{"id":258083,"date":"2025-07-17T21:32:03","date_gmt":"2025-07-17T21:32:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258083"},"modified":"2025-07-17T21:32:05","modified_gmt":"2025-07-17T21:32:05","slug":"four-ice-cubes-at-exactly-0-%e2%88%98c-with-a-total-mass-of-54-0-g-are-combined-with-140-g-of-water-at-90-%e2%88%98c-in-an-insulated-container","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/four-ice-cubes-at-exactly-0-%e2%88%98c-with-a-total-mass-of-54-0-g-are-combined-with-140-g-of-water-at-90-%e2%88%98c-in-an-insulated-container\/","title":{"rendered":"Four ice cubes at exactly 0 \u2218C with a total mass of 54.0 g are combined with 140 g of water at 90 \u2218C in an insulated container"},"content":{"rendered":"\n<p>Four ice cubes at exactly 0 \u2218C with a total mass of 54.0 g are combined with 140 g of water at 90 \u2218C in an insulated container. (\u0394H\u2218fus =6.02 kJ\/mol , cwater =4.18J\/g\u22c5\u2218C ) If no heat is lost to the surroundings, what is the final temperature of the mixture?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we need to consider the energy balance in the system, where heat lost by the hot water is equal to the heat gained by the ice cubes as they melt and warm up to the final temperature.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Heat required to melt the ice cubes<\/h3>\n\n\n\n<p>We begin by calculating how much heat is required to melt the ice cubes. The heat required to melt ice is given by the formula:Qmelt=n\u22c5\u0394HfusQ_{\\text{melt}} = n \\cdot \\Delta H_{\\text{fus}}Qmelt\u200b=n\u22c5\u0394Hfus\u200b<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nnn is the number of moles of ice,<\/li>\n\n\n\n<li>\u0394Hfus\\Delta H_{\\text{fus}}\u0394Hfus\u200b is the molar heat of fusion (6.02 kJ\/mol).<\/li>\n<\/ul>\n\n\n\n<p>First, we calculate the number of moles of ice:n=miceMice=54.0\u2009g18.0\u2009g\/mol=3.00\u2009moln = \\frac{m_{\\text{ice}}}{M_{\\text{ice}}} = \\frac{54.0 \\, \\text{g}}{18.0 \\, \\text{g\/mol}} = 3.00 \\, \\text{mol}n=Mice\u200bmice\u200b\u200b=18.0g\/mol54.0g\u200b=3.00mol<\/p>\n\n\n\n<p>So, the heat required to melt the ice is:Qmelt=3.00\u2009mol\u00d76.02\u2009kJ\/mol=18.06\u2009kJ=18060\u2009JQ_{\\text{melt}} = 3.00 \\, \\text{mol} \\times 6.02 \\, \\text{kJ\/mol} = 18.06 \\, \\text{kJ} = 18060 \\, \\text{J}Qmelt\u200b=3.00mol\u00d76.02kJ\/mol=18.06kJ=18060J<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Heat required to warm the melted water<\/h3>\n\n\n\n<p>After the ice melts, it will warm up from 0 \u2218C to the final temperature TfT_fTf\u200b. The heat required for this is:Qwarm=mwater\u22c5cwater\u22c5\u0394TQ_{\\text{warm}} = m_{\\text{water}} \\cdot c_{\\text{water}} \\cdot \\Delta TQwarm\u200b=mwater\u200b\u22c5cwater\u200b\u22c5\u0394T<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>mwater=54.0\u2009gm_{\\text{water}} = 54.0 \\, \\text{g}mwater\u200b=54.0g (mass of the melted ice),<\/li>\n\n\n\n<li>cwater=4.18\u2009J\/g\u22c5\u2218Cc_{\\text{water}} = 4.18 \\, \\text{J\/g\u22c5\u2218C}cwater\u200b=4.18J\/g\u22c5\u2218C,<\/li>\n\n\n\n<li>\u0394T=Tf\u22120\u2009\u2218C\\Delta T = T_f &#8211; 0 \\, \\text{\u2218C}\u0394T=Tf\u200b\u22120\u2218C.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the heat required to warm the melted water is:Qwarm=54.0\u2009g\u00d74.18\u2009J\/g\u22c5\u2218C\u00d7Tf=225.72\u2009Tf\u2009JQ_{\\text{warm}} = 54.0 \\, \\text{g} \\times 4.18 \\, \\text{J\/g\u22c5\u2218C} \\times T_f = 225.72 \\, T_f \\, \\text{J}Qwarm\u200b=54.0g\u00d74.18J\/g\u22c5\u2218C\u00d7Tf\u200b=225.72Tf\u200bJ<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Heat lost by the hot water<\/h3>\n\n\n\n<p>Now, the hot water will cool down from 90 \u2218C to the final temperature TfT_fTf\u200b. The heat lost by the hot water is:Qlost=mhot&nbsp;water\u22c5cwater\u22c5\u0394TQ_{\\text{lost}} = m_{\\text{hot water}} \\cdot c_{\\text{water}} \\cdot \\Delta TQlost\u200b=mhot&nbsp;water\u200b\u22c5cwater\u200b\u22c5\u0394T<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>mhot\u00a0water=140\u2009gm_{\\text{hot water}} = 140 \\, \\text{g}mhot\u00a0water\u200b=140g,<\/li>\n\n\n\n<li>cwater=4.18\u2009J\/g\u22c5\u2218Cc_{\\text{water}} = 4.18 \\, \\text{J\/g\u22c5\u2218C}cwater\u200b=4.18J\/g\u22c5\u2218C,<\/li>\n\n\n\n<li>\u0394T=90\u2009\u2218C\u2212Tf\\Delta T = 90 \\, \\text{\u2218C} &#8211; T_f\u0394T=90\u2218C\u2212Tf\u200b.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the heat lost by the hot water is:Qlost=140\u2009g\u00d74.18\u2009J\/g\u22c5\u2218C\u00d7(90\u2212Tf)=585.2\u2009(90\u2212Tf)\u2009JQ_{\\text{lost}} = 140 \\, \\text{g} \\times 4.18 \\, \\text{J\/g\u22c5\u2218C} \\times (90 &#8211; T_f) = 585.2 \\, (90 &#8211; T_f) \\, \\text{J}Qlost\u200b=140g\u00d74.18J\/g\u22c5\u2218C\u00d7(90\u2212Tf\u200b)=585.2(90\u2212Tf\u200b)J<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Energy balance equation<\/h3>\n\n\n\n<p>According to the principle of conservation of energy, the heat lost by the hot water equals the heat gained by the ice cubes. So, we can write the energy balance as:Qlost=Qmelt+QwarmQ_{\\text{lost}} = Q_{\\text{melt}} + Q_{\\text{warm}}Qlost\u200b=Qmelt\u200b+Qwarm\u200b<\/p>\n\n\n\n<p>Substituting the expressions we derived:585.2\u2009(90\u2212Tf)=18060+225.72\u2009Tf585.2 \\, (90 &#8211; T_f) = 18060 + 225.72 \\, T_f585.2(90\u2212Tf\u200b)=18060+225.72Tf\u200b<\/p>\n\n\n\n<p>Expanding both sides:585.2\u00d790\u2212585.2\u2009Tf=18060+225.72\u2009Tf585.2 \\times 90 &#8211; 585.2 \\, T_f = 18060 + 225.72 \\, T_f585.2\u00d790\u2212585.2Tf\u200b=18060+225.72Tf\u200b52668\u2212585.2\u2009Tf=18060+225.72\u2009Tf52668 &#8211; 585.2 \\, T_f = 18060 + 225.72 \\, T_f52668\u2212585.2Tf\u200b=18060+225.72Tf\u200b<\/p>\n\n\n\n<p>Rearranging terms to solve for TfT_fTf\u200b:52668\u221218060=585.2\u2009Tf+225.72\u2009Tf52668 &#8211; 18060 = 585.2 \\, T_f + 225.72 \\, T_f52668\u221218060=585.2Tf\u200b+225.72Tf\u200b34608=810.92\u2009Tf34608 = 810.92 \\, T_f34608=810.92Tf\u200bTf=34608810.92\u224842.7\u2009\u2218CT_f = \\frac{34608}{810.92} \\approx 42.7 \\, \\text{\u2218C}Tf\u200b=810.9234608\u200b\u224842.7\u2218C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Temperature<\/h3>\n\n\n\n<p>The final temperature of the mixture is approximately <strong>42.7 \u2218C<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-866.jpeg\" alt=\"\" class=\"wp-image-258084\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Four ice cubes at exactly 0 \u2218C with a total mass of 54.0 g are combined with 140 g of water at 90 \u2218C in an insulated container. (\u0394H\u2218fus =6.02 kJ\/mol , cwater =4.18J\/g\u22c5\u2218C ) If no heat is lost to the surroundings, what is the final temperature of the mixture? The Correct Answer and [&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-258083","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258083","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=258083"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258083\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258083"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258083"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258083"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}