MCQs of blackbody & Kirchhoff's law

1. By, Jayam chemistry adda BLACKBODY & KIRCHHOFF'S LAW It discusses multiple choice questions with clarified answer descriptions.

2. QUESTION-1 Answer What is the essence of Kirchhoff's law? Good absorbers are good radiatiors Explanation The blackbody absorbs 100% incident Excellent absorbers are good reflectors. 1 light falling on its surface. Hence, a perfect blackbody is an ideal absorber of Poor reflectors are poor emitters. 2 thermal energy. And on heating, it emits all absorbed radiations. Good absorbers are good radiators. 3 Visit: Jayam chemistry adda

3. QUESTION-2 A solid enclosure with the lamp black coated surface is at a temperature of T in the dark room. What is the maximum energy emitted by the blackbody at the wavelength λ on heating? Answer It varies inversely with the temperature of the enclosure Explanation The energy density of radiated It varies directly with the temperature of the enclosure 1 blackbody radiations depends only on the enclosure's temperature. The It varies inversely with the temperature of the enclosure 2 radiation wavelength emitted with maximum intensity is inversely It is independent of the enclosure's temperature 3 proportional to the body's absolute temperature by Wien's law. Visit: Jayam chemistry adda

4. QUESTION-3 A Mercury absorption spectrum shows a blue spectral line at 461 nm in thermal equilibrium conditions. Then what is the wavelength of the blue emission line in the Mercury emission spectrum? Answer 461 nm Explanation The position of the spectral line in both the absorption and emission 442 nm 1 spectrums of Mercury is the same in thermal equilibrium conditions 2 221 nm following Kirchhoff's law. Emissivity=absorptivity (at a fixed 461 nm 3 wavelength and temperature) Visit: Jayam chemistry adda

5. QUESTION-4 Answer Why decorated china bowls glow in the darkroom when hot? The dark-colored decorations of china bowls emit thermal radiation in the darkroom after heating Explanation A decorated china utensil on heating to a China bowls decoration seems to shine in the darkroom 1 high temperature and then taken to a darkroom glows brightly. The dark-colored The dark-colored decorations of china bowls emit thermal radiation in the darkroom after heating. 2 bowl decorations are excellent heat absorbers resembling blackbodies. And The material used to make china bowls glows in the darkroom 3 they emit all absorbed thermal radiations that shine brightly in the darkroom. Visit: Jayam chemistry adda

6. QUESTION-5 Answer Why is daytime hot in desert regions? Desert sand is a typical blackbody that is a good heat absorber. Explanation Desert sand evaporates water quickly, so we feel hot. Desert sand absorbs heat to a large 1 extent. Hence, it is hot in the daytime in deserts. At night time, the sand radiates Desert sand is a typical blackbody that is a good heat absorber. 2 all absorbed heat to the atmosphere. The desert fills with sand. And it is a good reflector of heat due to the greenhouse effect. Consequently, the nighttime is cold in 3 deserts. It follows Kirchhoff's law. Visit: Jayam chemistry adda

7. QUESTION-6 By using Kirchhoff's law of thermal radiations, solve this problem. Answer What is the absorbing power of a blackbody at 45 degrees Celsius temperature and 5400 nm wavelength if it has an emissive power of 4.6284 watts per square meter? 1 Explanation A perfect blackbody absorbing 2.334 1 power is equal to one at a particular wavelength in thermal equilibrium 2 1.298 conditions. 1 3 Visit: Jayam chemistry adda

8. QUESTION-7 A rigid rectangular box has two metallic balls at 30 degrees Celsius temperature. Among them, one is white, and the other is black colored. The wavelength of incident light for the rectangular box is 910 nm. What is the absorbing power of a white ball if its emissive power is 120 watts per square meter? And the emissive power of the black ball is 600 watts per square meter. Answer 0.2 Explanation Emissive power of the white ball = 120 watts per square meter 2 1 Emissive power of the black ball = 600 watts per square meter 2 2.22 Formula to calculate the absorbing power of the white ball is; 0.2 3 aλ = eλ/Eλ aλ = 120/600 = 0.2 Visit: Jayam chemistry adda

9. QUESTION-8 Consider a tiny iron ball in a blackbody enclosure. The temperature of the iron ball and the blackbody enclosure is separate. What is the direction of heat flow? Answer Heat flows from a hot body to a cold body Explanation The temperature of the blackbody Heat flows from a hot body to a cold body. 1 enclosure and the iron ball is different, which means they are not in thermal Heat flows from a low to the higher temperature body 2 equilibrium conditions. So, heat will flow from a higher temperature to a lower Heat does not flow in thermal equilibrium conditions. 3 temperature body. Visit: Jayam chemistry adda

10. QUESTION-9 The following are examples of hot bodies identical in every respect in thermal equilibrium conditions. Which one among them cools last? Answer A solid sphere Explanation The rate of thermal emissions of a A solid cube 1 blackbody varies directly with its surface area. Objects with a large surface area A solid sphere 2 have high emissive power. The solid sphere has less surface area than the 3 A solid cone other hot bodies. Hence, it cools last. Visit: Jayam chemistry adda

11. QUESTION-10 Answer What is the number of photons of light radiation when its frequency is 2 Hz? And the radiant energy is 39.75 joule. Explanation The light radiation frequency = 2 Hz Formula to calculate number of photons of light is 1 E=nhν 2 29 ∞ 3 Visit: Jayam chemistry adda

12. QUESTION-11 Answer What happens when heating a shining silver pot with black spots on its surface to a high temperature, then cooling it in a darkroom? The silver pot seems dull or invisible. The black spots shine dazzlingly. Explanation The glowing silver pot on heating to a high temperature will become a poor absorber of thermal energy. At the same time, the dark The silver pot becomes bright after heating, and the black spots become dull. 1 spots are effective heat absorbers and take up most of the supplied heat. While cooling The silver pot seems dull or invisible. The black spots shine dazzlingly. 2 them in the dark room, the black specks shine luminously by emitting all absorbed Both silver metal and dark spots turn bright red on heating. 3 heat radiations. They act as blackbodies following Kirchhoff's law. Visit: Jayam chemistry adda

13. QUESTION-12 Answer A cricketer wearing a black dress feels more heat in the sun time than the one in a white dress. Why? Black color dress is a good absorber of heat Explanation Black color dress is a good absorber of heat. In sunny weather, the cricketer wearing a Black color dress is a good absorber of heat. 1 black dress feels more heat due to thermal radiation absorption. While on the contrary, It is false. Both cricketers feel the same heat. 2 white color is a good reflector of heat and reflects off all thermal radiation that falls on The white dress absorbs heat and keeps him chill and fresh. 3 it. So, the white dress cricketer feels less heat than the other. It follows Kirchhoff’s law. Visit: Jayam chemistry adda

14. QUESTION-13 Answer Which surfaces are good reflectors of infrared radiations? Polished surfaces Explanation Shiny surfaces are poor absorbers and emitters of thermal radiations following Polished surfaces 1 Kirchhoff's law. Due to their small surface area, they became poor absorbers. As a result, 2 Rough surfaces good reflectors of heat radiation. They reflect off most of the radiations that fall on their Black coarse surfaces 3 surface. So, they are good reflectors of infrared radiations. Visit: Jayam chemistry adda

15. QUESTION-14 Answer The morning sun is not as hot as the midday sun. Why? The sun's rays travel a longer distance through the atmosphere in the morning. Explanation By inverse square law, the intensity of the thermal radiation is inversely proportional to Heat rays travel slowly in the morning. 1 the square of the distance between the light The sun's rays travel a longer distance through the atmosphere in the morning. source and the observer. Thermal radiations 2 are less intense for distant emitting sources. As the sun's rays travel a longer distance through The earth is farther away from the sun in the morning. 3 the atmosphere in the morning are less hot than the midday sunlight. Visit: Jayam chemistry adda

16. QUESTION-15 The emissive powers of a gold foil and the blackbody are 3.5 and 4.5 watts per square meter at 67 degrees centigrade and 908 nm. What is the emissivity of the gold foil? Answer 0.77 Explanation The emissive power of a gold foil=3.5 watts per square meter The emissive power of the blackbody=4.5 1 1 watts per square meter The formula to calculate the emissivity of 2 0.67 the gold foil is; Emissivity=emissive power of the gold 0.77 3 foil/emissive power of the blackbody Emissivity=3.5/4.5=0.77 Visit: Jayam chemistry adda

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