{"id":182828,"date":"2025-01-15T05:24:01","date_gmt":"2025-01-15T05:24:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182828"},"modified":"2025-01-15T05:24:05","modified_gmt":"2025-01-15T05:24:05","slug":"calculate-the-specific-heat-of-solution-in-joules-for-dissolving-8-00-g-of-nh4no3-in-100ml-water-by-using-the-following","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/15\/calculate-the-specific-heat-of-solution-in-joules-for-dissolving-8-00-g-of-nh4no3-in-100ml-water-by-using-the-following\/","title":{"rendered":"Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following"},"content":{"rendered":"\n<p>Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following equation: 0=qrxn + qsoln. Given: mass of calorimeter cup:2.96g\u00c3\u201a\u00c2 Given: mass of water and calorimeter cup: 102.63g Given: mass of ammonium nitrate: 8.02g given : initial temperature of water :22.1 Celsius\u00c3\u201a\u00c2 Given: Delta T: 6.1 Celsius\u00c3\u201a\u00c2 Given: the coffe cup calorimeter absorbs no heat so qcal is negligible.\u00c3\u201a\u00c2 Given: The specific heat of the solution is the same as pure water which is 4.184J\/g Celsius. \u00c3\u201a\u00c2<br>Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following equation: 0=qrxn + qsoln.<\/p>\n\n\n\n<p>Given: mass of calorimeter cup:2.96g\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>Given: mass of water and calorimeter cup: 102.63g<\/p>\n\n\n\n<p>Given: mass of ammonium nitrate: 8.02g<\/p>\n\n\n\n<p>given : initial temperature of water :22.1 Celsius\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>Given: Delta T: 6.1 Celsius\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>Given: the coffe cup calorimeter absorbs no heat so qcal is negligible.\u00c3\u201a\u00c2<br>Given: The specific heat of the solution is the same as pure water which is 4.184J\/g Celsius.<\/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 heat of the solution for dissolving ( \\text{NH}_4\\text{NO}_3 ), let&#8217;s break it down step by step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understand the relationship<\/h3>\n\n\n\n<p>The total heat exchanged is:<br>[<br>q_{\\text{rxn}} + q_{\\text{soln}} = 0<br>]<br>Since ( q_{\\text{rxn}} ) is the heat released or absorbed during the reaction, it is equal in magnitude but opposite in sign to ( q_{\\text{soln}} ). Thus:<br>[<br>q_{\\text{rxn}} = -q_{\\text{soln}}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Define ( q_{\\text{soln}} )<\/h3>\n\n\n\n<p>[<br>q_{\\text{soln}} = m_{\\text{soln}} \\cdot c_{\\text{soln}} \\cdot \\Delta T<br>]<br>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( m_{\\text{soln}} ): Total mass of the solution (water + ( \\text{NH}_4\\text{NO}_3 ))<\/li>\n\n\n\n<li>( c_{\\text{soln}} ): Specific heat capacity (same as water, ( 4.184 \\, \\text{J\/g}^\\circ\\text{C} ))<\/li>\n\n\n\n<li>( \\Delta T ): Change in temperature (( 6.1^\\circ\\text{C} ))<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate total mass of solution<\/h3>\n\n\n\n<p>The total mass is:<br>[<br>m_{\\text{soln}} = \\text{mass of water} + \\text{mass of NH}<em>4\\text{NO}_3 ] Given: [ \\text{Mass of water} = \\text{Mass of water and calorimeter cup} &#8211; \\text{Mass of calorimeter cup} = 102.63 \\, \\text{g} &#8211; 2.96 \\, \\text{g} = 99.67 \\, \\text{g} ] [ m<\/em>{\\text{soln}} = 99.67 \\, \\text{g} + 8.02 \\, \\text{g} = 107.69 \\, \\text{g}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate ( q_{\\text{soln}} )<\/h3>\n\n\n\n<p>Substitute values into the equation:<br>[<br>q_{\\text{soln}} = 107.69 \\, \\text{g} \\cdot 4.184 \\, \\frac{\\text{J}}{\\text{g}^\\circ\\text{C}} \\cdot 6.1^\\circ\\text{C}<br>]<br>[<br>q_{\\text{soln}} = 2744.2 \\, \\text{J}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Calculate ( q_{\\text{rxn}} )<\/h3>\n\n\n\n<p>[<br>q_{\\text{rxn}} = -q_{\\text{soln}}<br>]<br>[<br>q_{\\text{rxn}} = -2744.2 \\, \\text{J}<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The solution\u2019s specific heat was assumed to be ( 4.184 \\, \\text{J\/g}^\\circ\\text{C} ), identical to water, per the problem statement.<\/li>\n\n\n\n<li>The total mass was calculated by adding the mass of water and ( \\text{NH}_4\\text{NO}_3 ).<\/li>\n\n\n\n<li>Using the formula ( q = mc\\Delta T ), we calculated the heat absorbed by the solution, which is positive because the solution absorbs heat.<\/li>\n\n\n\n<li>The reaction releases the same amount of heat but with a negative sign.<\/li>\n<\/ol>\n\n\n\n<p>Hence, the specific heat of the solution corresponds to the energy exchanged during the dissolution, which is ( q_{\\text{soln}} = 2744.2 \\, \\text{J} ).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the specific heat of solution in Joules for dissolving 8.00 g of NH4NO3 in 100mL water by using the following equation: 0=qrxn + qsoln. Given: mass of calorimeter cup:2.96g\u00c3\u201a\u00c2 Given: mass of water and calorimeter cup: 102.63g Given: mass of ammonium nitrate: 8.02g given : initial temperature of water :22.1 Celsius\u00c3\u201a\u00c2 Given: Delta T: [&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-182828","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182828","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=182828"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182828\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}