{"id":181711,"date":"2025-01-11T09:37:10","date_gmt":"2025-01-11T09:37:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181711"},"modified":"2025-01-11T09:37:12","modified_gmt":"2025-01-11T09:37:12","slug":"the-freezing-point-of-mercury-is-38-8-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/11\/the-freezing-point-of-mercury-is-38-8-c\/","title":{"rendered":"The freezing point of mercury is -38.8 \u00b0C"},"content":{"rendered":"\n<p>The freezing point of mercury is -38.8 \u00b0C. What quantity of energy, in joules, is released to the surroundings if 1.00 mL of mercury is cooled from 23.0\u00b0C to -38.8\u00b0C and then frozen to a solid? (The density of liquid mercury is 13.6 g\/cm3. Its specific heat capacity is 0.140 J\/g\u00b7K and its heat of fusion is 11.4 J\/g\/)<\/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 find the quantity of energy released to the surroundings, we need to calculate the energy released in two stages: first, when mercury is cooled from 23.0\u00b0C to its freezing point (-38.8\u00b0C), and then when it freezes into a solid at that temperature.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Energy released during cooling<\/h3>\n\n\n\n<p>The energy released when a substance is cooled is given by the formula:<\/p>\n\n\n\n<p>[<br>q = m \\cdot c \\cdot \\Delta T<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(q) is the heat energy released (in joules),<\/li>\n\n\n\n<li>(m) is the mass of mercury (in grams),<\/li>\n\n\n\n<li>(c) is the specific heat capacity of mercury (in J\/g\u00b7K),<\/li>\n\n\n\n<li>(\\Delta T) is the change in temperature (in \u00b0C or K).<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Find the mass of mercury<\/h4>\n\n\n\n<p>The volume of mercury is given as 1.00 mL, and the density of mercury is 13.6 g\/cm\u00b3. Since the density of mercury is 13.6 g\/mL, the mass (m) is:<\/p>\n\n\n\n<p>[<br>m = \\text{density} \\times \\text{volume} = 13.6 \\, \\text{g\/mL} \\times 1.00 \\, \\text{mL} = 13.6 \\, \\text{g}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Change in temperature<\/h4>\n\n\n\n<p>The temperature change (\\Delta T) is the difference between the initial temperature (23.0\u00b0C) and the freezing point (-38.8\u00b0C):<\/p>\n\n\n\n<p>[<br>\\Delta T = 23.0\u00b0C &#8211; (-38.8\u00b0C) = 23.0 + 38.8 = 61.8 \\, \\text{\u00b0C}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Calculate the heat released during cooling<\/h4>\n\n\n\n<p>Now, we can calculate the energy released during cooling:<\/p>\n\n\n\n<p>[<br>q = m \\cdot c \\cdot \\Delta T = 13.6 \\, \\text{g} \\cdot 0.140 \\, \\text{J\/g\u00b7K} \\cdot 61.8 \\, \\text{K}<br>]<\/p>\n\n\n\n<p>[<br>q = 13.6 \\cdot 0.140 \\cdot 61.8 = 118.8 \\, \\text{J}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Energy released during freezing<\/h3>\n\n\n\n<p>The energy released during freezing is calculated using the heat of fusion, given by:<\/p>\n\n\n\n<p>[<br>q = m \\cdot \\Delta H_f<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(\\Delta H_f) is the heat of fusion of mercury (11.4 J\/g).<\/li>\n<\/ul>\n\n\n\n<p>[<br>q = 13.6 \\, \\text{g} \\cdot 11.4 \\, \\text{J\/g} = 154.4 \\, \\text{J}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Total energy released<\/h3>\n\n\n\n<p>The total energy released to the surroundings is the sum of the energy released during cooling and the energy released during freezing:<\/p>\n\n\n\n<p>[<br>q_{\\text{total}} = 118.8 \\, \\text{J} + 154.4 \\, \\text{J} = 273.2 \\, \\text{J}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer<\/h3>\n\n\n\n<p>The total energy released to the surroundings is <strong>273.2 J<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>To calculate the total energy released when mercury is cooled and frozen, we used two key concepts:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Heat transfer during cooling<\/strong>: The heat energy released when a substance is cooled is determined by its mass, specific heat capacity, and the change in temperature.<\/li>\n\n\n\n<li><strong>Latent heat during freezing<\/strong>: The energy released during phase change (from liquid to solid) is determined by the substance&#8217;s heat of fusion, which represents the amount of energy required to freeze 1 gram of a substance.<\/li>\n<\/ol>\n\n\n\n<p>These calculations account for both the cooling of the mercury to its freezing point and its subsequent phase change to a solid. The total energy released is the sum of these two processes.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The freezing point of mercury is -38.8 \u00b0C. What quantity of energy, in joules, is released to the surroundings if 1.00 mL of mercury is cooled from 23.0\u00b0C to -38.8\u00b0C and then frozen to a solid? (The density of liquid mercury is 13.6 g\/cm3. Its specific heat capacity is 0.140 J\/g\u00b7K and its heat of [&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-181711","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181711","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=181711"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181711\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181711"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181711"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}