-Boltzmann Law which calculates radiant power using the emissivity factor \( \epsilon \). It modifies how much power a real body emits compared to a perfect black body:
  • Higher emissivity means more energy is emitted.
  • Lower emissivity means less energy is emitted.
In the original problem, changing the radiator paint affected its emissivity, thus affecting the radiant power if the temperature stayed constant. Emissivity impacts many practical situations, such as energy efficiency in buildings or material selection in thermal wear.
Radiant Power
Radiant power is the energy emitted by an object in the form of radiation. For heating and thermal calculations, it is essential to understand how much power is discharged over a surface.Using the Stefan-Boltzmann Law:\[P = \epsilon \sigma A T^4\]This formula indicates radiant power \( P \) depends on:
  • Emissivity \( \epsilon \)
  • Stefan-Boltzmann constant \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4 \)
  • Surface area \( A \)
  • Temperature \( T \) raised to the fourth power
In the problem, equating the radiated power before and after changing the paint shows how slight changes in emissivity require temperature adjustments to maintain constant radiant power levels. Understanding this relationship helps in designing systems involving heat radiation efficiently.
Kelvin Temperature Conversion
Converting temperatures to Kelvin is a fundamental step in scientific calculations, where absolute temperature scales are required.Kelvin is used because it starts from absolute zero, the point where all thermal motion ceases. Here is how you convert from Celsius to Kelvin:
  • Add 273.15 to the Celsius temperature.
For instance, converting \( 62^{\circ} \text{C} \) to Kelvin involves:\[T_K = 62 + 273.15 = 335.15 \, \text{K}\]In the problem, initial and resulting radiator temperatures are converted to Kelvin for calculation purposes. Returning to Celsius for practical interpretation concludes the process. Understanding this conversion aids in tackling various thermal physics problems and ensures clarity when comparing scientific data.

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Most popular questions from this chapter

A cubical piece of heat-shield tile from the space shuttle measures \(0.10 \mathrm{m}\) on a side and has a thermal conductivity of \(0.065 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) The outer surface of the tile is heated to a temperature of \(1150^{\circ} \mathrm{C},\) while the inner surface is maintained at a temperature of \(20.0^{\circ} \mathrm{C} .\) (a) How much heat flows from the outer to the inner surface of the tile in five minutes? (b) If this amount of heat were transferred to two liters \((2.0 \mathrm{kg})\) of liquid water, by how many Celsius degrees would the temperature of the water rise?

The block in the drawing has dimensions \(L_{0} \times 2 L_{0} \times 3 L_{0},\) where \(L_{0}=0.30 \mathrm{m} .\) The block has a thermal conductivity of \(250 \mathrm{J} /\left(\mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right) .\) In drawings \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C},\) heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is \(35^{\circ} \mathrm{C}\) and that of the cooler surface is \(19^{\circ} \mathrm{C} .\) Determine the heat that flows in \(5.0 \mathrm{s}\) for each case.

A person is standing outdoors in the shade where the temperature is \(28^{\circ} \mathrm{C} .\) (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is \(160 \mathrm{cm}^{2}\) and its emissivity is \(0.85 .\) (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were \(0.65 ?\)

A wall in a house contains a single window. The window consists of a single pane of glass whose area is \(0.16 \mathrm{m}^{2}\) and whose thickness is \(2.0 \mathrm{mm}\). Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are \(18 \mathrm{m}^{2}\) and \(0.10 \mathrm{m},\) respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

One half of a kilogram of liquid water at \(273 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is placed outside on a day when the temperature is \(261 \mathrm{K}\left(-12^{\circ} \mathrm{C}\right) .\) Assume that the heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is \(0.60 .\) Consider two cases: when the surface area of the water is (a) \(0.035 \mathrm{m}^{2}\) (as it might be in a cup) and (b) \(1.5 \mathrm{m}^{2}\) (as it could be if the water were spilled out to form a thin sheet). Concepts: (i) In case (a) is the heat that must be removed to freeze the water less than, greater than, or the same as in case (b)? (ii) The loss of heat by radiation depends on the temperature of the radiating object. Does the temperature of the water change as it freezes? (iii) The water both loses and gains heat by radiation. How, then, can heat transfer by radiation lead to freezing of the water? (iv) Will it take longer for the water to freeze in case (a) when the area is smaller or in case (b) when the area is larger? Calculations: For each case, (a) and (b), determine the time it takes the water to freeze into ice at \(0^{\circ} \mathrm{C}\).
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