{"id":182117,"date":"2025-01-13T10:31:55","date_gmt":"2025-01-13T10:31:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182117"},"modified":"2025-01-13T10:31:58","modified_gmt":"2025-01-13T10:31:58","slug":"the-specific-weight-of-water-at-ordinary-pressure-and-temperature-is-62-4-lb-ft3-9-81-kn-m3-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/the-specific-weight-of-water-at-ordinary-pressure-and-temperature-is-62-4-lb-ft3-9-81-kn-m3-2\/","title":{"rendered":"The specific weight of water at ordinary pressure and temperature is 62.4 lb\/ft3 (9.81 kN\/m3 )"},"content":{"rendered":"\n<p>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. Calculate the density specific weight and density of mercury<br>The specific weight of water at ordinary pressure and<br>temperature is 62.4 lb\/ft3 (9.81 kN\/m3<br>). The specific<br>gravity of mercury is 13.55. Calculate the density<br>specific weight and density of mercury<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the specific weight and density of mercury based on the given data, we need to use the concept of <strong>specific gravity<\/strong> and the specific weight of water.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Specific Weight of Mercury<\/strong><\/h3>\n\n\n\n<p><strong>Specific weight<\/strong> ((\\gamma)) is the weight per unit volume and is given by the equation:<\/p>\n\n\n\n<p>[<br>\\gamma = \\text{Specific Gravity} \\times \\gamma_{\\text{water}}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\gamma_{\\text{water}}) is the specific weight of water.<\/li>\n\n\n\n<li>The specific gravity (SG) of mercury is given as <strong>13.55<\/strong>.<\/li>\n\n\n\n<li>The specific weight of water at ordinary temperature and pressure is <strong>62.4 lb\/ft\u00b3<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Now, we can calculate the specific weight of mercury:<\/p>\n\n\n\n<p>[<br>\\gamma_{\\text{mercury}} = 13.55 \\times 62.4 \\, \\text{lb\/ft}^3<br>]<\/p>\n\n\n\n<p>[<br>\\gamma_{\\text{mercury}} = 844.92 \\, \\text{lb\/ft}^3<br>]<\/p>\n\n\n\n<p>Thus, the <strong>specific weight of mercury<\/strong> is approximately <strong>844.92 lb\/ft\u00b3<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Density of Mercury<\/strong><\/h3>\n\n\n\n<p>The <strong>density<\/strong> ((\\rho)) is defined as the mass per unit volume and is related to the specific weight by the equation:<\/p>\n\n\n\n<p>[<br>\\rho = \\frac{\\gamma}{g}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\gamma) is the specific weight.<\/li>\n\n\n\n<li>(g) is the acceleration due to gravity, which is <strong>32.174 ft\/s\u00b2<\/strong> in imperial units.<\/li>\n<\/ul>\n\n\n\n<p>Substituting the values:<\/p>\n\n\n\n<p>[<br>\\rho_{\\text{mercury}} = \\frac{844.92 \\, \\text{lb\/ft}^3}{32.174 \\, \\text{ft\/s}^2}<br>]<\/p>\n\n\n\n<p>[<br>\\rho_{\\text{mercury}} \\approx 26.23 \\, \\text{lb\/ft}^3<br>]<\/p>\n\n\n\n<p>Thus, the <strong>density of mercury<\/strong> is approximately <strong>26.23 lb\/ft\u00b3<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Density in SI Units<\/strong><\/h3>\n\n\n\n<p>To convert this to SI units (kg\/m\u00b3), we use the conversion factor 1 lb\/ft\u00b3 = 16.0185 kg\/m\u00b3.<\/p>\n\n\n\n<p>[<br>\\rho_{\\text{mercury}} = 26.23 \\, \\text{lb\/ft}^3 \\times 16.0185 \\, \\text{kg\/m}^3<br>]<\/p>\n\n\n\n<p>[<br>\\rho_{\\text{mercury}} \\approx 420.8 \\, \\text{kg\/m}^3<br>]<\/p>\n\n\n\n<p>Thus, the <strong>density of mercury<\/strong> is approximately <strong>420.8 kg\/m\u00b3<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Specific weight of mercury<\/strong>: 844.92 lb\/ft\u00b3<\/li>\n\n\n\n<li><strong>Density of mercury<\/strong>: 26.23 lb\/ft\u00b3 (or 420.8 kg\/m\u00b3)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Specific gravity<\/strong> is the ratio of the density of a substance to the density of water. Mercury\u2019s specific gravity is given as 13.55, meaning it is 13.55 times denser than water.<\/li>\n\n\n\n<li>The specific weight is the product of the specific gravity and the specific weight of water.<\/li>\n\n\n\n<li>To calculate the density from the specific weight, we divide the specific weight by the acceleration due to gravity.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>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. Calculate the density specific weight and density of mercuryThe specific weight of water at ordinary pressure andtemperature is 62.4 lb\/ft3 (9.81 kN\/m3). The specificgravity of mercury is 13.55. Calculate the densityspecific weight [&hellip;]<\/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-182117","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182117","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=182117"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182117\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182117"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182117"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}