Quiz on blackbody & Kirchhoff's law-chemistry learners
MCQs with answers on blackbody and Kirchhoff's law
Gustav Kirchhoff, a physicist, and a creator of the ideal
thermal radiation absorber & emitter, formulated two radiation laws to
explain the absorbing and emissive powers of terrestrial objects. They are
Kirchhoff's laws of thermal radiation. This blog post discusses the
multiple-choice questions and answers to Kirchhoff's law and blackbody.
Table of contents:
Multiple choice questions and answers on Kirchhoff’s law of thermal radiation
True or false questions on Kirchhoff’s law
Reasoning questions on Kirchhoff’s law
Video description of Kirchhoff’s law
Numerical problems of
Kirchhoff’s law
PowerPoint notes of Kirchhoff’s law
Match the following
quiz on Kirchhoff’s law of thermal radiation
Our e-book:
To download an e-book on MCQ & answers on blackbody & its radiation, click on the provided link.
Multiple choice questions and answers of Kirchhoff’s law:
1. Which of the following law states that "good
absorbers are the good emitters of heat energy"?
A. Kirchhoff’s law
B. Planck’s law
C. Wein’s displacement law
D. James Clerk’s electromagnetic induction
Answer: Kirchhoff’s law
Explanation:
Kirchhoff's law states that a substance that absorbs 100%
incident light can emit all absorbed radiations at a particular wavelength in
thermal equilibrium conditions.
2. What happens when heat radiates from one part of the
medium to another?
A. the medium temperature rises
B. the medium temperature falls
C. the medium remains unaffected
D. It changes its form
Answer: the medium remains unaffected
Explanation:
Radiation is the method of transmission of heat from one
point to another without heating the intervening medium.
3. Which region of the electromagnetic spectrum does the
emitted black body radiation belong to at room temperature?
A. Ultraviolet region
B. Infrared region
C. Visible light
D. Cosmic region
Answer: Infrared region
Explanation:
At room temperature, most of the emitted black body
radiations fall in the infrared region of the electromagnetic spectrum.
4. Who predicted that the emissivity of a substance is equal
to its absorptivity?
A. Gustav Kirchhoff
B. J. H. Jeans
C. James Clerk Maxwell
D. Max Planck
Answer: Gustav Kirchhoff
Explanation:
Kirchhoff’s law states that emissivity equals the
absorptivity of a substance in thermal equilibrium.
5. What is the absorbing power of a perfect blackbody?
A. 1
B. 0
C. infinity
D. 100
Answer: 1
Explanation:
An ideal blackbody's surface absorbs all wavelength incident
radiations. Hence, its absorptivity is equal to one.
6. In which part of the visible spectrum does the emission
spectrum of sunlight is peaked?
A. Yellow-green
B. Yellow-red
C. Red-orange
D. Red-blue
Answer: Yellow-green
Explanation:
The sun has an effective temperature of 5800 K. It produces
an emission spectrum that peaks in the yellow-green part of the visible region.
7. The radiation inside a constant temperature enclosure
depends upon
A. Wavelength
B. Frequency
C. Temperature
D. Nature of the substance
Answer: Temperature
Explanation:
According to Kirchhoff’s law of thermal radiation, the black
body curve is characteristic of thermal light. It depends only on the
temperature of the walls of the cavity.
8. Which law states the continuous thermal energy exchange
between the body and the surrounding?
A. Stefan's law
B. Kirchhoff's law
C. Prevost's theory
D. Wien's law
Answer: Prevost's theory
Explanation:
Prevost's theory states that the hotness and coldness of a
body indicate a rise or fall in its temperature due to thermal energy exchange
with the surrounding. So, the body that absorbs more heat from the surrounding
becomes hot. The body that releases more heat to the surrounding is cooled.
Moreover, it elucidated continuous heat exchange between the body and the
surrounding at all finite temperatures.
9. At a given temperature and wavelength conditions, the
ratio between the emissive power to the absorbing power of a body
is____________
A. a constant value
B. a variable quantity
C. equal to zero
D. unpredictable number
Answer: a constant value
Explanation:
Kirchhoff's law states that the ratio of emissive and
absorbing powers of a material object is constant at every wavelength in
thermal equilibrium. And the emissive power of a perfect blackbody at that
wavelength and temperature gives its value.
10. What does an ice block do when placed in a room of 25
degrees centigrade temperature?
A. radiates less heat but absorbs more
B. radiates more heat than it absorbs
C. radiates as much heat as it absorbs
D. does not radiate and absorb heat
Answer: radiates less heat but absorbs more
Explanation:
An ice block placed in the room is at zero degrees Celsius
temperature. But, the room temperature is 25 degrees Celsius. So, heat flows
from the surroundings to the ice. It implies ice absorbs more heat than it
radiates.
11. Why are blue stars hotter than red stars?
A. With the increase in temperature, the color of the object
seems dull
B. The wavelength of the body increases with the rise in
temperature
C. It depends upon the distance of stars from the earth
D. The frequency of black body radiation increases with the
temperature rise.
Answer: the frequency of black body radiation increases with
the temperature rise.
Explanation:
Above 500 degrees Celsius, the radiated black body
radiations move towards the visible region of the electromagnetic spectrum. So,
the human eye can perceive them.
Initially, the radiations appear in dull red. An increase in
temperature further changes the radiation color to yellow, orange, and green.
It implies that objects with higher temperatures emit blue
light. And those with low temperatures emit red light.
12. Why are objects black in color?
A. They absorb all wavelength incident light.
B. They reflect black color more than the other colors
C. They have no color and seem black.
D. They absorb white light, so they appear black.
Answer: They absorb all wavelength incident light
Explanation:
A body's surface appears black when it absorbs all colored
radiations of all wavelengths.
13. Which law explains the continuous frequency spectrum of
the black body?
A. Planck’s law
B. Kirchhoff’s law
C. Rayleigh-Jeans law
D. Wein’s displacement law
Answer: Planck’s law
Explanation:
Black body radiations exhibit a continuous spectrum of
frequency of radiations that depend only on the body’s temperature is known
Planck spectrum.
14. Which law states that blackbody emissions increase
enormously in the ultraviolet region when the temperature rises?
A. Planck’s law
B. Kirchhoff’s law
C. Rayleigh-Jeans law
D. Wein’s displacement law
Answer: Rayleigh-Jeans law
Explanation:
Rayleigh's law follows classical physics assumptions. The Rayleigh-Jeans
law explained the variation of light intensity with wavelength at a particular
temperature (T).
This law states that the light intensity increases
enormously unbounded at shorter wavelengths of the black body radiations.
It causes excess radiant energy emission by the black body
in the ultraviolet region. It is named ultraviolet catastrophe.
15. Who discovered the term black body?
A. Gustav Kirchhoff
B. Max Planck
C. Isaac Newton
D. J.H. Jeans
Answer: Gustav Kirchhoff
Explanation:
Gustav Kirchhoff introduced the term black body in 1860. He
studied about radiated thermal energies of substances.
Answer whether the following statement is true or false:
1. The emissivity of a perfect blackbody is a universal
constant at room temperature.
A. True
B. False
Answer:
False
Explanation:
The emissivity of a perfect blackbody is always equal to one
at a particular wavelength in thermal equilibrium conditions.
2. The value of Kirchhoff's constant is 2.8 x 10-3
mK.
A. True
B. False
Answer:
False
Explanation:
Kirchhoff didn't introduce any constant quantity to measure
absorbing and emissive powers of a material object. Instead, he utilized the
emissive power of the perfect blackbody, and it varies with radiation
wavelength in thermal equilibrium conditions.
3. An object with zero emissivity exists in the universe.
A. True
B. False
Answer:
True
Explanation:
By Kirchhoff's law, the emissivity of ordinary objects
varies between 0 to 1. So, there is a terrestrial object with zero emissivity.
4. The human body is a partial blackbody that can always be
in thermal equilibrium with its surrounding.
A. True
B. False
Answer:
False
Explanation:
A human body temperature varies approximately between 36 to
38 degrees centigrade. When the room temperature limit lies between 36 to 38
degrees Celsius, the human body is in thermal equilibrium with its
surroundings. Due to climate changes, some regions on earth exhibit drastic
temperature changes that the human body cannot acquire.
For example- In places like New Delhi, the room temperature
range can reach 4 or 5 degrees Celsius in the winter season. In such scenarios,
a human body cannot be thermal equilibrium with its surrounding.
5. Kirchhoff proved that a blackbody emits thermal radiation
continuously on heating.
A. True
B. False
Answer:
False
Explanation:
Kirchhoff's theory is silent about the mode of thermal
energy emissions on heating. But by classical physics, the blackbody radiation
emission is continuous. And it led to an ultraviolet catastrophe.
Reasoning questions on Kirchhoff's law of thermal radiation:
Read the assertion and explanation statements carefully. And pick the correct option from the following. The options are common to all
questions.
A. Both assertion and explanation are correct
B. Assertion is true. And the explanation is false
C. Assertion is false. And the explanation is correct.
D. Both assertion and explanation are false.
Question-1:
Assertion:
Kirchhoff experimented with black and white colored
identical metallic balls in the open air instead of a closed solid enclosure.
Explanation:
The black-colored sphere absorbs more thermal radiation than
the white ball.
Answer:
Both the assertion and explanation are correct
Clarification:
When the identical metallic balls are in an unenclosed free
space, they are not in thermal equilibrium with the surrounding. As a result,
the black ball becomes a superior thermal energy absorber than the white ball.
Question-2:
Assertion:
The emissivity of Lamp black is 0.98
Explanation:
It will increase with the rise in the blackbody enclosure's
temperature
Answer:
The assertion is true. And the explanation is false.
Clarification:
The emissivity of any substance is a fixed numerical value
specific to it. And it is independent of the surrounding temperature. But, the
emissive power of the material varies with the enclosure's temperature.
Question-3:
Assertion:
A graphite-coated closed rigid chamber absorbs sunlight at
room temperature conditions.
Explanation:
The sun will absorb emitted blackbody radiations from the
solid enclosure
Answer:
The assertion is true. And the explanation is false.
Clarification:
The graphite-coated closed solid enclosure acts as a partial
blackbody that can absorb a copious amount of sunlight at room temperature. And
it emits the same quantity of energy to the surrounding with which it is in
thermal equilibrium. Obviously, the rigid solid chamber cannot be in thermal
equilibrium with the sun at room temperature.
Question-4:
Assertion:
The emissive power of a perfect blackbody is a universal
constant in thermal equilibrium following Kirchhoff's law
Explanation:
It is specific to each radiation wavelength at constant
temperature conditions
Answer:
Both the assertion and explanation are correct
Clarification:
The emissive power of a perfect blackbody at temperature T
gives the energy radiated in a vacuum per unit of time per unit area. And it is
a universal constant value specific to each radiation wavelength.
Question-5:
Assertion:
The emissivity of a poor absorber is equal to its
absorptivity at temperature T.
Explanation:
Kirchhoff's law only applies to perfect thermal energy
absorbers.
Answer:
The assertion is true. And the explanation is false.
Clarification:
Kirchhoff's law applies to all terrestrial objects and aids
measure their absorbing and emissive powers. Following this law, the absorbing
power of a poor absorber is equal to its emissivity at thermal equilibrium.
Video description of Kirchhoff's law:
Gustav Kirchhoff's hypothetical blackbody absorbs all
wavelength radiations that fall on its surface at temperature T. Consequently,
Kirchhoff depicted it as a perfect absorber of radiant energies. Not only
absorption, but the unreal body emits all absorbed light on heating. And it
became an ideal emitter of thermal radiation. The first image of the video
describes the same.
As a result, Kirchhoff had used the blackbody's emissive
power in his mathematical formula as a universal constant to measure the terrestrial
object's absorbing and emissive powers.
Kirchhoff's first radiation formula:
eλ⁄aλ=EλWhere,
Eλ =
Emissive power of a perfect blackbody
eλ = Emissive power of an object
aλ = Absorbing power of an object
Except for blackbody, all other non-ideal objects absorb
specific wavelength light that depends upon their structure and composition.
Nevertheless, they emit all absorbed radiations on heating. Hence, their
absorbing powers and emissivities are equal in quantity at thermal equilibrium.
It follows the second thermodynamics law.
Kirchhoff's second radiation formula:
aλ = e
Where,
aλ = Absorbing power of an object at the wavelength
e = Emissivity of the object
Apart from this, Kirchhoff proved that a perfect radiation
absorber is a perfect emitter. Also, a poor absorber is a poor emitter.
In the video, the black-colored sphere is a superior
absorber, and the white-colored circle is a poor absorber of thermal energies.
Here is the video link to Kirchhoff's law. Just have a look at it for a better understanding.
Mind map of Kirchhoff’s law:
This mind map briefly outlines Kirchhoff law with the two
radiation formulas.
Kirchhoff formulated an empirical equation to determine the
emissive power of ordinary objects from the known values of their absorbing
capacity and the blackbody’s emissive power
A blackbody is an assumed hypothetical closed hollow solid
chamber with an opaque surface. And it is a complete absorber and emitter of
the whole electromagnetic spectrum.
Hence, the emissive power of any material object is equal to
the product of its absorbing ability and the blackbody’s emissive power at that
particular wavelength and temperature
Kirchhoff’s second radiation law determines an object’s
emissivity from the known values of its absorbing power.
And both are equal for every radiation wavelength at thermal
equilibrium. Hence, it obeys the second thermodynamics law.
Moreover, the perfect blackbody has the highest absorbing
power and emissivity value of 1.
Additionally, Kirchhoff’s radiation laws help to measure thermal energy exchanges of earthly bodies and are also helpful in spectroscopy.
Numerical problems of Kirchhoff's law:
1. A metal beaker has absorbing power of 500 and emissive
power of 250 watts per square meter at 30 degrees centigrade and 400 nm. What
is the emissive power of a blackbody at the same temperature and wavelength?
Answer:
Absorbing power of the metal beaker = 500
Emissive power of the metal beaker = 250 watts per square
meter
The formula to calculate the emissive power of a blackbody is;
eλ⁄aλ=EλEλ = 250/500 = 1/2 = 0.5 watts per square meter
2. A blackbody has emissive power of 270 watts per square
meter at 35 nm in thermal equilibrium. What is the absorbing power of a copper
ball if it has emissive power of 130 watts per square meter?
Answer:
Emissive power of blackbody = 270 watts per square meter
Emissive power of the copper ball = 130 watts per square
meter
The formula to calculate the absorbing power of copper ball
is;
aλ
= 130/270 = 0.48148
3. A sodium bulb has an absorptivity of 0.6 at 42 degrees
centigrade. What is its emissivity at the same temperature?
Answer:
The absorptivity of sodium bulb = 0.6
As the temperature of the sodium bulb is constant, it is in
a thermal equilibrium state. By using Kirchhoff’s law, we can say that;
The absorptivity and emissivity of a body are numerically
equal in thermal equilibrium conditions.
The emissivity of the sodium bulb = 0.6
4. What is the emissivity of an iron rod if it has an
emissive power of 2.25 watts per square meter? The emissive power of a
blackbody is 0.75 watts per square meter at the same temperature and wavelength
conditions.
Answer:
Emissive power of iron rod = 2.25 watts per square meter
Emissive power of blackbody = 0.75 watts per square meter
Emissivity of iron rod = 2.25/0.75 = 3
5. A gold piece has an emissivity of 0.625 at 1200 degrees
centigrade and 254 nm. Then what is the emissivity of a perfect blackbody at
the same temperature and wavelength conditions?
Answer:
A perfect blackbody is a good absorber and emitter of
electromagnetic light at all wavelengths in thermal equilibrium conditions. Its
emissivity and absorptivity values are equal to one.
Hence, the emissivity of a perfect blackbody at 1200 degrees
centigrade and at 254 nm is one.
6. The absorbing power of a metal cone is 5 at 450 nm in
room temperature conditions. And the emissive power of a perfect blackbody is
20 watts per square meter for the same wavelength and temperature. What is the
emissive power of the metal cone?
Answer:
Absorbing power of metal cone = 5
The emissive power of perfect blackbody = 20 watts per
square meter
The formula to calculate the emissive power of a metal cone
is
eλ
= 5 x 20 = 100 watts per square meter
PowerPoint notes of Kirchhoff's law:
Kirchhoff's law
streamlined thermal energy exchanges between the system and the surrounding
with an ideal hollow solid named blackbody. Further, it framed mathematical
relations to relate an ordinary object's emissive and absorbing abilities with
the perfect blackbody's emissive power. The PowerPoint presentation discusses
phrases connected with Kirchhoff's law together with thermal radiation
formulas.
Here is a link to the PowerPoint presentation of Kirchhoff's law.
Our e-book:
To download PowerPoint notes on Kirchhoff's law, visit our e-book store, "Jayam chemistry adda."
Match the following:
| Column-A | Column-B |
|---|---|
| 1. Charcoal | A. A fixed number |
| 2. Graphite | B. A perfect radiation emitter |
| 3. Emissive power | C. A poor thermal energy absorber |
| 4. Emissivity | D. Watts per square meter |
| 5. A hard hollow chamber | E. 0.97 |