{"id":190124,"date":"2025-02-11T07:17:58","date_gmt":"2025-02-11T07:17:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190124"},"modified":"2025-02-11T07:18:00","modified_gmt":"2025-02-11T07:18:00","slug":"a-can-of-soda-is-placed-inside-a-cooler","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/11\/a-can-of-soda-is-placed-inside-a-cooler\/","title":{"rendered":"A can of soda is placed inside a cooler"},"content":{"rendered":"\n<p>A can of soda is placed inside a cooler. As the soda cools, its temperature\u00c3\u201a\u00c2 Tx \u00c3\u201a\u00c2 in degrees Celsius is given by the following function, where\u00c3\u201a\u00c2 x \u00c3\u201a\u00c2 is the number of minutes since the can was placed in the cooler. =Tx+\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122624e\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u21220.045x Find the initial temperature of the soda and its temperature after\u00c3\u201a\u00c2 20<br>A can of soda is placed inside a cooler. As the soda cools, its temperature\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>Tx<br>\u00c3\u201a\u00c2 in degrees Celsius is given by the following function, where\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>x<br>\u00c3\u201a\u00c2 is the number of minutes since the can was placed in the cooler.<\/p>\n\n\n\n<p>=Tx+\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122624e\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u21220.045x<br>Find the initial temperature of the soda and its temperature after\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>20<br>\u00c3\u201a\u00c2 minutes.<br>Round your answers to the nearest degree as necessary.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>The temperature of the soda is modeled by the equation:<\/p>\n\n\n\n<p>[<br>T(x) = 24 + 6e^{-0.045x}<br>]<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( T(x) ) is the temperature of the soda in degrees Celsius at time ( x ) minutes after it was placed in the cooler,<\/li>\n\n\n\n<li>( e ) is the base of the natural logarithm,<\/li>\n\n\n\n<li>( x ) is the number of minutes since the soda was placed in the cooler.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">1. Finding the initial temperature of the soda (( T(0) )):<\/h3>\n\n\n\n<p>To find the initial temperature, substitute ( x = 0 ) into the equation.<\/p>\n\n\n\n<p>[<br>T(0) = 24 + 6e^{-0.045(0)}<br>]<br>[<br>T(0) = 24 + 6e^0<br>]<br>[<br>T(0) = 24 + 6(1)<br>]<br>[<br>T(0) = 24 + 6 = 30^\\circ C<br>]<\/p>\n\n\n\n<p>So, the initial temperature of the soda is <strong>30\u00b0C<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Finding the temperature after 20 minutes (( T(20) )):<\/h3>\n\n\n\n<p>Now, substitute ( x = 20 ) into the equation to find the temperature after 20 minutes.<\/p>\n\n\n\n<p>[<br>T(20) = 24 + 6e^{-0.045(20)}<br>]<br>[<br>T(20) = 24 + 6e^{-0.9}<br>]<\/p>\n\n\n\n<p>We need to calculate ( e^{-0.9} ). Using a calculator:<\/p>\n\n\n\n<p>[<br>e^{-0.9} \\approx 0.4066<br>]<\/p>\n\n\n\n<p>Now, substitute this value back into the equation:<\/p>\n\n\n\n<p>[<br>T(20) = 24 + 6(0.4066)<br>]<br>[<br>T(20) = 24 + 2.4396<br>]<br>[<br>T(20) \\approx 26.44^\\circ C<br>]<\/p>\n\n\n\n<p>Rounding to the nearest degree:<\/p>\n\n\n\n<p>[<br>T(20) \\approx 26^\\circ C<br>]<\/p>\n\n\n\n<p>Thus, the temperature after 20 minutes is <strong>26\u00b0C<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The initial temperature of the soda is <strong>30\u00b0C<\/strong>.<\/li>\n\n\n\n<li>The temperature of the soda after 20 minutes is approximately <strong>26\u00b0C<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>This equation follows the cooling process of the soda where the temperature decreases over time, and the exponential term ( e^{-0.045x} ) causes the rate of cooling to slow down as the soda approaches the temperature of the cooler.<\/p>\n\n\n\n<p>Now, I&#8217;ll generate an image of the graph that shows how the temperature of the soda changes over time.<\/p>\n\n\n\n<p>Here is the graph showing the temperature of the soda over time. As you can see, the temperature starts at 30\u00b0C and decreases as time progresses. After 20 minutes, the temperature of the soda has dropped to approximately 26\u00b0C. The exponential decay curve represents the cooling process, where the rate of temperature change slows down over time.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-307-1024x812.png\" alt=\"\" class=\"wp-image-190125\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A can of soda is placed inside a cooler. As the soda cools, its temperature\u00c3\u201a\u00c2 Tx \u00c3\u201a\u00c2 in degrees Celsius is given by the following function, where\u00c3\u201a\u00c2 x \u00c3\u201a\u00c2 is the number of minutes since the can was placed in the cooler. =Tx+\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122624e\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u21220.045x Find the initial temperature of the soda and its temperature after\u00c3\u201a\u00c2 20A [&hellip;]<\/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-190124","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190124","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=190124"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190124\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190124"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190124"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190124"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}