{"id":217766,"date":"2025-05-22T12:53:31","date_gmt":"2025-05-22T12:53:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217766"},"modified":"2025-05-22T12:53:34","modified_gmt":"2025-05-22T12:53:34","slug":"the-viscosity-of-water-at-20-0-c-68-0-f-is-1-008-cp-centipoises","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/the-viscosity-of-water-at-20-0-c-68-0-f-is-1-008-cp-centipoises\/","title":{"rendered":"The viscosity of water at 20 0 C (68 0 F) is 1.008 cp (centipoises)."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The viscosity of water at 20 0 C (68 0 F) is 1.008 cp (centipoises).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Compute the absolute viscosity in lb-sec\/ft2.<br>If the specific gravity at 20 0 C is 0.998, compute the kinematic viscosity in ft2\/sec.<br>Use:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">1 poise = 1 dyne-sec\/cm2<br>1 lb = 444,800 dynes<br>1 ft = 30.48 cm<br>Please provide a brief step explanation since a solution has been already provided, but it isn&#8217;t clear how they arrived at the final solution. Thank you!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">Sure! Let\u2019s break down the problem and clarify each step in detail to understand how to convert viscosity units and calculate kinematic viscosity.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Viscosity of water at 20\u00b0C = 1.008 cp (centipoises)<\/li>\n\n\n\n<li>Specific gravity (SG) at 20\u00b0C = 0.998<\/li>\n\n\n\n<li>1 poise = 1 dyne\u00b7sec\/cm\u00b2<\/li>\n\n\n\n<li>1 lb = 444,800 dynes<\/li>\n\n\n\n<li>1 ft = 30.48 cm<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert centipoises (cp) to absolute viscosity in lb\u00b7sec\/ft\u00b2<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>What is centipoise?<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 poise = 1 dyne\u00b7sec\/cm\u00b2<\/li>\n\n\n\n<li>1 centipoise = 0.01 poise = 0.01 dyne\u00b7sec\/cm\u00b2<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>1.008 \\text{ cp} = 1.008 \\times 0.01 = 0.01008 \\text{ poise} = 0.01008 \\text{ dyne\u00b7sec\/cm}^2<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Convert dyne\u00b7sec\/cm\u00b2 to lb\u00b7sec\/ft\u00b2<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 lb = 444,800 dynes<\/li>\n\n\n\n<li>1 ft = 30.48 cm \u2192 so $1 \\text{ ft}^2 = (30.48)^2 \\text{ cm}^2 = 929.0304 \\text{ cm}^2$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">The goal: convert from dyne\u00b7sec\/cm\u00b2 to lb\u00b7sec\/ft\u00b2.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 dyne\u00b7sec\/cm\u00b2 = ?<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Since 1 lb = 444,800 dynes, and 1 ft\u00b2 = 929.0304 cm\u00b2:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>1 \\text{ dyne\u00b7sec\/cm}^2 = \\frac{1}{444,800} \\text{ lb\u00b7sec\/cm}^2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But we want per ft\u00b2:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{To get lb\u00b7sec\/ft}^2 = \\text{dyne\u00b7sec\/cm}^2 \\times \\frac{1 \\text{ lb}}{444,800 \\text{ dynes}} \\times \\frac{929.0304 \\text{ cm}^2}{1 \\text{ ft}^2}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since dyne\u00b7sec\/cm\u00b2 is per cm\u00b2, to convert to per ft\u00b2 we multiply by cm\u00b2\/ft\u00b2:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Absolute viscosity} = 0.01008 \\times \\frac{929.0304}{444,800} \\quad \\text{lb\u00b7sec\/ft}^2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate the factor:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\frac{929.0304}{444,800} = 0.002087<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Absolute viscosity} = 0.01008 \\times 0.002087 = 2.10 \\times 10^{-5} \\text{ lb\u00b7sec\/ft}^2<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Compute kinematic viscosity $\\nu$ in ft\u00b2\/sec<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Definition:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\nu = \\frac{\\mu}{\\rho}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\nu$ = kinematic viscosity (ft\u00b2\/sec)<\/li>\n\n\n\n<li>$\\mu$ = absolute viscosity (lb\u00b7sec\/ft\u00b2)<\/li>\n\n\n\n<li>$\\rho$ = density (lb\/ft\u00b3)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Calculate density $\\rho$:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Specific gravity (SG) = $\\frac{\\rho_{\\text{water}}}{\\rho_{\\text{reference}}}$<\/li>\n\n\n\n<li>At 20\u00b0C, water density = 62.4 lb\/ft\u00b3 (approximate)<\/li>\n\n\n\n<li>So,<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\rho = SG \\times 62.4 = 0.998 \\times 62.4 = 62.28 \\text{ lb\/ft}^3<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Calculate kinematic viscosity:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\nu = \\frac{2.10 \\times 10^{-5} \\text{ lb\u00b7sec\/ft}^2}{62.28 \\text{ lb\/ft}^3} = 3.37 \\times 10^{-7} \\text{ ft}^2\/\\text{sec}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary of final answers:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Absolute viscosity $\\mu = 2.10 \\times 10^{-5}$ lb\u00b7sec\/ft\u00b2<\/li>\n\n\n\n<li>Kinematic viscosity $\\nu = 3.37 \\times 10^{-7}$ ft\u00b2\/sec<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">This problem is about unit conversion and understanding the physical relationship between absolute and kinematic viscosity.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Viscosity in Centipoises:<\/strong> Viscosity is often given in poise or centipoise in the CGS system (centimeter-gram-second). One poise is 1 dyne\u00b7sec\/cm\u00b2. Since the given viscosity is in centipoise, you multiply by 0.01 to convert to poise.<\/li>\n\n\n\n<li><strong>Absolute Viscosity Conversion:<\/strong> To convert the viscosity to imperial units (lb\u00b7sec\/ft\u00b2), convert dynes to pounds and cm\u00b2 to ft\u00b2. The key conversion factor is that 1 lb equals 444,800 dynes and 1 ft equals 30.48 cm, so 1 ft\u00b2 = (30.48)\u00b2 cm\u00b2. This lets us scale the units accordingly.<\/li>\n\n\n\n<li><strong>Specific Gravity and Density:<\/strong> The specific gravity relates the fluid density to water\u2019s density. Since the density of water at 20\u00b0C is about 62.4 lb\/ft\u00b3, multiply by the specific gravity to find the actual density.<\/li>\n\n\n\n<li><strong>Kinematic Viscosity:<\/strong> Kinematic viscosity is absolute viscosity divided by density, giving a measure of the fluid\u2019s resistance to flow under gravity. The units ft\u00b2\/sec correspond to how quickly the fluid spreads or flows.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">This detailed breakdown clarifies how the unit conversions and calculations come together to provide the absolute and kinematic viscosity in imperial units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">If you want, I can walk through the math again with more detail or help with related problems!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-63.png\" alt=\"\" class=\"wp-image-217767\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The viscosity of water at 20 0 C (68 0 F) is 1.008 cp (centipoises). Compute the absolute viscosity in lb-sec\/ft2.If the specific gravity at 20 0 C is 0.998, compute the kinematic viscosity in ft2\/sec.Use: 1 poise = 1 dyne-sec\/cm21 lb = 444,800 dynes1 ft = 30.48 cmPlease provide a brief step explanation since [&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-217766","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217766","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=217766"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217766\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217766"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217766"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217766"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}