{"id":181559,"date":"2025-01-11T02:36:37","date_gmt":"2025-01-11T02:36:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181559"},"modified":"2025-01-11T02:36:39","modified_gmt":"2025-01-11T02:36:39","slug":"thermal-energy-name-figure-2-9b-temperature-conversion-formulas-conversion","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/11\/thermal-energy-name-figure-2-9b-temperature-conversion-formulas-conversion\/","title":{"rendered":"Thermal Energy Name: Figure 2-9b Temperature Conversion Formulas Conversion"},"content":{"rendered":"\n

Thermal Energy Name: Figure 2-9b Temperature Conversion Formulas Conversion Formula Example Celsius to Kelvin K=C+ 273 21\u00b0C = 294 K Kelvin to Celsius C=K-273 313 K = 40\u00b0C Fahrenheit to Celsius C = (F-32) x 59 89 “F= 31.7\u00b0C Celsius to Fahrenheit F=(CX9\/5) + 32 50 \u00b0C = 122\u00b0F Questions 1. Convert the following numbers from degrees Fahrenheit to degrees Celsius. a. 0\u00b0F \u00b0C d. 98.6\u00b0 F (body temperature) b. 32\u00b0 F (freezing point) “C e. 100\u00b0 F c. 70\u00b0 F (room temperature) \u00b0C f. 212\u00b0 F (boiling point) \u00b0C \u00b0C \u00b0C “F F F 2. Convert the following numbers from degrees Celsius to degrees Fahrenheit. d. 98.6\u00b0C a. 0\u00b0 C (freezing point) c. 100\u00b0C (boiling point) b. 32\u00b0C F f. 212\u00b0C c. 70\u00b0C OF 3. Complete the chart using the correct conversion formula b. 339 Kelvin to Celsius e. 17 Celsius to Kelvin a. 250 Kelvin to Celsius d. 55 Celsius to Kelvin c. 89.5 Fahrenheit to Celsius d. 383 Kelvin to Fahrenheit 4. The weather forecaster predicts that today’s high will be 70. Which temperature scale is being used? What would be the corresponding temperature on the other two scales? S. “I was so cold yesterday that the temperature only reached 275.” Which temperature scale is being used? What would be the corresponding temperature on the other two scales? 6. “Today’s temperature of 42 in Chicago set a record high for the month of August.” Which temperature scale is being used? What would be the corresponding temperature on the other two scales?<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

1. Convert the following numbers from degrees Fahrenheit to degrees Celsius.<\/h3>\n\n\n\n

a. 0\u00b0F<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=0F = 0: C=(0\u221232)\u00d759=\u221232\u00d759=\u22121609\u2248\u221217.78\u00b0CC = \\frac{(0 – 32) \\times 5}{9} = \\frac{-32 \\times 5}{9} = \\frac{-160}{9} \\approx -17.78\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: -17.78\u00b0C<\/p>\n\n\n\n

b. 32\u00b0F (freezing point)<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=32F = 32: C=(32\u221232)\u00d759=0\u00b0CC = \\frac{(32 – 32) \\times 5}{9} = 0\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: 0\u00b0C<\/p>\n\n\n\n

c. 70\u00b0F (room temperature)<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=70F = 70: C=(70\u221232)\u00d759=38\u00d759=1909\u224821.11\u00b0CC = \\frac{(70 – 32) \\times 5}{9} = \\frac{38 \\times 5}{9} = \\frac{190}{9} \\approx 21.11\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: 21.11\u00b0C<\/p>\n\n\n\n

d. 98.6\u00b0F (body temperature)<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=98.6F = 98.6: C=(98.6\u221232)\u00d759=66.6\u00d759=3339\u224837\u00b0CC = \\frac{(98.6 – 32) \\times 5}{9} = \\frac{66.6 \\times 5}{9} = \\frac{333}{9} \\approx 37\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: 37\u00b0C<\/p>\n\n\n\n

e. 100\u00b0F<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=100F = 100: C=(100\u221232)\u00d759=68\u00d759=3409\u224837.78\u00b0CC = \\frac{(100 – 32) \\times 5}{9} = \\frac{68 \\times 5}{9} = \\frac{340}{9} \\approx 37.78\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: 37.78\u00b0C<\/p>\n\n\n\n

f. 212\u00b0F (boiling point)<\/strong>
Using the formula: C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}<\/p>\n\n\n\n

Substitute F=212F = 212: C=(212\u221232)\u00d759=180\u00d759=9009=100\u00b0CC = \\frac{(212 – 32) \\times 5}{9} = \\frac{180 \\times 5}{9} = \\frac{900}{9} = 100\u00b0C<\/p>\n\n\n\n

Answer<\/strong>: 100\u00b0C<\/p>\n\n\n\n

\n\n\n\n

2. Convert the following numbers from degrees Celsius to degrees Fahrenheit.<\/h3>\n\n\n\n

a. 0\u00b0C (freezing point)<\/strong>
Using the formula: F=(C\u00d795)+32F = \\left( C \\times \\frac{9}{5} \\right) + 32<\/p>\n\n\n\n

Substitute C=0C = 0: F=(0\u00d795)+32=32\u00b0FF = (0 \\times \\frac{9}{5}) + 32 = 32\u00b0F<\/p>\n\n\n\n

Answer<\/strong>: 32\u00b0F<\/p>\n\n\n\n

b. 32\u00b0C<\/strong>
Using the formula: F=(C\u00d795)+32F = \\left( C \\times \\frac{9}{5} \\right) + 32<\/p>\n\n\n\n

Substitute C=32C = 32: F=(32\u00d795)+32=57.6+32=89.6\u00b0FF = (32 \\times \\frac{9}{5}) + 32 = 57.6 + 32 = 89.6\u00b0F<\/p>\n\n\n\n

Answer<\/strong>: 89.6\u00b0F<\/p>\n\n\n\n

c. 100\u00b0C (boiling point)<\/strong>
Using the formula: F=(C\u00d795)+32F = \\left( C \\times \\frac{9}{5} \\right) + 32<\/p>\n\n\n\n

Substitute C=100C = 100: F=(100\u00d795)+32=180+32=212\u00b0FF = (100 \\times \\frac{9}{5}) + 32 = 180 + 32 = 212\u00b0F<\/p>\n\n\n\n

Answer<\/strong>: 212\u00b0F<\/p>\n\n\n\n

d. 98.6\u00b0C<\/strong>
Using the formula: F=(C\u00d795)+32F = \\left( C \\times \\frac{9}{5} \\right) + 32<\/p>\n\n\n\n

Substitute C=98.6C = 98.6: F=(98.6\u00d795)+32=177.48+32=209.48\u00b0FF = (98.6 \\times \\frac{9}{5}) + 32 = 177.48 + 32 = 209.48\u00b0F<\/p>\n\n\n\n

Answer<\/strong>: 209.48\u00b0F<\/p>\n\n\n\n

\n\n\n\n

3. Complete the chart using the correct conversion formula.<\/h3>\n\n\n\n
    \n
  • a. 250 Kelvin to Celsius<\/strong>
    Use the formula C=K\u2212273C = K – 273:<\/li>\n<\/ul>\n\n\n\n

    C=250\u2212273=\u221223\u00b0CC = 250 – 273 = -23\u00b0C<\/p>\n\n\n\n

    Answer<\/strong>: -23\u00b0C<\/p>\n\n\n\n

      \n
    • b. 339 Kelvin to Celsius<\/strong>
      Use the formula C=K\u2212273C = K – 273:<\/li>\n<\/ul>\n\n\n\n

      C=339\u2212273=66\u00b0CC = 339 – 273 = 66\u00b0C<\/p>\n\n\n\n

      Answer<\/strong>: 66\u00b0C<\/p>\n\n\n\n

        \n
      • c. 89.5\u00b0F to Celsius<\/strong>
        Use the formula C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}:<\/li>\n<\/ul>\n\n\n\n

        C=(89.5\u221232)\u00d759=57.5\u00d759=287.59\u224831.94\u00b0CC = \\frac{(89.5 – 32) \\times 5}{9} = \\frac{57.5 \\times 5}{9} = \\frac{287.5}{9} \\approx 31.94\u00b0C<\/p>\n\n\n\n

        Answer<\/strong>: 31.94\u00b0C<\/p>\n\n\n\n

          \n
        • d. 383 Kelvin to Fahrenheit<\/strong>
          First, convert to Celsius:<\/li>\n<\/ul>\n\n\n\n

          C=383\u2212273=110\u00b0CC = 383 – 273 = 110\u00b0C<\/p>\n\n\n\n

          Then convert Celsius to Fahrenheit: F=(110\u00d795)+32=198+32=230\u00b0FF = (110 \\times \\frac{9}{5}) + 32 = 198 + 32 = 230\u00b0F<\/p>\n\n\n\n

          Answer<\/strong>: 230\u00b0F<\/p>\n\n\n\n

            \n
          • e. 17\u00b0C to Kelvin<\/strong>
            Use the formula K=C+273K = C + 273:<\/li>\n<\/ul>\n\n\n\n

            K=17+273=290KK = 17 + 273 = 290 K<\/p>\n\n\n\n

            Answer<\/strong>: 290 K<\/p>\n\n\n\n

              \n
            • f. 55\u00b0C to Kelvin<\/strong>
              Use the formula K=C+273K = C + 273:<\/li>\n<\/ul>\n\n\n\n

              K=55+273=328KK = 55 + 273 = 328 K<\/p>\n\n\n\n

              Answer<\/strong>: 328 K<\/p>\n\n\n\n

              \n\n\n\n

              4. The weather forecaster predicts that today\u2019s high will be 70\u00b0F.<\/h3>\n\n\n\n

              The temperature scale being used is Fahrenheit. To convert to Celsius and Kelvin:
              Using the formula C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}: C=(70\u221232)\u00d759=21.11\u00b0CC = \\frac{(70 – 32) \\times 5}{9} = 21.11\u00b0C<\/p>\n\n\n\n

              Then convert Celsius to Kelvin: K=21.11+273=294.11KK = 21.11 + 273 = 294.11 K<\/p>\n\n\n\n

              Answer<\/strong>: 70\u00b0F, 21.11\u00b0C, 294.11 K<\/p>\n\n\n\n

              \n\n\n\n

              5. “I was so cold yesterday that the temperature only reached 275.”<\/h3>\n\n\n\n

              The temperature scale being used is Kelvin. To convert to Celsius and Fahrenheit:
              Using the formula C=K\u2212273C = K – 273: C=275\u2212273=2\u00b0CC = 275 – 273 = 2\u00b0C<\/p>\n\n\n\n

              Then convert Celsius to Fahrenheit: F=(2\u00d795)+32=35.6\u00b0FF = (2 \\times \\frac{9}{5}) + 32 = 35.6\u00b0F<\/p>\n\n\n\n

              Answer<\/strong>: 275 K, 2\u00b0C, 35.6\u00b0F<\/p>\n\n\n\n

              \n\n\n\n

              6. “Today\u2019s temperature of 42 in Chicago set a record high for the month of August.”<\/h3>\n\n\n\n

              The temperature scale being used is Fahrenheit. To convert to Celsius and Kelvin:
              Using the formula C=(F\u221232)\u00d759C = \\frac{(F – 32) \\times 5}{9}: C=(42\u221232)\u00d759=5.56\u00b0CC = \\frac{(42 – 32) \\times 5}{9} = 5.56\u00b0C<\/p>\n\n\n\n

              Then convert Celsius to Kelvin: K=5.56+273=278.56KK = 5.56 + 273 = 278.56 K<\/p>\n\n\n\n

              Answer<\/strong>: 42\u00b0F, 5.56\u00b0C, 278.56 K<\/p>\n\n\n\n

              \n\n\n\n

              These temperature conversion formulas are essential for comparing temperatures in different scales, whether for scientific, everyday, or meteorological use. They help to convert temperatures in one scale to another, making it easier to understand the relationship between Celsius, Fahrenheit, and Kelvin.<\/p>\n","protected":false},"excerpt":{"rendered":"

              Thermal Energy Name: Figure 2-9b Temperature Conversion Formulas Conversion Formula Example Celsius to Kelvin K=C+ 273 21\u00b0C = 294 K Kelvin to Celsius C=K-273 313 K = 40\u00b0C Fahrenheit to Celsius C = (F-32) x 59 89 “F= 31.7\u00b0C Celsius to Fahrenheit F=(CX9\/5) + 32 50 \u00b0C = 122\u00b0F Questions 1. Convert the following numbers […]<\/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-181559","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181559","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=181559"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181559\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181559"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181559"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181559"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}