{"id":584,"date":"2018-02-12T11:46:23","date_gmt":"2018-02-12T03:46:23","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=584"},"modified":"2018-02-12T11:46:23","modified_gmt":"2018-02-12T03:46:23","slug":"converting-temperature-scales-celsius-scale-fahrenheit-scale-kelvin-scale","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/converting-temperature-scales-celsius-scale-fahrenheit-scale-kelvin-scale.htm","title":{"rendered":"Converting temperature scales (Celsius scale Fahrenheit scale Kelvin scale)","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

9 Converting temperature scales (Celsius scale Fahrenheit scale Kelvin scale)<\/strong><\/p>\n

1. 50 <\/span><\/span>o<\/span><\/span><\/sup>C = ….. <\/span><\/span>o<\/span><\/span><\/sup>F ?<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

At standard atmospheric pressure<\/a>, the freezing point of water is 0 <\/span><\/span>o<\/span><\/span><\/sup>C on the Celsius scale<\/a> and 32 <\/span><\/span>o<\/span><\/span><\/sup>F on the Fahrenheit scale. At standard atmospheric pressure, the boiling point of water is 100 <\/span><\/span>o<\/span><\/span><\/sup>C on the Celsius scale and 212 <\/span><\/span>o<\/span><\/span><\/sup>F on the Fahrenheit scale.<\/span><\/span><\/p>\n

0<\/span><\/span> o<\/span><\/span><\/sup>C = 32 <\/span><\/span>o<\/span><\/span><\/sup>F and 1<\/span><\/span>00 <\/span><\/span>o<\/span><\/span><\/sup>C = 212 <\/span><\/span>o<\/span><\/span><\/sup>F. A change of 5 C<\/span><\/span>o <\/span><\/span><\/sup>= a change of 9 F<\/span><\/span>o<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

For a Celsius scale, the distance between <\/span><\/span>0 <\/span><\/span>o<\/span><\/span><\/sup>C and 100 <\/span><\/span>o<\/span><\/span><\/sup>C divided into 100 equal intervals. For a Fahrenheit scale, the distance between 0<\/span><\/span> o<\/span><\/span><\/sup>C and 100<\/span><\/span> o<\/span><\/span><\/sup>C divided into 180 equal intervals. <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (180\/100) T<\/span><\/span>o<\/span><\/span><\/sup>C + 32<\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9\/5) T<\/span><\/span>o<\/span><\/span><\/sup>C + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9\/5) 50 + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9) 10<\/span><\/span> + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = 90<\/span><\/span> + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = 122 <\/span><\/span><\/p>\n

50 <\/span><\/span>o<\/span><\/span><\/sup>C = 122 <\/span><\/span>o<\/span><\/span><\/sup>F<\/span><\/span><\/p>\n

<\/p>\n

2. 86 <\/span><\/span>o<\/span><\/span><\/sup>F = ….. <\/span><\/span>o<\/span><\/span><\/sup>C ?<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (100\/180)(T<\/span><\/span>o<\/span><\/span><\/sup>F – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(T<\/span><\/span>o<\/span><\/span><\/sup>F – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(86 – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(54) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5)(6) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = 30<\/span><\/span><\/p>\n

86 <\/span><\/span>o<\/span><\/span><\/sup>F = 30 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

3. 50<\/span><\/span>o<\/span><\/span><\/sup>C = ….. K ?<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

T = T <\/span><\/span>o<\/span><\/span><\/sup>C + 273<\/span><\/span><\/p>\n

T = 50 + 273<\/span><\/span><\/p>\n

T = 323 <\/span><\/span><\/p>\n

50 <\/span><\/span>o<\/span><\/span><\/sup>C = <\/span><\/span>323 K<\/span><\/span><\/p>\n

<\/p>\n

4. 212<\/span><\/span>o<\/span><\/span><\/sup>F = ….. K ?<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (100\/180)(T<\/span><\/span>o<\/span><\/span><\/sup>F – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(T<\/span><\/span>o<\/span><\/span><\/sup>F – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(212 – 32) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5\/9)(180) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = (5)(20) <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>C = 100<\/span><\/span><\/p>\n

212 <\/span><\/span>o<\/span><\/span><\/sup>F = 100 <\/span><\/span>o<\/span><\/span><\/sup>C + 273<\/span><\/span><\/p>\n

212 <\/span><\/span>o<\/span><\/span><\/sup>F = 373 <\/span><\/span>K<\/span><\/span><\/p>\n

 <\/p>\n

5. x <\/span><\/span>o<\/span><\/span><\/sup>C = x <\/span><\/span>o<\/span><\/span><\/sup>F<\/span><\/span><\/p>\n

x = ….. ?<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

1 : Converting Celsius scale into Fahrenheit scale<\/i><\/span><\/span><\/p>\n

\"Converting<\/p>\n

2 : Converting Fahrenheit scale into Celsius scale<\/i><\/span><\/span><\/p>\n

\"Converting<\/p>\n\n

6. 122\u00b0F = \u2026.. Celsius<\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

The conversion between the two temperature scales can be written :<\/span><\/span><\/p>\n

T<\/span><\/span>C<\/span><\/sub> = 5\/9 (T<\/span><\/span>F<\/span><\/sub> \u2013 32) <\/span><\/span><\/p>\n

T<\/i><\/span><\/span>C<\/i><\/span><\/sub> = Temperature<\/a> in Celsius, T<\/i><\/span><\/span>F<\/i><\/span><\/sub> = temperature in Fahrenheit<\/i><\/span><\/span><\/p>\n

The temperature in Celsius :<\/span><\/span><\/p>\n

T<\/span><\/span>C<\/span><\/sub> = 5\/9 (122 \u2013 32) = T<\/span><\/span>C<\/span><\/sub> = 5\/9 (90) = 5 (10) <\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 50 <\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

7<\/span><\/span>. <\/span><\/span>The figure below shows <\/span><\/span>the temperature measurement of <\/span><\/span><\/span>a <\/span><\/span><\/span>liquid with the Fahrenheit scale thermometer! If the temperature of the liquid is measured using a Celsius scale thermometer, then <\/span><\/span><\/span>what is <\/span><\/span><\/span>the liquid temperatur<\/span><\/span><\/span>e.<\/span><\/span><\/span><\/p>\n

Known :\"Converting<\/u><\/span><\/span><\/p>\n

Fahrenheit <\/span><\/span>scale <\/span><\/span>(T<\/span><\/span>F<\/span><\/span><\/sub>) = 95<\/span><\/span>o<\/span><\/span><\/sup>F<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Celsius scale<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

At a pressure of 1 atm, <\/span>the freezing point of water <\/span>is <\/span>0 \u00b0C while the Fahrenheit scale is 32 <\/span>o<\/span><\/sup>F. Conversely, <\/span>t<\/span>he boiling point of water <\/span>for the C<\/span>elsius <\/span>scale is 100 <\/span>o<\/span><\/sup>C while the Fahrenheit scale <\/span>is <\/span>212 <\/span>o<\/span><\/sup>F. <\/span><\/span><\/span><\/p>\n

On the Celsius scale, between 0 \u00b0C and 100 \u00b0 C there are 100 \u00b0 while on the Fahrenheit scale between 32 \u00b0F to 212 \u00b0F there is 180\u00b0.<\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 100\/180 (T<\/span><\/span>F<\/span><\/sub> \u2013 32) <\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 5\/9 (T<\/span><\/span>F<\/span><\/sub> \u2013 32)<\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 5\/9 (<\/span><\/span>95<\/span><\/span> \u2013 32)<\/span><\/span><\/p>\n

T<\/span><\/span>C<\/sub> <\/span><\/span>= 5\/9 (<\/span><\/span>63<\/span><\/span>)<\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 315\/9<\/span><\/span><\/p>\n

T<\/span><\/span>C <\/span><\/sub>= 35<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

8. Based on figure below, determine the temperature P on the Celsius thermometer.<\/span><\/span><\/span><\/p>\n

Solution<\/span><\/span><\/p>\n

TC<\/sub> = 100\/180 (TF<\/sub> \u2013 32) \"Converting<\/span><\/span><\/p>\n

TC<\/sub> = 5\/9 (TF <\/sub>\u2013 32)<\/span><\/span><\/p>\n

TC<\/sub> = 5\/9 (104 \u2013 32)<\/span><\/span><\/p>\n

TC <\/sub>= 5\/9 (72)<\/span><\/span><\/p>\n

TC <\/sub>= 360\/9<\/span><\/span><\/p>\n

TC<\/sub> = 40 o<\/sup>C<\/span><\/span><\/p>\n

9<\/span><\/span>. <\/span><\/span>If the temperature of Celsius scale as shown in figure below, determine the temperature of Fahrenheit scale as shown in figure below.<\/span><\/span><\/p>\n

Solution :<\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (180\/100) T<\/span><\/span>o<\/span><\/span><\/sup>C + 32\"Converting<\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9\/5) T<\/span><\/span>o<\/span><\/span><\/sup>C + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9\/5) 60 + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = (9) 12<\/span><\/span> + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = 108<\/span><\/span> + 32 <\/span><\/span><\/p>\n

T<\/span><\/span>o<\/span><\/span><\/sup>F = 140 <\/span><\/span><\/p>\n

    \n
  1. Converting temperature scales<\/a><\/li>\n
  2. Linear expansion<\/a><\/li>\n
  3. Area expansion<\/a><\/li>\n
  4. Volume expansion<\/a><\/li>\n
  5. Heat<\/a><\/li>\n
  6. Mechanical equivalent of heat<\/a><\/li>\n
  7. Specific heat and heat capacity<\/a><\/li>\n
  8. Latent heat, heat of fusion, heat of vaporization<\/a><\/li>\n
  9. Energy conservation for heat transfer<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    9 Converting temperature scales (Celsius scale Fahrenheit scale Kelvin scale) 1. 50 oC = ….. oF ? Solution At standard atmospheric pressure, the freezing point of water is 0 oC on the Celsius scale and 32 oF on the Fahrenheit scale. At standard atmospheric pressure, the boiling point of water is 100 oC on the … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Converting temperature scales (Celsius scale Fahrenheit scale Kelvin scale)","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-584","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/584","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=584"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/584\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}