The specific heat of mercury is 0.140 J\/g \u00b0C. Calculate the heat necessary to raise the temperature of one mole of liquid mercury by 36 \u00c3\u201a\u00c2\u00b0C.
1-\u00c3\u201a\u00c2 The specific heat of mercury is 0.140 J\/g\u00c3\u201a\u00c2\u00b0C. Calculate the heat necessary to raise the temperature of one mole of liquid mercury by 36 \u00c3\u201a\u00c2\u00b0C.<\/p>\n\n\n\n
\u00c3\u201a\u00c2 a) What if the heat of vaporization of liquid Freon-12, CCl2F2, is 19.7 kJ\/mole.\u00c3\u201a\u00c2 Calculate the energy required to vaporize 39.2 g of this compound<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve the problem, we\u2019ll break it into two parts:<\/p>\n\n\n\n We know the following information:<\/p>\n\n\n\n The formula for the heat required to change the temperature of a substance is:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n Substitute the known values for mercury:<\/p>\n\n\n\n [ [ Thus, the heat required to raise the temperature of one mole of liquid mercury by 36\u00b0C is 1005.31 J<\/strong>.<\/p>\n\n\n\n The heat of vaporization of Freon-12 (( \\Delta H_{\\text{vap}} )) is given as 19.7 kJ\/mol.<\/p>\n\n\n\n We need to calculate the energy required to vaporize 39.2 g of Freon-12.<\/p>\n\n\n\n First, calculate the number of moles of Freon-12. The molar mass of Freon-12 (CCl\u2082F\u2082) is:<\/p>\n\n\n\n [ Now, calculate the moles of Freon-12 in 39.2 g:<\/p>\n\n\n\n [ Now, use the heat of vaporization to find the total energy required to vaporize 0.324 moles of Freon-12:<\/p>\n\n\n\n [ [ Thus, the energy required to vaporize 39.2 g of Freon-12 is 6.39 kJ<\/strong>.<\/p>\n\n\n\n In both cases, understanding how to use specific heat capacity and heat of vaporization to find the energy needed for temperature change or phase transition is essential. The principles apply to a variety of substances beyond mercury and Freon-12, demonstrating the wide use of thermodynamic concepts.<\/p>\n","protected":false},"excerpt":{"rendered":" The specific heat of mercury is 0.140 J\/g \u00b0C. Calculate the heat necessary to raise the temperature of one mole of liquid mercury by 36 \u00c3\u201a\u00c2\u00b0C.1-\u00c3\u201a\u00c2 The specific heat of mercury is 0.140 J\/g\u00c3\u201a\u00c2\u00b0C. Calculate the heat necessary to raise the temperature of one mole of liquid mercury by 36 \u00c3\u201a\u00c2\u00b0C. \u00c3\u201a\u00c2 a) What if […]<\/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-189493","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189493","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=189493"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189493\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189493"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189493"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part 1: Calculating Heat Necessary to Raise the Temperature of Mercury<\/h3>\n\n\n\n
\n
q = m \\cdot c \\cdot \\Delta T
]<\/p>\n\n\n\n\n
q_{\\text{Hg}} = 200.59 \\, \\text{g} \\cdot 0.140 \\, \\text{J\/g\u00b0C} \\cdot 36 \\, \\text{\u00b0C}
]<\/p>\n\n\n\n
q_{\\text{Hg}} = 1005.31 \\, \\text{J}
]<\/p>\n\n\n\n
\n\n\n\nPart 2: Energy Required to Vaporize 39.2 g of Freon-12<\/h3>\n\n\n\n
M_{\\text{Freon-12}} = 120.91 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{moles of Freon-12} = \\frac{39.2 \\, \\text{g}}{120.91 \\, \\text{g\/mol}} = 0.324 \\, \\text{mol}
]<\/p>\n\n\n\n
q_{\\text{Freon-12}} = 0.324 \\, \\text{mol} \\cdot 19.7 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\n
q_{\\text{Freon-12}} = 6.39 \\, \\text{kJ}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
\n
For mercury, we used the specific heat capacity formula, which gives the amount of heat needed to change the temperature of a substance. Since the mass and specific heat capacity are known, we can directly apply them to calculate the energy. In this case, raising the temperature of one mole (200.59 g) of mercury by 36\u00b0C required 1005.31 J<\/strong> of energy.<\/li>\n\n\n\n
The heat of vaporization is the amount of energy required to convert one mole of liquid into gas at its boiling point without changing its temperature. The calculation for Freon-12 involved determining the number of moles in 39.2 g and then multiplying by the heat of vaporization. For 0.324 moles of Freon-12, the energy required to vaporize it was 6.39 kJ<\/strong>.<\/li>\n<\/ol>\n\n\n\n