{"id":220932,"date":"2025-05-29T04:08:52","date_gmt":"2025-05-29T04:08:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220932"},"modified":"2025-05-29T04:08:54","modified_gmt":"2025-05-29T04:08:54","slug":"the-molar-heat-capacity-of-liquid-water-is-75-3-j-mol-k-if-37-5-g-of-water-is-cooled-from-42-0-to-7-0c-what-is-q-for-the-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/the-molar-heat-capacity-of-liquid-water-is-75-3-j-mol-k-if-37-5-g-of-water-is-cooled-from-42-0-to-7-0c-what-is-q-for-the-water\/","title":{"rendered":"The molar heat capacity of liquid water is 75.3 J\/mol K. If 37.5 g of water is cooled from 42.0 to 7.0\u00b0C, what is q for the water"},"content":{"rendered":"\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-501.png\" alt=\"\" class=\"wp-image-220933\"\/><\/figure>\n\n\n\n<p>The molar heat capacity of liquid water is 75.3 J\/mol K. If 37.5 g of water is cooled from 42.0 to 7.0\u00b0C, what is q for the water?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>-98.83 kJ<\/li>\n\n\n\n<li>5.49 J<\/li>\n\n\n\n<li>98.83 J<\/li>\n\n\n\n<li>-5.49 kJ<\/li>\n<\/ul>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we use the formula for heat transfer: q=n\u22c5Cm\u22c5\u0394Tq = n \\cdot C_m \\cdot \\Delta T<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>qq is the heat (in joules),<\/li>\n\n\n\n<li>nn is the number of moles of water,<\/li>\n\n\n\n<li>CmC_m is the <strong>molar heat capacity<\/strong> of water (given as 75.3 J\/mol\u00b7K),<\/li>\n\n\n\n<li>\u0394T\\Delta T is the change in temperature (final &#8211; initial).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate moles of water:<\/strong> Mass\u00a0of\u00a0water=37.5\u00a0g,Molar\u00a0mass\u00a0of\u00a0water=18.015\u00a0g\/mol\\text{Mass of water} = 37.5\\ \\text{g},\\quad \\text{Molar mass of water} = 18.015\\ \\text{g\/mol} n=37.5\u00a0g18.015\u00a0g\/mol\u22482.082\u00a0moln = \\frac{37.5\\ \\text{g}}{18.015\\ \\text{g\/mol}} \\approx 2.082\\ \\text{mol}<\/li>\n\n\n\n<li><strong>Calculate temperature change:<\/strong> \u0394T=Tfinal\u2212Tinitial=7.0\u2218C\u221242.0\u2218C=\u221235.0\u2218C=\u221235.0\u00a0K\\Delta T = T_{\\text{final}} &#8211; T_{\\text{initial}} = 7.0^\\circ C &#8211; 42.0^\\circ C = -35.0^\\circ C = -35.0\\ \\text{K}<\/li>\n\n\n\n<li><strong>Calculate heat qq:<\/strong> q=2.082\u00a0mol\u22c575.3\u00a0J\/mol\\cdotpK\u22c5(\u221235.0\u00a0K)=\u22125491.6\u00a0Jq = 2.082\\ \\text{mol} \\cdot 75.3\\ \\text{J\/mol\u00b7K} \\cdot (-35.0\\ \\text{K}) = -5491.6\\ \\text{J}<\/li>\n\n\n\n<li><strong>Convert to kilojoules:<\/strong> q=\u22125491.6\u00a0J=\u22125.49\u00a0kJq = -5491.6\\ \\text{J} = -5.49\\ \\text{kJ}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>\u2714\ufe0f -5.49 kJ<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The water is <strong>cooling down<\/strong>, which means it is <strong>losing heat<\/strong>, so qq is negative. We used the <strong>molar heat capacity<\/strong> because the question provides it in units of J\/mol\u00b7K. By converting the mass of water to moles and multiplying by the molar heat capacity and temperature change, we determine the total energy lost by the water during the cooling process.<\/p>\n\n\n\n<p>Among the choices:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u201c-5.49 kJ\u201d is the correct value and sign,<\/li>\n\n\n\n<li>The other options are either in the wrong units (J instead of kJ), have the wrong magnitude, or have the wrong sign.<\/li>\n<\/ul>\n\n\n\n<p>This calculation is crucial in thermodynamics and chemistry for understanding how substances absorb or release energy with temperature changes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner5-101.jpeg\" alt=\"\" class=\"wp-image-220934\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The molar heat capacity of liquid water is 75.3 J\/mol K. If 37.5 g of water is cooled from 42.0 to 7.0\u00b0C, what is q for the water? The Correct Answer and Explanation is: To solve this problem, we use the formula for heat transfer: q=n\u22c5Cm\u22c5\u0394Tq = n \\cdot C_m \\cdot \\Delta T Where: Step-by-step: [&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-220932","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220932","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=220932"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220932\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220932"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220932"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}