{"id":231379,"date":"2025-06-11T03:58:42","date_gmt":"2025-06-11T03:58:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231379"},"modified":"2025-06-11T03:58:45","modified_gmt":"2025-06-11T03:58:45","slug":"suppose-it-is-winter-and-the-system-is-filled-with-methanol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/suppose-it-is-winter-and-the-system-is-filled-with-methanol\/","title":{"rendered":"Suppose it is winter and the system is filled with methanol."},"content":{"rendered":"\n<p>Suppose it is winter and the system is filled with methanol. The density of methanol is 0.80 g\/cm3. What would be the increase in temperature of the methanol if it absorbed 836.0 kJ of heat<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Based on the data, the increase in temperature of the methanol would be&nbsp;<strong>413.0 \u00b0C<\/strong>.<\/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>The relationship between heat energy, mass, specific heat, and temperature change is described by the specific heat formula. This principle states that the amount of heat absorbed or released by a substance is directly proportional to its mass and the change in its temperature. The formula is expressed as:<\/p>\n\n\n\n<p><strong>Q = mc\u0394T<\/strong><\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Q<\/strong>\u00a0is the heat energy absorbed or released (in Joules, J).<\/li>\n\n\n\n<li><strong>m<\/strong>\u00a0is the mass of the substance (in grams, g).<\/li>\n\n\n\n<li><strong>c<\/strong>\u00a0is the specific heat capacity of the substance (in J\/g\u00b0C). This is a constant value unique to each material, representing the energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For methanol, the value is approximately\u00a0<strong>2.53 J\/g\u00b0C<\/strong>.<\/li>\n\n\n\n<li><strong>\u0394T<\/strong>\u00a0(delta T) is the change in temperature (in \u00b0C), which we need to find.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Determine the mass of the methanol.<\/strong><br>The problem does not provide the mass directly but gives the density. To make the problem solvable, we must assume a standard volume, such as 1.0 liter (1000 cm\u00b3). Using the density, we can calculate the mass.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume (V) = 1000 cm\u00b3<\/li>\n\n\n\n<li>Density (\u03c1) = 0.80 g\/cm\u00b3<\/li>\n\n\n\n<li>Mass (m) = V \u00d7 \u03c1 = 1000 cm\u00b3 \u00d7 0.80 g\/cm\u00b3 =\u00a0<strong>800 g<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Convert the heat energy to Joules.<\/strong><br>The heat absorbed (Q) is given in kilojoules (kJ). We must convert it to Joules to match the units of the specific heat capacity.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Q = 836.0 kJ \u00d7 1000 J\/kJ =\u00a0<strong>836,000 J<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Calculate the temperature increase (\u0394T).<\/strong><br>Rearrange the specific heat formula to solve for \u0394T:<\/p>\n\n\n\n<p><strong>\u0394T = Q \/ (mc)<\/strong><\/p>\n\n\n\n<p>Now, substitute the known values into the equation:<\/p>\n\n\n\n<p>\u0394T = 836,000 J \/ (800 g \u00d7 2.53 J\/g\u00b0C)<br>\u0394T = 836,000 \/ 2024<br><strong>\u0394T = 413.0 \u00b0C<\/strong><\/p>\n\n\n\n<p>Therefore, the absorption of 836.0 kJ of heat would cause the temperature of 1.0 L of methanol to increase by 413.0 \u00b0C.thumb_upthumb_down<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-887.jpeg\" alt=\"\" class=\"wp-image-231380\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Suppose it is winter and the system is filled with methanol. The density of methanol is 0.80 g\/cm3. What would be the increase in temperature of the methanol if it absorbed 836.0 kJ of heat The Correct Answer and Explanation is: Based on the data, the increase in temperature of the methanol would be&nbsp;413.0 \u00b0C. [&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-231379","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231379","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=231379"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231379\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231379"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231379"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231379"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}