{"id":43177,"date":"2025-06-29T14:25:33","date_gmt":"2025-06-29T14:25:33","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43177"},"modified":"2025-06-29T14:25:35","modified_gmt":"2025-06-29T14:25:35","slug":"calculating-specific-heat-worksheet-q-mca%cb%86%e2%80%a0t-where-q-heat-energy-m-mass-and-a%cb%86%e2%80%a0t-change-in-temperature","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculating-specific-heat-worksheet-q-mca%cb%86%e2%80%a0t-where-q-heat-energy-m-mass-and-a%cb%86%e2%80%a0t-change-in-temperature\/","title":{"rendered":"Calculating Specific Heat Worksheet Q = mc\u00e2\u02c6\u2020T, where Q = heat energy, m = mass, and \u00e2\u02c6\u2020T = change in temperature."},"content":{"rendered":"\n<p>Calculating Specific Heat Worksheet Q = mc\u00e2\u02c6\u2020T, where Q = heat energy, m = mass, and \u00e2\u02c6\u2020T = change in temperature. Remember, \u00e2\u02c6\u2020T = (Tfinal \u00e2\u20ac\u201c Tinitial). Show all work and proper units. 1) A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25\u00c2\u00b0C to 175\u00c2\u00b0C. Calculate the specific heat capacity of iron. 2) How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22\u00c2\u00b0C to 55\u00c2\u00b0C, if the specific heat of aluminum is 0.90 J\/g\u00c2\u00b0C? 3) Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 67,500 joules of heat, and its temperature changes from 32\u00c2\u00b0C to 57\u00c2\u00b0C. 4) 100.0 g of 4.0\u00c2\u00b0C water is heated until its temperature is 37\u00c2\u00b0C. If the specific heat of water is 4.18 J\/g\u00c2\u00b0C, calculate the amount of heat energy needed to cause this rise in temperature. 5) 25.0 g of mercury is heated from 25\u00c2\u00b0C to 155\u00c2\u00b0C, and absorbs 455 joules of heat in the process. Calculate the specific heat capacity of mercury. 6) What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 47.3 J of heat and the temperature rises 15.0\u00c2\u00b0C? 7) What mass of water will change its temperature by 30\u00c2\u00b0C when 525 J of heat is added to it? The specific heat of water is 4.18 J\/g\u00c2\u00b0C.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Formula:<\/strong>Q=mc\u0394TQ = mc\\Delta TQ=mc\u0394T<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>QQQ = heat energy (joules),<\/li>\n\n\n\n<li>mmm = mass (grams),<\/li>\n\n\n\n<li>ccc = specific heat capacity (J\/g\u00b0C),<\/li>\n\n\n\n<li>\u0394T\\Delta T\u0394T = change in temperature (\u00b0C), where \u0394T=Tfinal\u2212Tinitial\\Delta T = T_{final} &#8211; T_{initial}\u0394T=Tfinal\u200b\u2212Tinitial\u200b.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">1) Specific Heat Capacity of Iron<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 15.75 g,<\/li>\n\n\n\n<li>Heat energy (QQQ) = 1086.75 J,<\/li>\n\n\n\n<li>Initial temperature (TinitialT_{initial}Tinitial\u200b) = 25\u00b0C,<\/li>\n\n\n\n<li>Final temperature (TfinalT_{final}Tfinal\u200b) = 175\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>First, find \u0394T\\Delta T\u0394T:\u0394T=Tfinal\u2212Tinitial=175\u00b0C\u221225\u00b0C=150\u00b0C\\Delta T = T_{final} &#8211; T_{initial} = 175\u00b0C &#8211; 25\u00b0C = 150\u00b0C\u0394T=Tfinal\u200b\u2212Tinitial\u200b=175\u00b0C\u221225\u00b0C=150\u00b0C<\/p>\n\n\n\n<p>Now, solve for ccc (specific heat):1086.75\u2009J=15.75\u2009g\u00d7c\u00d7150\u00b0C1086.75 \\, J = 15.75 \\, g \\times c \\times 150\u00b0C1086.75J=15.75g\u00d7c\u00d7150\u00b0Cc=1086.7515.75\u00d7150=0.459\u2009J\/g\u00b0Cc = \\frac{1086.75}{15.75 \\times 150} = 0.459 \\, J\/g\u00b0Cc=15.75\u00d71501086.75\u200b=0.459J\/g\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2) Heat Energy for Aluminum<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 10.0 g,<\/li>\n\n\n\n<li>Specific heat of aluminum (ccc) = 0.90 J\/g\u00b0C,<\/li>\n\n\n\n<li>Initial temperature (TinitialT_{initial}Tinitial\u200b) = 22\u00b0C,<\/li>\n\n\n\n<li>Final temperature (TfinalT_{final}Tfinal\u200b) = 55\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Find \u0394T\\Delta T\u0394T:\u0394T=55\u00b0C\u221222\u00b0C=33\u00b0C\\Delta T = 55\u00b0C &#8211; 22\u00b0C = 33\u00b0C\u0394T=55\u00b0C\u221222\u00b0C=33\u00b0C<\/p>\n\n\n\n<p>Now, solve for QQQ (heat energy):Q=m\u00d7c\u00d7\u0394T=10.0\u2009g\u00d70.90\u2009J\/g\u00b0C\u00d733\u00b0CQ = m \\times c \\times \\Delta T = 10.0 \\, g \\times 0.90 \\, J\/g\u00b0C \\times 33\u00b0CQ=m\u00d7c\u00d7\u0394T=10.0g\u00d70.90J\/g\u00b0C\u00d733\u00b0CQ=297\u2009JQ = 297 \\, JQ=297J<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3) Specific Heat Capacity of Wood<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 1500.0 g,<\/li>\n\n\n\n<li>Heat energy (QQQ) = 67,500 J,<\/li>\n\n\n\n<li>Initial temperature (TinitialT_{initial}Tinitial\u200b) = 32\u00b0C,<\/li>\n\n\n\n<li>Final temperature (TfinalT_{final}Tfinal\u200b) = 57\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Find \u0394T\\Delta T\u0394T:\u0394T=57\u00b0C\u221232\u00b0C=25\u00b0C\\Delta T = 57\u00b0C &#8211; 32\u00b0C = 25\u00b0C\u0394T=57\u00b0C\u221232\u00b0C=25\u00b0C<\/p>\n\n\n\n<p>Now, solve for ccc (specific heat):67,500\u2009J=1500.0\u2009g\u00d7c\u00d725\u00b0C67,500 \\, J = 1500.0 \\, g \\times c \\times 25\u00b0C67,500J=1500.0g\u00d7c\u00d725\u00b0Cc=67,5001500.0\u00d725=1.80\u2009J\/g\u00b0Cc = \\frac{67,500}{1500.0 \\times 25} = 1.80 \\, J\/g\u00b0Cc=1500.0\u00d72567,500\u200b=1.80J\/g\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">4) Heat Energy for Water<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 100.0 g,<\/li>\n\n\n\n<li>Specific heat of water (ccc) = 4.18 J\/g\u00b0C,<\/li>\n\n\n\n<li>Initial temperature (TinitialT_{initial}Tinitial\u200b) = 4.0\u00b0C,<\/li>\n\n\n\n<li>Final temperature (TfinalT_{final}Tfinal\u200b) = 37\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Find \u0394T\\Delta T\u0394T:\u0394T=37\u00b0C\u22124.0\u00b0C=33\u00b0C\\Delta T = 37\u00b0C &#8211; 4.0\u00b0C = 33\u00b0C\u0394T=37\u00b0C\u22124.0\u00b0C=33\u00b0C<\/p>\n\n\n\n<p>Now, solve for QQQ (heat energy):Q=m\u00d7c\u00d7\u0394T=100.0\u2009g\u00d74.18\u2009J\/g\u00b0C\u00d733\u00b0CQ = m \\times c \\times \\Delta T = 100.0 \\, g \\times 4.18 \\, J\/g\u00b0C \\times 33\u00b0CQ=m\u00d7c\u00d7\u0394T=100.0g\u00d74.18J\/g\u00b0C\u00d733\u00b0CQ=13,794\u2009JQ = 13,794 \\, JQ=13,794J<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">5) Specific Heat Capacity of Mercury<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 25.0 g,<\/li>\n\n\n\n<li>Heat energy (QQQ) = 455 J,<\/li>\n\n\n\n<li>Initial temperature (TinitialT_{initial}Tinitial\u200b) = 25\u00b0C,<\/li>\n\n\n\n<li>Final temperature (TfinalT_{final}Tfinal\u200b) = 155\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Find \u0394T\\Delta T\u0394T:\u0394T=155\u00b0C\u221225\u00b0C=130\u00b0C\\Delta T = 155\u00b0C &#8211; 25\u00b0C = 130\u00b0C\u0394T=155\u00b0C\u221225\u00b0C=130\u00b0C<\/p>\n\n\n\n<p>Now, solve for ccc (specific heat):455\u2009J=25.0\u2009g\u00d7c\u00d7130\u00b0C455 \\, J = 25.0 \\, g \\times c \\times 130\u00b0C455J=25.0g\u00d7c\u00d7130\u00b0Cc=45525.0\u00d7130=0.141\u2009J\/g\u00b0Cc = \\frac{455}{25.0 \\times 130} = 0.141 \\, J\/g\u00b0Cc=25.0\u00d7130455\u200b=0.141J\/g\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">6) Specific Heat Capacity of Silver<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass (mmm) = 55.00 g,<\/li>\n\n\n\n<li>Heat energy (QQQ) = 47.3 J,<\/li>\n\n\n\n<li>Temperature rise (\u0394T\\Delta T\u0394T) = 15.0\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Now, solve for ccc (specific heat):47.3\u2009J=55.00\u2009g\u00d7c\u00d715.0\u00b0C47.3 \\, J = 55.00 \\, g \\times c \\times 15.0\u00b0C47.3J=55.00g\u00d7c\u00d715.0\u00b0Cc=47.355.00\u00d715.0=0.0574\u2009J\/g\u00b0Cc = \\frac{47.3}{55.00 \\times 15.0} = 0.0574 \\, J\/g\u00b0Cc=55.00\u00d715.047.3\u200b=0.0574J\/g\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">7) Mass of Water with Heat Energy<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Heat energy (QQQ) = 525 J,<\/li>\n\n\n\n<li>Specific heat of water (ccc) = 4.18 J\/g\u00b0C,<\/li>\n\n\n\n<li>Temperature change (\u0394T\\Delta T\u0394T) = 30\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Now, solve for mass (mmm):Q=m\u00d7c\u00d7\u0394TQ = m \\times c \\times \\Delta TQ=m\u00d7c\u00d7\u0394T525\u2009J=m\u00d74.18\u2009J\/g\u00b0C\u00d730\u00b0C525 \\, J = m \\times 4.18 \\, J\/g\u00b0C \\times 30\u00b0C525J=m\u00d74.18J\/g\u00b0C\u00d730\u00b0Cm=5254.18\u00d730=4.19\u2009gm = \\frac{525}{4.18 \\times 30} = 4.19 \\, gm=4.18\u00d730525\u200b=4.19g<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Results:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Specific heat of iron = 0.459 J\/g\u00b0C<\/li>\n\n\n\n<li>Heat needed for aluminum = 297 J<\/li>\n\n\n\n<li>Specific heat of wood = 1.80 J\/g\u00b0C<\/li>\n\n\n\n<li>Heat energy for water = 13,794 J<\/li>\n\n\n\n<li>Specific heat of mercury = 0.141 J\/g\u00b0C<\/li>\n\n\n\n<li>Specific heat of silver = 0.0574 J\/g\u00b0C<\/li>\n\n\n\n<li>Mass of water = 4.19 g<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-431.jpeg\" alt=\"\" class=\"wp-image-43178\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-431.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-431-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-431-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-431-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculating Specific Heat Worksheet Q = mc\u00e2\u02c6\u2020T, where Q = heat energy, m = mass, and \u00e2\u02c6\u2020T = change in temperature. Remember, \u00e2\u02c6\u2020T = (Tfinal \u00e2\u20ac\u201c Tinitial). Show all work and proper units. 1) A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25\u00c2\u00b0C to 175\u00c2\u00b0C. Calculate the [&hellip;]<\/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-43177","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43177","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=43177"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43177\/revisions"}],"predecessor-version":[{"id":43179,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43177\/revisions\/43179"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43177"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43177"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}