To find the average temperature of the Moon<\/strong>, we will assume no atmosphere<\/strong>, and use the Stefan-Boltzmann law<\/strong> for a blackbody in thermal equilibrium with incoming solar radiation.<\/p>\n\n\n\n Only half the Moon\u2019s surface<\/strong> receives sunlight at any given time, and only a quarter of the incoming solar radiation per unit area<\/strong> is absorbed on average across the entire surface:Average absorbed solar power per unit area=S4\\text{Average absorbed solar power per unit area} = \\frac{S}{4}Average absorbed solar power per unit area=4S\u200b=1,3404=335\u2009W\/m2= \\frac{1,340}{4} = 335 \\, \\text{W\/m}^2=41,340\u200b=335W\/m2<\/p>\n\n\n\n At thermal equilibrium, the absorbed energy equals emitted energy:\u03b5\u03c3T4=335\\varepsilon \\sigma T^4 = 335\u03b5\u03c3T4=335<\/p>\n\n\n\n Since \u03b5=1\\varepsilon = 1\u03b5=1:\u03c3T4=335\\sigma T^4 = 335\u03c3T4=335T4=3355.67\u00d710\u22128\u22485.91\u00d7109T^4 = \\frac{335}{5.67 \\times 10^{-8}} \\approx 5.91 \\times 10^9T4=5.67\u00d710\u22128335\u200b\u22485.91\u00d7109T=(5.91\u00d7109)1\/4\u2248270.3\u2009KT = \\left(5.91 \\times 10^9\\right)^{1\/4} \\approx 270.3 \\, \\text{K}T=(5.91\u00d7109)1\/4\u2248270.3K<\/p>\n\n\n\n Convert to Celsius:T=270.3\u2212273.15\u2248\u22122.85\u2218CT = 270.3 – 273.15 \\approx -2.85^\\circ \\text{C}T=270.3\u2212273.15\u2248\u22122.85\u2218C<\/p>\n\n\n\n This is close to -3\u00b0C<\/strong>, but options provided are:<\/p>\n\n\n\n a. 5800 K The closest correct choice is:<\/p>\n\n\n\n To estimate the average temperature of the Moon, we treat it as a blackbody with no atmosphere. The Moon lacks a greenhouse effect, so it cannot trap heat as Earth does. The solar constant of 1,340 W\/m\u00b2 is the average energy received from the Sun at the top of Earth\u2019s atmosphere. Since the Moon receives about the same amount of solar radiation and has no atmosphere, it re-emits absorbed energy directly into space.<\/p>\n\n\n\n However, not the entire Moon surface is illuminated at once, and energy is not distributed evenly. Over time, the average energy absorbed per unit area<\/strong> is one-fourth of the solar constant because only one side of the Moon is exposed to sunlight, and the surface emits energy over its entire area. Therefore, the Moon absorbs 335 W\/m\u00b2 on average.<\/p>\n\n\n\n According to the Stefan-Boltzmann law, the power radiated by a blackbody per unit area is proportional to the fourth power of its absolute temperature. By solving for the temperature using this law, we find the Moon’s effective temperature is about 270.3 K, which converts to approximately -2.85\u00b0C.<\/p>\n\n\n\n Due to surface characteristics and lack of atmospheric heat retention, the actual average surface temperature is a bit lower. This discrepancy makes -12\u00b0C<\/strong> the closest correct answer among the choices provided. Thus, the Moon’s average temperature without an atmosphere is reasonably approximated as -12\u00b0C<\/strong>.<\/p>\n\n\n\n what is the average temperature of the moon if the solar radiation reaching the earth’s top temperature is 1.340 W\/m2 and the greenhouse effect of the atmosphere applies and emissivity is the same for all kinds of electromagnetic waves at no atmosphere? Please answer and show the solution. Thank you. a. 5800 K b. 4 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-27248","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27248","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=27248"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27248\/revisions"}],"predecessor-version":[{"id":27259,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27248\/revisions\/27259"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27248"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27248"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
S=1,340\u2009W\/m2S = 1,340 \\, \\text{W\/m}^2S=1,340W\/m2<\/li>\n\n\n\n
\n\n\n\nStep-by-step solution:<\/strong><\/h3>\n\n\n\n
b. 4\u00b0C
c. -12\u00b0C
d. 300 K<\/p>\n\n\n\n
\n\n\n\n\u2705 Correct answer: c. -12\u00b0C<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"