{"id":625,"date":"2018-02-14T06:36:28","date_gmt":"2018-02-13T22:36:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=625"},"modified":"2018-02-14T06:36:28","modified_gmt":"2018-02-13T22:36:28","slug":"latent-heat-heat-of-fusion-heat-of-vaporization-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/latent-heat-heat-of-fusion-heat-of-vaporization-problems-and-solutions.htm","title":{"rendered":"Latent heat Heat of fusion Heat of vaporization \u2013 Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

4 Latent heat Heat of fusion Heat of vaporization \u2013 Problems and Solutions<\/strong><\/p>\n

1. Calculate the amount of heat<\/a> added to 1 gram gold to change phase from solid to liquid. The heat of fusion<\/a> for gold is 64.5 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg. <\/span><\/span><\/p>\n

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

Mass<\/a> (m) = 1 gram = 1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg <\/span><\/span><\/p>\n

Heat of fusion (L<\/span><\/span>F<\/span><\/sub>) = 64.5 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Heat (Q)<\/span><\/span><\/p>\n

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

Q = m L<\/span><\/span>F<\/span><\/sub><\/p>\n

Q = (1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg)(64.5 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg)<\/span><\/span><\/p>\n

Q = 64.5 Joule<\/span><\/span><\/p>\n

<\/p>\n

2. Calculate the amount of heat released by 1 gram mercury to change phase from liquid to solid. Heat of fusion for mercury is 11.8 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg. <\/span><\/span><\/p>\n

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

Mass (m) = 1 gram = 1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg <\/span><\/span><\/p>\n

Heat of fusion (L<\/span><\/span>F<\/span><\/sub>) = 11.8 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Heat (Q)<\/span><\/span><\/p>\n

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

Q = m L<\/span><\/span>F<\/span><\/sub><\/p>\n

Q = (1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg)(11.8 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg)<\/span><\/span><\/p>\n

Q = 11.8 Joule<\/span><\/span><\/p>\n

3. Determine the amount of heat absorbed by 1 kg water to change phase from liquid to vapor (steam). Heat of vaporization for water = 2256 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg<\/span><\/span><\/p>\n

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

Mass (m) = 1 kg <\/span><\/span><\/p>\n

Heat of vaporization (L<\/span><\/span>V<\/span><\/sub>) = <\/span><\/span>2256 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Heat (Q)<\/span><\/span><\/p>\n

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

Q = m L<\/span><\/span>V<\/span><\/sub><\/p>\n

Q = (1 kg)(<\/span><\/span>2256 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg)<\/span><\/span><\/p>\n

Q = 2256 x 10<\/span><\/span>3<\/span><\/span><\/sup> Joule<\/span><\/span><\/p>\n

<\/p>\n

4. Determine the amount of heat released by nitrogen to change phase from vapor to liquid. Heat of vaporization for nitrogen <\/span><\/span>= <\/span><\/span>200 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg<\/span><\/span><\/p>\n

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

Mass (m) = 1 gram = 1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg <\/span><\/span><\/p>\n

Heat of vaporization (L<\/span><\/span>V<\/span><\/sub>) = <\/span><\/span>200 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span> Heat (Q)<\/span><\/span><\/p>\n

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

Q = m L<\/span><\/span>V<\/span><\/sub><\/p>\n

Q = (1 x 10<\/span><\/span>-3<\/span><\/span><\/sup> kg)(<\/span><\/span>200 x 10<\/span><\/span>3<\/span><\/span><\/sup> J\/kg)<\/span><\/span><\/p>\n

Q = 200 Joule<\/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":"

    4 Latent heat Heat of fusion Heat of vaporization \u2013 Problems and Solutions 1. Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 103 J\/kg. Known : Mass (m) = 1 gram = 1 x 10-3 kg Heat … 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":"Latent heat heat of fusion heat of vaporization \u2013 problems and solutions","_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-625","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\/625","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=625"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/625\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=625"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=625"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}