{"id":8456,"date":"2025-05-25T10:12:34","date_gmt":"2025-05-25T10:12:34","guid":{"rendered":"https:\/\/yaveni.com\/blog\/?p=8456"},"modified":"2025-05-25T10:12:38","modified_gmt":"2025-05-25T10:12:38","slug":"the-unit-weight-of-water-at-50-f-is-62-4-pcf","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-unit-weight-of-water-at-50-f-is-62-4-pcf\/","title":{"rendered":"The unit weight of water at 50 \u00b0F is 62.4 pcf"},"content":{"rendered":"\n<p>The unit weight of water at 50 \u00b0F is 62.4 pcf. If the volume of the vessel that contains water is 3.50 ft A. What will be the change in its volume when it is heated to 160\u00b0F (Unit weight of water is 61 pcf 160\u00b0F) B. Compute the percentage change of volume. C. What weight of water must be removed to maintain the original volume.<\/p>\n\n\n\n<p><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>Let&#8217;s analyze the problem step by step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Unit weight of water at 50 \u00b0F = 62.4 pcf<\/li>\n\n\n\n<li>Unit weight of water at 160 \u00b0F = 61 pcf<\/li>\n\n\n\n<li>Initial volume V1=3.50\u2009ft3V_1 = 3.50 \\, \\text{ft}^3<\/li>\n\n\n\n<li>Temperature change from 50 \u00b0F to 160 \u00b0F<\/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\">A. What will be the change in volume when heated from 50\u00b0F to 160\u00b0F?<\/h3>\n\n\n\n<p><strong>Concept:<\/strong><\/p>\n\n\n\n<p>The unit weight \u03b3\\gamma is defined as: \u03b3=WV\\gamma = \\frac{W}{V}<\/p>\n\n\n\n<p>where WW is the weight and VV is the volume.<\/p>\n\n\n\n<p>If the weight remains constant and the unit weight changes (due to temperature), then the volume must change accordingly.<\/p>\n\n\n\n<p>Since weight WW is constant: W=\u03b31V1=\u03b32V2W = \\gamma_1 V_1 = \\gamma_2 V_2<\/p>\n\n\n\n<p>Rearranged for V2V_2: V2=\u03b31\u03b32V1V_2 = \\frac{\\gamma_1}{\\gamma_2} V_1<\/p>\n\n\n\n<p>Substitute values: V2=62.461\u00d73.50=1.02295\u00d73.50=3.5803\u2009ft3V_2 = \\frac{62.4}{61} \\times 3.50 = 1.02295 \\times 3.50 = 3.5803 \\, \\text{ft}^3<\/p>\n\n\n\n<p><strong>Change in volume, \u0394V\\Delta V:<\/strong> \u0394V=V2\u2212V1=3.5803\u22123.50=0.0803\u2009ft3\\Delta V = V_2 &#8211; V_1 = 3.5803 &#8211; 3.50 = 0.0803 \\, \\text{ft}^3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">B. Compute the percentage change of volume.<\/h3>\n\n\n\n<p>%&nbsp;change=\u0394VV1\u00d7100=0.08033.50\u00d7100=2.29%\\%\\text{ change} = \\frac{\\Delta V}{V_1} \\times 100 = \\frac{0.0803}{3.50} \\times 100 = 2.29\\%<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">C. What weight of water must be removed to maintain the original volume of 3.50 ft\u00b3 at 160\u00b0F?<\/h3>\n\n\n\n<p>At 160 \u00b0F, if the volume is to remain the same V=3.50\u2009ft3V = 3.50 \\, \\text{ft}^3, but the unit weight decreases to 61 pcf, the weight must change accordingly. W2=\u03b32\u00d7V=61\u00d73.50=213.5\u2009lbW_2 = \\gamma_2 \\times V = 61 \\times 3.50 = 213.5 \\, \\text{lb}<\/p>\n\n\n\n<p>Original weight at 50 \u00b0F: W1=\u03b31\u00d7V1=62.4\u00d73.50=218.4\u2009lbW_1 = \\gamma_1 \\times V_1 = 62.4 \\times 3.50 = 218.4 \\, \\text{lb}<\/p>\n\n\n\n<p>Weight to be removed: Wremoved=W1\u2212W2=218.4\u2212213.5=4.9\u2009lbW_{\\text{removed}} = W_1 &#8211; W_2 = 218.4 &#8211; 213.5 = 4.9 \\, \\text{lb}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Summary of answers:<\/h2>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Part<\/th><th>Result<\/th><\/tr><\/thead><tbody><tr><td>A<\/td><td>Volume increases by 0.0803 ft\u00b3<\/td><\/tr><tr><td>B<\/td><td>Percentage volume change = 2.29%<\/td><\/tr><tr><td>C<\/td><td>Weight to be removed = 4.9 lb<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Explanation (around 300 words):<\/h2>\n\n\n\n<p>When water is heated, its density decreases, causing the volume to expand if the weight remains constant. The unit weight of water is directly proportional to its density, and the decrease in unit weight from 62.4 pcf at 50 \u00b0F to 61 pcf at 160 \u00b0F indicates that water expands upon heating.<\/p>\n\n\n\n<p>The problem gives the initial volume of water at 50 \u00b0F and asks for the volume change after heating to 160 \u00b0F. Since the weight stays the same, we use the relationship between weight, volume, and unit weight. At constant weight, the volume changes inversely with the unit weight. As the unit weight decreases, volume must increase to keep weight constant.<\/p>\n\n\n\n<p>Calculating the new volume at 160 \u00b0F showed a 2.29% increase in volume (0.0803 ft\u00b3) compared to the original 3.50 ft\u00b3. This expansion is typical for liquids as they heat.<\/p>\n\n\n\n<p>If it\u2019s necessary to keep the volume fixed at 3.50 ft\u00b3 despite the temperature increase, some water must be removed because the water expands and would occupy more space otherwise. Calculating the weight of water for the fixed volume at 160 \u00b0F yields a smaller weight (213.5 lb) compared to the original (218.4 lb). The difference (4.9 lb) is the amount of water to remove to maintain the original volume.<\/p>\n\n\n\n<p>This problem exemplifies thermal expansion of liquids and its impact on volume and weight relationships. It\u2019s important in practical applications such as fluid storage and transportation, where temperature changes can cause volume changes that affect system design or operation. Understanding these principles helps in maintaining desired system parameters by compensating for expansion or contraction.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The unit weight of water at 50 \u00b0F is 62.4 pcf. If the volume of the vessel that contains water is 3.50 ft A. What will be the change in its volume when it is heated to 160\u00b0F (Unit weight of water is 61 pcf 160\u00b0F) B. Compute the percentage change of volume. C. What [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-8456","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/8456","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=8456"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/8456\/revisions"}],"predecessor-version":[{"id":8457,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/8456\/revisions\/8457"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=8456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=8456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=8456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}