{"id":16914,"date":"2025-06-11T21:13:38","date_gmt":"2025-06-11T21:13:38","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16914"},"modified":"2025-06-11T21:13:41","modified_gmt":"2025-06-11T21:13:41","slug":"determine-the-height-that-water-at-60-f-will-rise-due-to-capillary-action-in-a-clean","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-height-that-water-at-60-f-will-rise-due-to-capillary-action-in-a-clean\/","title":{"rendered":"Determine the height that water at 60 F will rise due to capillary action in a clean"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Determine the height that water at 60 F will rise due to capillary action in a clean, 3\/8-in-diameter glass tube. What will be the height if the diameter is reduced to 0.0117 in?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the height that water will rise in a capillary tube due to <strong>capillary action<\/strong>, we use the <strong>Jurin\u2019s Law<\/strong>: h=4\u03b3cos\u2061\u03b8\u03c1gdh = \\frac{4 \\gamma \\cos\\theta}{\\rho g d}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>hh = height of capillary rise (m or in)<\/li>\n\n\n\n<li>\u03b3\\gamma = surface tension of the liquid (N\/m)<\/li>\n\n\n\n<li>\u03b8\\theta = contact angle (\u00b0) (for water and clean glass, \u2248 0\u00b0, so cos\u2061\u03b8=1\\cos\\theta = 1)<\/li>\n\n\n\n<li>\u03c1\\rho = density of the liquid (kg\/m\u00b3)<\/li>\n\n\n\n<li>gg = acceleration due to gravity (9.81 m\/s\u00b2)<\/li>\n\n\n\n<li>dd = inner diameter of the tube (m)<\/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\">Constants for water at 60\u00b0F (15.6\u00b0C):<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Surface tension \u03b3=0.07275\u2009N\/m\\gamma = 0.07275 \\, \\text{N\/m}<\/li>\n\n\n\n<li>Density \u03c1=999\u2009kg\/m3\\rho = 999 \\, \\text{kg\/m}^3<\/li>\n\n\n\n<li>Contact angle \u03b8=0\u2218\\theta = 0^\\circ, so cos\u2061\u03b8=1\\cos\\theta = 1<\/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\">Case 1: 3\/8 in diameter (0.375 in = 0.009525 m)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">h=4(0.07275)(1)(999)(9.81)(0.009525)=0.29193.35=0.00312\u2009m=3.12\u2009mmh = \\frac{4 (0.07275)(1)}{(999)(9.81)(0.009525)} = \\frac{0.291}{93.35} = 0.00312 \\, \\text{m} = 3.12 \\, \\text{mm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Case 2: 0.0117 in diameter (0.00029718 m)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">h=4(0.07275)(999)(9.81)(0.00029718)=0.2912.91=0.100\u2009m=100.0\u2009mmh = \\frac{4 (0.07275)}{(999)(9.81)(0.00029718)} = \\frac{0.291}{2.91} = 0.100 \\, \\text{m} = 100.0 \\, \\text{mm}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>3\/8-in tube<\/strong>: 3.12\u2009mm\\boxed{3.12 \\, \\text{mm}}<\/li>\n\n\n\n<li><strong>0.0117-in tube<\/strong>: 100.0\u2009mm\\boxed{100.0 \\, \\text{mm}}<\/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\">Explanation (\u2248300 words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity. This phenomenon is driven by surface tension and the adhesive forces between the liquid and the solid surface of the tube. Jurin\u2019s Law describes the height to which a liquid will rise or fall in a capillary tube and shows that capillary height is <strong>inversely proportional<\/strong> to the tube diameter.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">When water contacts a clean glass surface, strong adhesive forces between the water molecules and glass cause the water to &#8220;climb&#8221; the tube, forming a concave meniscus. The narrower the tube, the greater the surface area of contact per unit volume, which increases the upward adhesive force relative to the weight of the water column.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this problem, water at 60\u00b0F (about 15.6\u00b0C) has a known surface tension and density. Using these values and assuming a contact angle of zero (perfect wetting), we applied Jurin\u2019s Law to determine the height of the water column for two tube diameters.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For the 3\/8-inch (9.525 mm) diameter tube, the capillary rise is quite small\u2014only about <strong>3.12 mm<\/strong>, because the diameter is relatively large. In contrast, when the diameter is reduced drastically to <strong>0.0117 in (0.297 mm)<\/strong>, the height increases significantly to <strong>100 mm<\/strong>. This illustrates how smaller tubes generate more substantial capillary action due to the increased dominance of surface forces over gravity.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Understanding this relationship is essential in fields like soil science, biology (e.g., water transport in plants), and engineering (e.g., inkjet printers, microfluidics), where controlling fluid behavior in small channels is crucial.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Determine the height that water at 60 F will rise due to capillary action in a clean, 3\/8-in-diameter glass tube. What will be the height if the diameter is reduced to 0.0117 in? The correct answer and explanation is: To determine the height that water will rise in a capillary tube due to capillary action, [&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-16914","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16914","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=16914"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16914\/revisions"}],"predecessor-version":[{"id":16915,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16914\/revisions\/16915"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16914"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16914"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}