The specific weight of water at ordinary pressure and temperature is 62.4 lb\/ft3 (9.81 kN\/m3). The specific gravity of mercury is 13.55. Compute the density of water and the specific weight and density of mercury<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The specific weight (\u03b3\\gamma) and density (\u03c1\\rho) are related by the equation: \u03b3=\u03c1\u22c5g\\gamma = \\rho \\cdot g<\/p>\n\n\n\n where \u03b3\\gamma is the specific weight, \u03c1\\rho is the density, and gg is the acceleration due to gravity. We will use the value of gg as 32.2\u2009ft\/s232.2 \\, \\text{ft\/s}^2 (standard in imperial units).<\/p>\n\n\n\n The density of water in imperial units can be calculated by rearranging the equation: \u03c1water=\u03b3waterg=62.4\u2009lb\/ft332.2\u2009ft\/s2\\rho_{\\text{water}} = \\frac{\\gamma_{\\text{water}}}{g} = \\frac{62.4 \\, \\text{lb\/ft}^3}{32.2 \\, \\text{ft\/s}^2}<\/p>\n\n\n\n This gives the density of water in lbm\/ft3\\text{lbm\/ft}^3, which is: \u03c1water=1.937\u2009lbm\/ft3\\rho_{\\text{water}} = 1.937 \\, \\text{lbm\/ft}^3<\/p>\n\n\n\n For SI units, the specific weight of water is 9.81\u2009kN\/m39.81 \\, \\text{kN\/m}^3, and using the standard gravitational acceleration g=9.81\u2009m\/s2g = 9.81 \\, \\text{m\/s}^2, the density of water is: \u03c1water=9.81\u2009kN\/m3\u00f79.81\u2009m\/s2=1\u2009kg\/m3\\rho_{\\text{water}} = 9.81 \\, \\text{kN\/m}^3 \\div 9.81 \\, \\text{m\/s}^2 = 1 \\, \\text{kg\/m}^3<\/p>\n\n\n\n Thus, the density of water is 62.4\u2009lbm\/ft362.4 \\, \\text{lbm\/ft}^3 or 1000\u2009kg\/m31000 \\, \\text{kg\/m}^3 in standard units.<\/p>\n\n\n\n Mercury’s specific gravity, SGHg=13.55SG_{\\text{Hg}} = 13.55, is the ratio of mercury’s density to water’s density. Since we know the density of water is 62.4\u2009lb\/ft362.4 \\, \\text{lb\/ft}^3, we can calculate the density of mercury using: \u03c1Hg=SGHg\u00d7\u03c1water=13.55\u00d762.4\u2009lbm\/ft3\\rho_{\\text{Hg}} = SG_{\\text{Hg}} \\times \\rho_{\\text{water}} = 13.55 \\times 62.4 \\, \\text{lbm\/ft}^3<\/p>\n\n\n\n Thus, \u03c1Hg=844.32\u2009lbm\/ft3\\rho_{\\text{Hg}} = 844.32 \\, \\text{lbm\/ft}^3<\/p>\n\n\n\n The specific weight of mercury in imperial units is: \u03b3Hg=\u03c1Hg\u00d7g=844.32\u2009lbm\/ft3\u00d732.2\u2009ft\/s2=27272.3\u2009lb\/ft3\\gamma_{\\text{Hg}} = \\rho_{\\text{Hg}} \\times g = 844.32 \\, \\text{lbm\/ft}^3 \\times 32.2 \\, \\text{ft\/s}^2 = 27272.3 \\, \\text{lb\/ft}^3<\/p>\n\n\n\n For SI units, using SGHg=13.55SG_{\\text{Hg}} = 13.55 and the density of water as 1000\u2009kg\/m31000 \\, \\text{kg\/m}^3, we find: \u03c1Hg=13.55\u00d71000=13,550\u2009kg\/m3\\rho_{\\text{Hg}} = 13.55 \\times 1000 = 13,550 \\, \\text{kg\/m}^3<\/p>\n\n\n\n The specific weight of mercury in SI units is: \u03b3Hg=\u03c1Hg\u00d7g=13,550\u2009kg\/m3\u00d79.81\u2009m\/s2=133,535.5\u2009N\/m3=133.5\u2009kN\/m3\\gamma_{\\text{Hg}} = \\rho_{\\text{Hg}} \\times g = 13,550 \\, \\text{kg\/m}^3 \\times 9.81 \\, \\text{m\/s}^2 = 133,535.5 \\, \\text{N\/m}^3 = 133.5 \\, \\text{kN\/m}^3<\/p>\n\n\n\n To compute these values, the relationship between specific weight and density is essential. Specific gravity is a dimensionless quantity representing the ratio of the density of a substance to the density of water. By knowing the specific gravity of mercury, we can calculate both its density and specific weight by using the density of water as a reference. Additionally, converting between imperial and SI units is straightforward once we establish the known conversion factors (e.g., 1 lb = 0.4536 kg, 1 ft = 0.3048 m, etc.).<\/p>\n","protected":false},"excerpt":{"rendered":" The specific weight of water at ordinary pressure and temperature is 62.4 lb\/ft3 (9.81 kN\/m3). The specific gravity of mercury is 13.55. Compute the density of water and the specific weight and density of mercury The Correct Answer and Explanation is : Problem Breakdown and Solution Given: Part 1: Density of Water The specific weight […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-182113","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182113","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=182113"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182113\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182113"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182113"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182113"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Problem Breakdown and Solution<\/h3>\n\n\n\n
Given:<\/h4>\n\n\n\n
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Part 1: Density of Water<\/strong><\/h4>\n\n\n\n
Part 2: Density and Specific Weight of Mercury<\/strong><\/h4>\n\n\n\n
Final Answers:<\/h3>\n\n\n\n
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Explanation:<\/h3>\n\n\n\n