{"id":11248,"date":"2015-07-14T19:12:00","date_gmt":"2015-07-14T19:12:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&#038;p=11248"},"modified":"2021-12-29T19:24:14","modified_gmt":"2021-12-29T19:24:14","slug":"use-newtons-law-of-cooling","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/use-newtons-law-of-cooling\/","title":{"raw":"Use Newton\u2019s Law of Cooling","rendered":"Use Newton\u2019s Law of Cooling"},"content":{"raw":"<p id=\"fs-id1165137854986\">Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/p>\r\n\r\n<div id=\"eip-809\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\left(t\\right)=a{e}^{kt}+{T}_{s}[\/latex]<\/div>\r\n<p id=\"fs-id1165137602810\">This formula is derived as follows:<\/p>\r\n\r\n<div id=\"eip-741\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill &amp; \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill &amp; \\text{Laws of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill &amp; \\text{Laws of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill &amp; \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{cases}[\/latex]<\/div>\r\n<div id=\"fs-id1165135629984\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">A General Note: Newton\u2019s Law of Cooling<\/h3>\r\n<p id=\"fs-id1165137644683\">The temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\r\n\r\n<div id=\"fs-id1165137660934\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<\/div>\r\nwhere\r\n<div id=\"fs-id1165137415924\" data-type=\"list\" data-list-type=\"bulleted\">\r\n<ul>\r\n\t<li data-type=\"item\"><em>t<\/em>\u00a0is time<\/li>\r\n\t<li data-type=\"item\"><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\r\n\t<li data-type=\"item\"><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137667467\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"eip-id2434562\">How To: Given a set of conditions, apply Newton\u2019s Law of Cooling.<\/h3>\r\n<ol id=\"fs-id1165137563116\" data-number-style=\"arabic\">\r\n\t<li>Set [latex]{T}_{s}[\/latex] equal to the <em data-effect=\"italics\">y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\r\n\t<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\r\n\t<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_07_05\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165135422904\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165135422906\" class=\"problem textbox shaded\" data-type=\"problem\">\r\n<h3 data-type=\"title\">Example 4: Using Newton\u2019s Law of Cooling<\/h3>\r\n<p id=\"fs-id1165135439846\">A cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex], and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137705784\" class=\"solution textbox shaded\" data-type=\"solution\">\r\n<h3>Solution<\/h3>\r\n<p id=\"fs-id1165137705786\">Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation<\/p>\r\n\r\n<div id=\"eip-id1165134081701\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/div>\r\n<p id=\"fs-id1165137563332\">We know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].<\/p>\r\n\r\n<div id=\"eip-id1165134353692\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}165=A{e}^{k0}+35\\hfill &amp; \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill &amp; \\text{Solve for }A.\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165137677897\">We were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.<\/p>\r\n\r\n<div id=\"eip-id1165134081505\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }150=130{e}^{k10}+35\\hfill &amp; \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill &amp; \\text{Subtract 35}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill &amp; \\text{Divide by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill &amp; \\text{Divide by the coefficient of }k.\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165134294911\">This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].<\/p>\r\n<p id=\"fs-id1165135194273\">Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\r\n\r\n<div id=\"eip-id1165135149883\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}70=130{e}^{-0.0123t}+35\\hfill &amp; \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill &amp; \\text{Subtract 35}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill &amp; \\text{Divide by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill &amp; \\text{Take the natural log of both sides}\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill &amp; \\text{Divide by the coefficient of }t.\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165135186349\">It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 4<\/h3>\r\n<p id=\"fs-id1165137686701\">A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?<\/p>\r\n<a href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a>\r\n\r\n<\/div>","rendered":"<p id=\"fs-id1165137854986\">Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/p>\n<div id=\"eip-809\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\left(t\\right)=a{e}^{kt}+{T}_{s}[\/latex]<\/div>\n<p id=\"fs-id1165137602810\">This formula is derived as follows:<\/p>\n<div id=\"eip-741\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill & \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill & \\text{Laws of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill & \\text{Laws of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill & \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{cases}[\/latex]<\/div>\n<div id=\"fs-id1165135629984\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">A General Note: Newton\u2019s Law of Cooling<\/h3>\n<p id=\"fs-id1165137644683\">The temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\n<div id=\"fs-id1165137660934\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<\/div>\n<p>where<\/p>\n<div id=\"fs-id1165137415924\" data-type=\"list\" data-list-type=\"bulleted\">\n<ul>\n<li data-type=\"item\"><em>t<\/em>\u00a0is time<\/li>\n<li data-type=\"item\"><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\n<li data-type=\"item\"><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137667467\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"eip-id2434562\">How To: Given a set of conditions, apply Newton\u2019s Law of Cooling.<\/h3>\n<ol id=\"fs-id1165137563116\" data-number-style=\"arabic\">\n<li>Set [latex]{T}_{s}[\/latex] equal to the <em data-effect=\"italics\">y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\n<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\n<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_07_05\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165135422904\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165135422906\" class=\"problem textbox shaded\" data-type=\"problem\">\n<h3 data-type=\"title\">Example 4: Using Newton\u2019s Law of Cooling<\/h3>\n<p id=\"fs-id1165135439846\">A cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex], and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?<\/p>\n<\/div>\n<div id=\"fs-id1165137705784\" class=\"solution textbox shaded\" data-type=\"solution\">\n<h3>Solution<\/h3>\n<p id=\"fs-id1165137705786\">Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation<\/p>\n<div id=\"eip-id1165134081701\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/div>\n<p id=\"fs-id1165137563332\">We know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].<\/p>\n<div id=\"eip-id1165134353692\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}165=A{e}^{k0}+35\\hfill & \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill & \\text{Solve for }A.\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165137677897\">We were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.<\/p>\n<div id=\"eip-id1165134081505\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }150=130{e}^{k10}+35\\hfill & \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill & \\text{Subtract 35}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill & \\text{Divide by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill & \\text{Divide by the coefficient of }k.\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165134294911\">This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].<\/p>\n<p id=\"fs-id1165135194273\">Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\n<div id=\"eip-id1165135149883\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}70=130{e}^{-0.0123t}+35\\hfill & \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill & \\text{Subtract 35}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill & \\text{Divide by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill & \\text{Take the natural log of both sides}\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill & \\text{Divide by the coefficient of }t.\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165135186349\">It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 4<\/h3>\n<p id=\"fs-id1165137686701\">A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/precalcone\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-11248\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":24,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-11248","chapter","type-chapter","status-publish","hentry"],"part":11222,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11248","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11248\/revisions"}],"predecessor-version":[{"id":16429,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11248\/revisions\/16429"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/parts\/11222"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/11248\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/media?parent=11248"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapter-type?post=11248"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/contributor?post=11248"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/license?post=11248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}