To find the heat of fusion of mercury<\/strong>, we will apply the principle of conservation of energy<\/strong>. The heat lost by the water and the aluminum calorimeter must equal the heat gained by the mercury as it melts and warms up.<\/p>\n\n\n\n Water:<\/strong> Qwater=m\u22c5c\u22c5\u0394T=1.2\u22c54186\u22c5(20.0\u221216.5)=1.2\u22c54186\u22c53.5=17,581.2\u2009JQ_{\\text{water}} = m \\cdot c \\cdot \\Delta T = 1.2 \\cdot 4186 \\cdot (20.0 – 16.5) = 1.2 \\cdot 4186 \\cdot 3.5 = 17,581.2 \\, \\text{J}Qwater\u200b=m\u22c5c\u22c5\u0394T=1.2\u22c54186\u22c5(20.0\u221216.5)=1.2\u22c54186\u22c53.5=17,581.2J<\/p>\n\n\n\n Aluminum calorimeter:<\/strong> QAl=0.50\u22c5900\u22c5(20.0\u221216.5)=0.50\u22c5900\u22c53.5=1,575\u2009JQ_{\\text{Al}} = 0.50 \\cdot 900 \\cdot (20.0 – 16.5) = 0.50 \\cdot 900 \\cdot 3.5 = 1,575 \\, \\text{J}QAl\u200b=0.50\u22c5900\u22c5(20.0\u221216.5)=0.50\u22c5900\u22c53.5=1,575J<\/p>\n\n\n\n Total heat lost:<\/strong> Qlost=17,581.2+1,575=19,156.2\u2009JQ_{\\text{lost}} = 17,581.2 + 1,575 = 19,156.2 \\, \\text{J}Qlost\u200b=17,581.2+1,575=19,156.2J<\/p>\n\n\n\n Mercury must:<\/p>\n\n\n\n Let LfL_fLf\u200b be the unknown heat of fusion in J\/kg.<\/p>\n\n\n\n Melting mercury:<\/strong> Qmelt=Lf\u22c51.0Q_{\\text{melt}} = L_f \\cdot 1.0Qmelt\u200b=Lf\u200b\u22c51.0<\/p>\n\n\n\n Heating mercury:<\/strong> Qwarm=1.0\u22c5140\u22c5(16.5\u2212(\u221239))=1.0\u22c5140\u22c555.5=7,770\u2009JQ_{\\text{warm}} = 1.0 \\cdot 140 \\cdot (16.5 – (-39)) = 1.0 \\cdot 140 \\cdot 55.5 = 7,770 \\, \\text{J}Qwarm\u200b=1.0\u22c5140\u22c5(16.5\u2212(\u221239))=1.0\u22c5140\u22c555.5=7,770J<\/p>\n\n\n\n Qmelt+Qwarm=QlostQ_{\\text{melt}} + Q_{\\text{warm}} = Q_{\\text{lost}}Qmelt\u200b+Qwarm\u200b=Qlost\u200b Lf+7,770=19,156.2L_f + 7,770 = 19,156.2Lf\u200b+7,770=19,156.2 Lf=19,156.2\u22127,770=11,386.2\u2009J\/kgL_f = 19,156.2 – 7,770 = 11,386.2 \\, \\text{J\/kg}Lf\u200b=19,156.2\u22127,770=11,386.2J\/kg<\/p>\n\n\n\n Lf=11,\u2009\u2063386\u2009J\/kg\\boxed{L_f = 11,\\!386 \\, \\text{J\/kg}}Lf\u200b=11,386J\/kg\u200b<\/p>\n\n\n\n This problem revolves around the principle of conservation of energy. When solid mercury is placed into a calorimeter containing water, the system reaches a thermal equilibrium. The initial temperatures are different. The mercury is at its melting point of \u221239\u2218C-39^\\circ\\text{C}\u221239\u2218C, while the water and aluminum calorimeter are at 20.0\u2218C20.0^\\circ\\text{C}20.0\u2218C. As the system reaches equilibrium, heat flows from the warmer substances (water and aluminum) to the colder substance (mercury).<\/p>\n\n\n\n The energy given up by the water and aluminum is used in two stages. First, it melts the mercury. Second, it warms the now-liquid mercury from \u221239\u2218C-39^\\circ\\text{C}\u221239\u2218C to the final temperature of 16.5\u2218C16.5^\\circ\\text{C}16.5\u2218C.<\/p>\n\n\n\n To find the heat of fusion, we calculated the heat lost by the water and the aluminum using the specific heat formula Q=mc\u0394TQ = mc\\Delta TQ=mc\u0394T. The total energy lost by the water and aluminum was about 19,156 joules. Then, we calculated how much energy is required to warm the mercury after melting, which came out to 7,770 joules.<\/p>\n\n\n\n Since the total energy gained by the mercury must equal the total energy lost by the other components, we subtracted the energy used for warming from the total heat lost to isolate the energy used just for melting. This difference gives us the latent heat of fusion for 1 kilogram of mercury, which came out to be approximately 11,386 joules per kilogram.<\/p>\n\n\n\n This value represents the energy required to convert solid mercury at its melting point into liquid mercury without changing its temperature.<\/p>\n\n\n\n The specific heat of liquid mercury is 140 J\/kg\u00c2\u00b0C. When 1.0 kg of solid mercury at its melting point of -39\u00c2\u00b0C is placed in a 0.50-kg aluminum calorimeter filled with 1.2 kg of water at 20.0\u00c2\u00b0C, the mercury melts and the final temperature of the combination is found to be 16.5\u00c2\u00b0C. What is the heat […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-39242","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39242","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=39242"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39242\/revisions"}],"predecessor-version":[{"id":39244,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39242\/revisions\/39244"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39242"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39242"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39242"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given Data:<\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Calculate heat lost by water and calorimeter<\/h3>\n\n\n\n
\n\n\n\nStep 2: Heat gained by mercury<\/h3>\n\n\n\n
\n
\n\n\n\nStep 3: Set up energy balance<\/h3>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-244.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"