{"id":180868,"date":"2025-01-09T06:46:22","date_gmt":"2025-01-09T06:46:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=180868"},"modified":"2025-01-09T06:46:24","modified_gmt":"2025-01-09T06:46:24","slug":"how-much-energy-is-needed-to-change-the-temperature-of-50-0-g-of-water-by-15-0c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/09\/how-much-energy-is-needed-to-change-the-temperature-of-50-0-g-of-water-by-15-0c\/","title":{"rendered":"How much energy is needed to change the temperature of 50.0 g of water by 15.0\u00b0C"},"content":{"rendered":"\n<p> How much energy is needed to change the temperature of 50.0 g of water by 15.0\u00b0C? 2. How many grams of water can be heated from 20.0 C to 75\u00b0C using 12500.0 Joules?<\/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<h3 class=\"wp-block-heading\">Problem 1: How much energy is needed to change the temperature of 50.0 g of water by 15.0\u00b0C?<\/h3>\n\n\n\n<p>To calculate the energy required, we can use the formula:<\/p>\n\n\n\n<p>[<br>Q = mc\\Delta T<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( Q ) = heat energy (in Joules)<\/li>\n\n\n\n<li>( m ) = mass of water (in grams)<\/li>\n\n\n\n<li>( c ) = specific heat capacity of water (( 4.18 \\, \\text{J\/g\u00b0C} ))<\/li>\n\n\n\n<li>( \\Delta T ) = change in temperature (in \u00b0C)<\/li>\n<\/ul>\n\n\n\n<p>Plugging the values into the formula:<\/p>\n\n\n\n<p>[<br>Q = (50.0 \\, \\text{g}) \\times (4.18 \\, \\text{J\/g\u00b0C}) \\times (15.0 \\, \\text{\u00b0C})<br>]<\/p>\n\n\n\n<p>[<br>Q = 3135 \\, \\text{J}<br>]<\/p>\n\n\n\n<p>So, <strong>3135 Joules<\/strong> of energy are needed to change the temperature of 50.0 g of water by 15.0\u00b0C.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Problem 2: How many grams of water can be heated from 20.0\u00b0C to 75.0\u00b0C using 12,500.0 Joules?<\/h3>\n\n\n\n<p>Again, we use the same formula:<\/p>\n\n\n\n<p>[<br>Q = mc\\Delta T<br>]<\/p>\n\n\n\n<p>Rearranging it to solve for mass ( m ):<\/p>\n\n\n\n<p>[<br>m = \\frac{Q}{c \\Delta T}<br>]<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( Q = 12,500.0 \\, \\text{J} )<\/li>\n\n\n\n<li>( c = 4.18 \\, \\text{J\/g\u00b0C} )<\/li>\n\n\n\n<li>( \\Delta T = 75.0\u00b0C &#8211; 20.0\u00b0C = 55.0\u00b0C )<\/li>\n<\/ul>\n\n\n\n<p>Now substitute the values:<\/p>\n\n\n\n<p>[<br>m = \\frac{12,500.0 \\, \\text{J}}{(4.18 \\, \\text{J\/g\u00b0C}) \\times (55.0 \\, \\text{\u00b0C})}<br>]<\/p>\n\n\n\n<p>[<br>m = \\frac{12,500.0 \\, \\text{J}}{229.9 \\, \\text{J\/g}}<br>]<\/p>\n\n\n\n<p>[<br>m \\approx 54.4 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, <strong>54.4 grams<\/strong> of water can be heated from 20.0\u00b0C to 75.0\u00b0C using 12,500.0 Joules of energy.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>In calorimetry, the amount of heat energy required to change the temperature of a substance is determined by the specific heat capacity, which is the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. Water has a relatively high specific heat capacity, meaning it takes a significant amount of energy to change its temperature.<\/p>\n\n\n\n<p>In both problems, we apply the formula ( Q = mc\\Delta T ), where ( Q ) is the heat energy, ( m ) is the mass of water, ( c ) is the specific heat capacity of water (4.18 J\/g\u00b0C), and ( \\Delta T ) is the change in temperature.<\/p>\n\n\n\n<p>In the first problem, we are given the mass and temperature change, so we simply multiply them together to find the energy required. In the second problem, we are given the energy and temperature change, and we rearrange the formula to solve for mass.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How much energy is needed to change the temperature of 50.0 g of water by 15.0\u00b0C? 2. How many grams of water can be heated from 20.0 C to 75\u00b0C using 12500.0 Joules? The Correct Answer and Explanation is : Problem 1: How much energy is needed to change the temperature of 50.0 g of [&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-180868","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180868","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=180868"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180868\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180868"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180868"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180868"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}