{"id":46799,"date":"2025-07-02T08:35:04","date_gmt":"2025-07-02T08:35:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46799"},"modified":"2025-07-02T08:35:07","modified_gmt":"2025-07-02T08:35:07","slug":"a-can-of-soda-is-placed-inside-a-cooler","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/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 T(x) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler. T(x) = -19 + 41e^(-0.45x) Find the initial temperature of the soda: Find its temperature after 20 minutes: Round your answers to the nearest degree as necessary.<\/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>To solve the given problem, we need to evaluate the temperature function at two different points: at the initial time (when x = 0) and after 20 minutes (when x = 20).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The Temperature Function:<\/h3>\n\n\n\n<p>The temperature of the soda is modeled by the function:T(x)=\u221219+41e\u22120.45xT(x) = -19 + 41e^{-0.45x}T(x)=\u221219+41e\u22120.45x<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>T(x)T(x)T(x) is the temperature of the soda at time xxx (in minutes),<\/li>\n\n\n\n<li>xxx is the number of minutes since the soda was placed in the cooler,<\/li>\n\n\n\n<li>eee is Euler&#8217;s number (approximately 2.71828),<\/li>\n\n\n\n<li>\u221219-19\u221219 represents a constant temperature shift,<\/li>\n\n\n\n<li>414141 is the initial temperature difference from \u221219-19\u221219,<\/li>\n\n\n\n<li>\u22120.45-0.45\u22120.45 is the rate of cooling of the soda over time.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Finding the Initial Temperature (at x=0x = 0x=0):<\/h3>\n\n\n\n<p>To find the initial temperature, substitute x=0x = 0x=0 into the temperature function:T(0)=\u221219+41e\u22120.45(0)T(0) = -19 + 41e^{-0.45(0)}T(0)=\u221219+41e\u22120.45(0)T(0)=\u221219+41e0T(0) = -19 + 41e^{0}T(0)=\u221219+41e0T(0)=\u221219+41(1)(because&nbsp;e0=1)T(0) = -19 + 41(1) \\quad \\text{(because } e^0 = 1\\text{)}T(0)=\u221219+41(1)(because&nbsp;e0=1)T(0)=\u221219+41=22\u2218CT(0) = -19 + 41 = 22^\\circ CT(0)=\u221219+41=22\u2218C<\/p>\n\n\n\n<p>So, the <strong>initial temperature<\/strong> of the soda is <strong>22\u00b0C<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Finding the Temperature After 20 Minutes (at x=20x = 20x=20):<\/h3>\n\n\n\n<p>Now, to find the temperature after 20 minutes, substitute x=20x = 20x=20 into the function:T(20)=\u221219+41e\u22120.45(20)T(20) = -19 + 41e^{-0.45(20)}T(20)=\u221219+41e\u22120.45(20)T(20)=\u221219+41e\u22129T(20) = -19 + 41e^{-9}T(20)=\u221219+41e\u22129T(20)\u2248\u221219+41(0.000123)(using&nbsp;a&nbsp;calculator&nbsp;for&nbsp;e\u22129\u22480.000123)T(20) \\approx -19 + 41(0.000123) \\quad \\text{(using a calculator for } e^{-9} \\approx 0.000123\\text{)}T(20)\u2248\u221219+41(0.000123)(using&nbsp;a&nbsp;calculator&nbsp;for&nbsp;e\u22129\u22480.000123)T(20)\u2248\u221219+0.00504T(20) \\approx -19 + 0.00504T(20)\u2248\u221219+0.00504T(20)\u2248\u221218.99496T(20) \\approx -18.99496T(20)\u2248\u221218.99496<\/p>\n\n\n\n<p>Rounding to the nearest degree, the temperature after 20 minutes is approximately <strong>-19\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 <strong>initial temperature<\/strong> of the soda is <strong>22\u00b0C<\/strong>.<\/li>\n\n\n\n<li>The temperature of the soda after 20 minutes is approximately <strong>-19\u00b0C<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>This result illustrates the cooling process, where the temperature quickly drops initially and then slows down as the soda approaches a cooler equilibrium.<\/p>\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\/07\/learnexams-banner5-186.jpeg\" alt=\"\" class=\"wp-image-46809\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-186.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-186-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-186-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-186-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A can of soda is placed inside a cooler. As the soda cools, its temperature T(x) in degrees Celsius is given by the following function, where x is the number of minutes since the can was placed in the cooler. T(x) = -19 + 41e^(-0.45x) Find the initial temperature of the soda: Find its temperature [&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-46799","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46799","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=46799"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46799\/revisions"}],"predecessor-version":[{"id":46810,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46799\/revisions\/46810"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46799"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46799"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46799"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}