{"id":2674,"date":"2016-08-24T15:38:41","date_gmt":"2016-08-24T15:38:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2674"},"modified":"2017-08-28T22:08:28","modified_gmt":"2017-08-28T22:08:28","slug":"specific-heat-calculations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/specific-heat-calculations\/","title":{"raw":"Specific Heat Calculations","rendered":"Specific Heat Calculations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-Y2VkMGMzMWQzMDAyODg5YjY4YmZhZDJjMzc4ZTJiMTI.-2d7\">\r\n \t<li>Perform specific heat calculations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\n<p id=\"x-ck12-NzNkMjFkM2MzNzFlNTYxNGNkNTQwOTQ4MmY0ODBlY2U.-mbl\"><span class=\"x-ck12-img-inline\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212608\/20140811155647533921.jpeg\" alt=\"Water has a very high heat capacity, which makes it useful for radiators\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NDU4Yjk1ZDEyYzU3ODE2NDlhMjA0MmUxNDVhM2ZiNTY.-tl5\"><strong>Does water have a high capacity for absorbing heat?<\/strong><\/p>\r\n<p id=\"x-ck12-MGZjYzdkNDg0YWRhYzFjYzcxNDA1Yjk3N2RiZTM1ZWI.-1hd\">Yes. In a car radiator, it serves to keep the engine cooler than it would otherwise run. (In the picture above, the radiator is the black object on the left.) As the water circulates through the engine, it absorbs heat from the engine block. When it passes through the radiator, the cooling fan and the exposure to the outside environment allow the water to cool somewhat before it makes another passage through the engine.<\/p>\r\n\r\n<\/div>\r\n<h3>Specific Heat Calculations<\/h3>\r\n<p id=\"x-ck12-ZmQxNWZhYjM5YjdhODdhNjJjYTVkMGNhZWNjNjY5YWU.-cyq\">The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat\u00a0 <img id=\"x-ck12-MTM2Njk1NzM2NjQ5OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/7d955896ae32b20b542111613f47fb0c.png\" alt=\"(q)\" width=\"21\" height=\"18\" \/> to specific heat <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/f53a0d02248f799b2d7a083545bc0114.png\" alt=\"(c_p)\" width=\"27\" height=\"20\" \/> , mass <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/f4b8deb9be9af0dbc0cca882803a4223.png\" alt=\"(m)\" width=\"28\" height=\"18\" \/> , and temperature change\u00a0 <img id=\"x-ck12-MTM2Njk1NzM2NjUwMA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/c1294e31ecbbf36e4b4111478d20fccc.png\" alt=\"(Delta{T})\" width=\"40\" height=\"18\" \/> is shown below.<\/p>\r\n<p id=\"x-ck12-je4\"><img id=\"x-ck12-MTM2Njk1NzM2NjUwMQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/e0a105c8b66e5ce9b5079f677bffafb9.png\" alt=\"q=c_p times m times Delta{T}\" width=\"136\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-NGY3YzcwNDEzNjM4YmY3YTk3NWM4MWQ3YmZjZGJhNTY.-cg5\">The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212611\/98aa9fec05a8a07dbbe4a3e51610f6b2.png\" alt=\"Delta{T}= T_f - T_i\" width=\"107\" height=\"18\" \/> , where\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/3cd35e5bc16e7e8954fa3ec245190b5d.png\" alt=\"T_f\" width=\"17\" height=\"18\" \/> is the final temperature and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/51817229f8380cba09312c9f2839275c.png\" alt=\"T_i\" width=\"14\" height=\"15\" \/> is the initial temperature.<\/p>\r\n\r\n<h4>Sample Problem: Calculating Specific Heat<\/h4>\r\n<p id=\"x-ck12-ZjJkZTZhOTY5ZDYyNmM1MTk2MWQxYjk4YjRhMGNhM2I.-zoi\">A 15.0 g piece of cadmium metal absorbs 134 J of heat while rising from 24.0\u00b0C to 62.7\u00b0C. Calculate the specific heat of cadmium.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-loc\"><em>Step 1: List the known quantities and plan the problem <\/em>.<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-6gr\"><span class=\"x-ck12-underline\">Known<\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OTNiY2M1ZDQ3ZTg1MWI3NGIzZjdjMmMwZDU2ZDRkNmY.-wgy\">\r\n \t<li>heat = <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212559\/deaa0b282c5a1d56df2ebf1d9a749ce0.png\" alt=\"q\" width=\"9\" height=\"12\" \/> = 134 J<\/li>\r\n \t<li>mass =\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> = 15.0 g<\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/13e7e32e5ff791c6c243e9b5a3ec0d6c.png\" alt=\"Delta{text{T}} = 62.7^circ text{C} - 24.0^circ text{C} = 38.7^circ text{C}\" width=\"253\" height=\"14\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-zny\"><span class=\"x-ck12-underline\">Unknown<\/span><\/p>\r\n\r\n<ul id=\"x-ck12-l4n\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212613\/33ec3ea0433c13c17f9a8d0033622c9b.png\" alt=\"c_p text{of cadmium}= ? text{J}\/ text{g}^circ text{C} \" width=\"193\" height=\"19\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NzM1NTA3MDgxYTE1NWNiOGQ4ODBhMjMwMTkyMGVmYzU.-tn8\">The specific heat equation can be rearranged to solve for the specific heat.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-fyj\"><em>Step 2: Solve <\/em>.<\/p>\r\n<p id=\"x-ck12-pz3\"><img id=\"x-ck12-MTM2Njk1NzM2NjUwMw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212614\/5c0bc0832f7652ebb094fab54a8dc73e.png\" alt=\"c_p=frac{q}{m times Delta{T}}=frac{134 text{ J}}{15.0 text{ g} times 38.7^circ text{C}}=0.231 text{ J\/g}^circ text{C}\" width=\"374\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-xsr\"><em>Step 3: Think about your result <\/em>.<\/p>\r\n<p id=\"x-ck12-NzA5ZGRmMWQ1OGU0OTc1ZTY3NTM5ZmU0MWE5YjQyYTQ.-wdm\">The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.<\/p>\r\n<p id=\"x-ck12-YzE1NWU1OGRmZmUzNDQ0OWViNzVjNmJmYTljYzUwM2Y.-tyi\">Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0 g sample of water at 23.52\u00b0C was cooled by the removal of 813 J of heat. The change in temperature can be calculated using the specific heat equation.<\/p>\r\n<p id=\"x-ck12-obl\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212615\/03e62eb2f9f15fb7152422be3f3ed906.png\" alt=\"Delta{T}=frac{q}{c_p times m}=frac{813 text{ J}}{4.18 text{ J\/g}^circ text{C} times 60.0 text{ g}}=3.24^circ text{C}\" width=\"365\" height=\"43\" \/><\/p>\r\n<p id=\"x-ck12-ZjAyNzM2ODc3MTJlNTNlOTRjN2JlZjdmM2E1OGE1NWE.-jo9\">Since the water was being cooled, the temperature decreases. The final temperature is:<\/p>\r\n<p id=\"x-ck12-nhd\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212616\/71a69d94537cef8f02b381624714658d.png\" alt=\"T_f=23.52^circ text{C} - 3.24^circ text{C}=20.28^circ text{C}\" width=\"262\" height=\"20\" \/><\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NmFlNzkzYWE4ZWRjNjIxNWNhN2NlMWM2NDc2Mjg4YWM.-re0\">\r\n \t<li>Specific heat calculations are illustrated.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-8k3\" dir=\"ltr\">Solve the problems found in <a href=\"https:\/\/web.archive.org\/web\/20150312203727\/http:\/\/www.sciencebugz.com\/chemistry\/chprbspheat.htm\" target=\"_blank\" rel=\"noopener\">Problems in Specific Heat.<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-NmYzMTA4MmRhMzdmNjFhMGM3NzM1NTZiZmQ3MjllYTY.-vs0\">\r\n \t<li>Do different materials have different specific heats?<\/li>\r\n \t<li>How does mass affect heat absorbed?<\/li>\r\n \t<li>If we know the specific heat of a material, can we determine how much heat is released under a given set of circumstances?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Glosary<\/h3>\r\n<strong style=\"line-height: 1.5;\">specific heat: <\/strong><span style=\"line-height: 1.5;\">The amount of energy required to raise the temperature of 1 gram of the substance by 1\u00b0C.<\/span>\r\n\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>Adrian Pingstone (Wikimedia: Arpingstone). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Flat.six.honda.valkyrie.arp.750pix.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Flat.six.honda.valkyrie.arp.750pix.jpg <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-Y2VkMGMzMWQzMDAyODg5YjY4YmZhZDJjMzc4ZTJiMTI.-2d7\">\n<li>Perform specific heat calculations.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p id=\"x-ck12-NzNkMjFkM2MzNzFlNTYxNGNkNTQwOTQ4MmY0ODBlY2U.-mbl\"><span class=\"x-ck12-img-inline\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212608\/20140811155647533921.jpeg\" alt=\"Water has a very high heat capacity, which makes it useful for radiators\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-NDU4Yjk1ZDEyYzU3ODE2NDlhMjA0MmUxNDVhM2ZiNTY.-tl5\"><strong>Does water have a high capacity for absorbing heat?<\/strong><\/p>\n<p id=\"x-ck12-MGZjYzdkNDg0YWRhYzFjYzcxNDA1Yjk3N2RiZTM1ZWI.-1hd\">Yes. In a car radiator, it serves to keep the engine cooler than it would otherwise run. (In the picture above, the radiator is the black object on the left.) As the water circulates through the engine, it absorbs heat from the engine block. When it passes through the radiator, the cooling fan and the exposure to the outside environment allow the water to cool somewhat before it makes another passage through the engine.<\/p>\n<\/div>\n<h3>Specific Heat Calculations<\/h3>\n<p id=\"x-ck12-ZmQxNWZhYjM5YjdhODdhNjJjYTVkMGNhZWNjNjY5YWU.-cyq\">The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Njk1NzM2NjQ5OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/7d955896ae32b20b542111613f47fb0c.png\" alt=\"(q)\" width=\"21\" height=\"18\" \/> to specific heat <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/f53a0d02248f799b2d7a083545bc0114.png\" alt=\"(c_p)\" width=\"27\" height=\"20\" \/> , mass <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/f4b8deb9be9af0dbc0cca882803a4223.png\" alt=\"(m)\" width=\"28\" height=\"18\" \/> , and temperature change\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Njk1NzM2NjUwMA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/c1294e31ecbbf36e4b4111478d20fccc.png\" alt=\"(Delta{T})\" width=\"40\" height=\"18\" \/> is shown below.<\/p>\n<p id=\"x-ck12-je4\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Njk1NzM2NjUwMQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212610\/e0a105c8b66e5ce9b5079f677bffafb9.png\" alt=\"q=c_p times m times Delta{T}\" width=\"136\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-NGY3YzcwNDEzNjM4YmY3YTk3NWM4MWQ3YmZjZGJhNTY.-cg5\">The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212611\/98aa9fec05a8a07dbbe4a3e51610f6b2.png\" alt=\"Delta{T}= T_f - T_i\" width=\"107\" height=\"18\" \/> , where\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/3cd35e5bc16e7e8954fa3ec245190b5d.png\" alt=\"T_f\" width=\"17\" height=\"18\" \/> is the final temperature and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/51817229f8380cba09312c9f2839275c.png\" alt=\"T_i\" width=\"14\" height=\"15\" \/> is the initial temperature.<\/p>\n<h4>Sample Problem: Calculating Specific Heat<\/h4>\n<p id=\"x-ck12-ZjJkZTZhOTY5ZDYyNmM1MTk2MWQxYjk4YjRhMGNhM2I.-zoi\">A 15.0 g piece of cadmium metal absorbs 134 J of heat while rising from 24.0\u00b0C to 62.7\u00b0C. Calculate the specific heat of cadmium.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-loc\"><em>Step 1: List the known quantities and plan the problem <\/em>.<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-6gr\"><span class=\"x-ck12-underline\">Known<\/span><\/p>\n<ul id=\"x-ck12-OTNiY2M1ZDQ3ZTg1MWI3NGIzZjdjMmMwZDU2ZDRkNmY.-wgy\">\n<li>heat = <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212559\/deaa0b282c5a1d56df2ebf1d9a749ce0.png\" alt=\"q\" width=\"9\" height=\"12\" \/> = 134 J<\/li>\n<li>mass =\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> = 15.0 g<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212612\/13e7e32e5ff791c6c243e9b5a3ec0d6c.png\" alt=\"Delta{text{T}} = 62.7^circ text{C} - 24.0^circ text{C} = 38.7^circ text{C}\" width=\"253\" height=\"14\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-zny\"><span class=\"x-ck12-underline\">Unknown<\/span><\/p>\n<ul id=\"x-ck12-l4n\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212613\/33ec3ea0433c13c17f9a8d0033622c9b.png\" alt=\"c_p text{of cadmium}= ? text{J}\/ text{g}^circ text{C}\" width=\"193\" height=\"19\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-NzM1NTA3MDgxYTE1NWNiOGQ4ODBhMjMwMTkyMGVmYzU.-tn8\">The specific heat equation can be rearranged to solve for the specific heat.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-fyj\"><em>Step 2: Solve <\/em>.<\/p>\n<p id=\"x-ck12-pz3\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Njk1NzM2NjUwMw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212614\/5c0bc0832f7652ebb094fab54a8dc73e.png\" alt=\"c_p=frac{q}{m times Delta{T}}=frac{134 text{ J}}{15.0 text{ g} times 38.7^circ text{C}}=0.231 text{ J\/g}^circ text{C}\" width=\"374\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-xsr\"><em>Step 3: Think about your result <\/em>.<\/p>\n<p id=\"x-ck12-NzA5ZGRmMWQ1OGU0OTc1ZTY3NTM5ZmU0MWE5YjQyYTQ.-wdm\">The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.<\/p>\n<p id=\"x-ck12-YzE1NWU1OGRmZmUzNDQ0OWViNzVjNmJmYTljYzUwM2Y.-tyi\">Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0 g sample of water at 23.52\u00b0C was cooled by the removal of 813 J of heat. The change in temperature can be calculated using the specific heat equation.<\/p>\n<p id=\"x-ck12-obl\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212615\/03e62eb2f9f15fb7152422be3f3ed906.png\" alt=\"Delta{T}=frac{q}{c_p times m}=frac{813 text{ J}}{4.18 text{ J\/g}^circ text{C} times 60.0 text{ g}}=3.24^circ text{C}\" width=\"365\" height=\"43\" \/><\/p>\n<p id=\"x-ck12-ZjAyNzM2ODc3MTJlNTNlOTRjN2JlZjdmM2E1OGE1NWE.-jo9\">Since the water was being cooled, the temperature decreases. The final temperature is:<\/p>\n<p id=\"x-ck12-nhd\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212616\/71a69d94537cef8f02b381624714658d.png\" alt=\"T_f=23.52^circ text{C} - 3.24^circ text{C}=20.28^circ text{C}\" width=\"262\" height=\"20\" \/><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NmFlNzkzYWE4ZWRjNjIxNWNhN2NlMWM2NDc2Mjg4YWM.-re0\">\n<li>Specific heat calculations are illustrated.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-8k3\" dir=\"ltr\">Solve the problems found in <a href=\"https:\/\/web.archive.org\/web\/20150312203727\/http:\/\/www.sciencebugz.com\/chemistry\/chprbspheat.htm\" target=\"_blank\" rel=\"noopener\">Problems in Specific Heat.<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-NmYzMTA4MmRhMzdmNjFhMGM3NzM1NTZiZmQ3MjllYTY.-vs0\">\n<li>Do different materials have different specific heats?<\/li>\n<li>How does mass affect heat absorbed?<\/li>\n<li>If we know the specific heat of a material, can we determine how much heat is released under a given set of circumstances?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox learning-objectives\">\n<h3>Glosary<\/h3>\n<p><strong style=\"line-height: 1.5;\">specific heat: <\/strong><span style=\"line-height: 1.5;\">The amount of energy required to raise the temperature of 1 gram of the substance by 1\u00b0C.<\/span><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>Adrian Pingstone (Wikimedia: Arpingstone). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Flat.six.honda.valkyrie.arp.750pix.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Flat.six.honda.valkyrie.arp.750pix.jpg <\/a>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":1507,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2674","chapter","type-chapter","status-publish","hentry"],"part":2338,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2674","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/1507"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2674\/revisions"}],"predecessor-version":[{"id":3622,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2674\/revisions\/3622"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2338"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2674\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2674"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2674"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2674"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2674"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}