{"id":194511,"date":"2025-02-24T08:26:50","date_gmt":"2025-02-24T08:26:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194511"},"modified":"2025-02-24T08:26:52","modified_gmt":"2025-02-24T08:26:52","slug":"two-20-0-g-ice-cubes-at-14-0-c-14-0-c-are-placed-into-295-g295-g-of-water-at-25-0-c-25-0-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/two-20-0-g-ice-cubes-at-14-0-c-14-0-c-are-placed-into-295-g295-g-of-water-at-25-0-c-25-0-c\/","title":{"rendered":"Two 20.0 g ice cubes at -14.0 \u00b0C-14.0 \u00b0C are placed into 295 g295 g of water at 25.0 \u00b0C.25.0 \u00b0C"},"content":{"rendered":"\n

Two 20.0 g ice cubes at -14.0 \u00b0C-14.0 \u00b0C are placed into 295 g295 g of water at 25.0 \u00b0C.25.0 \u00b0C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, ??f,Tf, of the water after all the ice melts.<\/p>\n\n\n\n

heat capacity of H2O(s)H2O(s)<\/p>\n\n\n\n

37.7 J\/(mol\u00b7K)37.7 J\/(mol\u00b7K)<\/p>\n\n\n\n

heat capacity of H2O(l)H2O(l)<\/p>\n\n\n\n

75.3 J\/(mol\u00b7K)75.3 J\/(mol\u00b7K)<\/p>\n\n\n\n

enthalpy of fusion of H2OH2O<\/p>\n\n\n\n

6.01 kJ\/mol6.01 kJ\/mol<\/p>\n\n\n\n

??f=Tf=<\/p>\n\n\n\n

\u00b0C<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

The final temperature of the water after all the ice melts is approximately 13.2\u00b0C<\/strong>.<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n
    \n
  1. Heat Required to Warm Ice from -14.0\u00b0C to 0\u00b0C<\/strong>
    Ice must first be warmed up to 0\u00b0C before melting. Using the specific heat capacity of ice:
    [
    q_1 = n_{\\text{ice}} \\times C_{\\text{ice}} \\times \\Delta T
    ]
    where:<\/li>\n<\/ol>\n\n\n\n
      \n
    • ( n_{\\text{ice}} = \\frac{40.0 \\text{ g}}{18.015 \\text{ g\/mol}} = 2.22 \\text{ moles} )<\/li>\n\n\n\n
    • ( C_{\\text{ice}} = 37.7 \\text{ J\/(mol\u00b7K)} )<\/li>\n\n\n\n
    • ( \\Delta T = 0 – (-14) = 14 \\text{ K} ) [
      q_1 = 2.22 \\times 37.7 \\times 14 = 1170.9 \\text{ J}
      ]<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Heat Required to Melt Ice at 0\u00b0C<\/strong>
        The heat needed to melt the ice completely:
        [
        q_2 = n_{\\text{ice}} \\times \\Delta H_{\\text{fusion}}
        ]
        where ( \\Delta H_{\\text{fusion}} = 6.01 \\text{ kJ\/mol} = 6010 \\text{ J\/mol} ). [
        q_2 = 2.22 \\times 6010 = 13342.2 \\text{ J}
        ] Total heat required to warm and melt the ice:
        [
        q_{\\text{total}} = q_1 + q_2 = 1170.9 + 13342.2 = 14513.1 \\text{ J}
        ]<\/li>\n\n\n\n
      2. Heat Lost by Water<\/strong>
        The water will release heat to the ice:
        [
        q_{\\text{water}} = n_{\\text{water}} \\times C_{\\text{water}} \\times (T_{\\text{initial, water}} – T_f)
        ]
        where:<\/li>\n<\/ol>\n\n\n\n
          \n
        • ( n_{\\text{water}} = \\frac{295.0 \\text{ g}}{18.015 \\text{ g\/mol}} = 16.38 \\text{ moles} )<\/li>\n\n\n\n
        • ( C_{\\text{water}} = 75.3 \\text{ J\/(mol\u00b7K)} )<\/li>\n\n\n\n
        • ( \\Delta T = (25 – T_f) ) The total heat available from the water:
          [
          q_{\\text{available}} = 16.38 \\times 75.3 \\times (25 – 0) = 30866.3 \\text{ J}
          ] Since ( q_{\\text{available}} > q_{\\text{total}} ), all ice melts, and the remaining heat increases the final temperature.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Solving for ( T_f )<\/strong>
            The remaining heat after melting:
            [
            q_{\\text{remaining}} = q_{\\text{available}} – q_{\\text{total}} = 30866.3 – 14513.1 = 16353.2 \\text{ J}
            ] The final temperature is found using:
            [
            T_f = \\frac{q_{\\text{remaining}}}{n_{\\text{water}} \\times C_{\\text{water}}}
            ] [
            T_f = \\frac{16353.2}{16.38 \\times 75.3} = 13.2\u00b0C
            ]<\/li>\n<\/ol>\n\n\n\n

            Thus, after all the ice melts, the equilibrium temperature of the system is 13.2\u00b0C<\/strong>.<\/p>\n\n\n\n

            Now, I’ll generate an image to illustrate this process.<\/p>\n\n\n\n

            \"\"<\/figure>\n\n\n\n

            The bar chart above illustrates the heat transfer process during the interaction of ice and water:<\/p>\n\n\n\n

              \n
            1. Heating Ice (-14\u00b0C to 0\u00b0C)<\/strong>: The ice requires 1170.9 J<\/strong> to warm up to 0\u00b0C.<\/li>\n\n\n\n
            2. Melting Ice at 0\u00b0C<\/strong>: The phase change from solid to liquid consumes 13,342.2 J<\/strong>.<\/li>\n\n\n\n
            3. Cooling Water (25\u00b0C to Final Temperature)<\/strong>: The water releases 16,353.2 J<\/strong>, bringing the final temperature of the system to 13.2\u00b0C<\/strong>.<\/li>\n<\/ol>\n\n\n\n

              This visualization helps to understand the energy exchange involved in reaching thermal equilibrium.<\/p>\n\n\n\n

              <\/p>\n","protected":false},"excerpt":{"rendered":"

              Two 20.0 g ice cubes at -14.0 \u00b0C-14.0 \u00b0C are placed into 295 g295 g of water at 25.0 \u00b0C.25.0 \u00b0C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, ??f,Tf, of the water after all the ice melts. heat capacity of H2O(s)H2O(s) 37.7 J\/(mol\u00b7K)37.7 J\/(mol\u00b7K) heat capacity of H2O(l)H2O(l) […]<\/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-194511","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194511","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=194511"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194511\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194511"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194511"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194511"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}