{"id":600,"date":"2018-02-13T11:13:11","date_gmt":"2018-02-13T03:13:11","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=600"},"modified":"2018-02-13T11:13:11","modified_gmt":"2018-02-13T03:13:11","slug":"heat-mass-specific-heat-the-change-in-temperature-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/heat-mass-specific-heat-the-change-in-temperature-problems-and-solutions.htm","title":{"rendered":"Heat Mass Specific heat The change in temperature \u2013 Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

9 Heat Mass Specific heat The change in temperature \u2013 Problems and Solutions<\/strong><\/p>\n

1. <\/span><\/span>A 2 kg lead is heated from <\/span><\/span>50<\/span><\/span>o<\/span><\/span><\/sup>C to 100<\/span><\/span>o<\/span><\/span><\/sup>C. The specific heat<\/a> of lead is <\/span><\/span>1<\/span><\/span>3<\/span><\/span>0 J.kg<\/span><\/span>-1 o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. <\/span><\/span>How much heat<\/a> is absorbed by the lead?<\/span><\/span><\/p>\n

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

Mass<\/a> (m) = 2 kg<\/span><\/span><\/p>\n

The specific heat <\/span><\/span>(c) = 1<\/span><\/span>3<\/span><\/span>0 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The change in temperature <\/span><\/span><\/a>(<\/span><\/span>\u0394T) = <\/span><\/span>100<\/span><\/span>o<\/span><\/span><\/sup>C \u2013 50<\/span><\/span>o<\/span><\/span><\/sup>C = 50<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

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

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

Q = m c <\/span><\/span>\u0394T<\/span><\/span><\/p>\n

Q = <\/span><\/span>heat<\/span><\/span>, m = mass, c = <\/span><\/span>the specific heat<\/span><\/span>, <\/span><\/span>\u0394T = <\/span><\/span>the change in temperature<\/span><\/span><\/i><\/p>\n

The heat absorbed by lead :<\/span><\/span><\/p>\n

Q = (2 kg)(1<\/span><\/span>3<\/span><\/span>0 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>)(50<\/span><\/span>o<\/span><\/span><\/sup>C)<\/span><\/span><\/p>\n

Q = (100)(1<\/span><\/span>3<\/span><\/span>0)<\/span><\/span><\/p>\n

Q = 1<\/span><\/span>3,<\/span><\/span>000 Joule<\/span><\/span><\/p>\n

Q = 1.<\/span><\/span>3<\/span><\/span> x 10<\/span><\/span>4<\/span><\/span><\/sup> Joule<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>The specific heat of copper is 390 J\/k<\/span><\/span>g<\/span><\/span> o<\/span><\/span><\/sup>C, <\/span><\/span>the change in temperature is <\/span><\/span>40<\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>If the copper absorbs 40 Joule of heat, what is the copper’s mass!<\/span><\/span><\/p>\n

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

The specific heat of copper <\/span><\/span>(c) = <\/span><\/span>390 J\/k<\/span><\/span>g<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The change in temperature <\/span><\/span>(<\/span><\/span>\u0394T) = <\/span><\/span>40<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Heat (<\/span><\/span>Q) = <\/span><\/span>40 J<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Mass (m) <\/span><\/span>of copper <\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394T<\/span><\/span><\/p>\n

40 J <\/span><\/span>= <\/span><\/span>(m)(<\/span><\/span>390 J\/k<\/span><\/span>g <\/span><\/span>o<\/span><\/span><\/sup>C)(40<\/span><\/span>o<\/span><\/span><\/sup>C)<\/span><\/span><\/p>\n

40<\/span><\/span> = (m)(<\/span><\/span>390 \/k<\/span><\/span>g)(40)<\/span><\/span><\/p>\n

40<\/span><\/span> = (m)(<\/span><\/span>390 \/kg<\/span><\/span>)(4)<\/span><\/span><\/p>\n

40<\/span><\/span> = (m)(<\/span><\/span>1560 \/kg<\/span><\/span>)<\/span><\/span><\/p>\n

m = <\/span><\/span>40<\/span><\/span> \/ <\/span><\/span>1560 <\/span><\/span><\/p>\n

m = <\/span><\/span>0.026 kg<\/span><\/span><\/p>\n

m = 26 gram<\/span><\/span><\/p>\n

3. <\/span><\/span>The initial temperature of 20 gram water is 30<\/span><\/span>o<\/span><\/span><\/sup>C. The specific heat of water is 1 cal g<\/span><\/span>-1<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. If water absorbs 300 calories of heat, determine the final temperature!<\/span><\/span><\/p>\n

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

Mass (m) = 20 gr<\/span><\/span><\/p>\n

The initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/sub>) = 30<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The specific heat of water <\/span><\/span>(c) = 1 cal gr<\/span><\/span>-1<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Heat <\/span><\/span>(Q) = 300 cal<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The final temperature of water<\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394T<\/span><\/span><\/p>\n

300 cal = (20 gr)(<\/span><\/span>1 cal gr<\/span><\/span>-1<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>)(<\/span><\/span>T<\/span><\/span>2<\/span><\/sub>-30<\/span><\/span>)<\/span><\/span><\/p>\n

300 = (20)(1)(<\/span><\/span>T2<\/sub>-30)<\/span><\/span><\/p>\n

300 = 20 (<\/span><\/span>T2<\/sub>-30)<\/span><\/span><\/p>\n

300 = 20<\/span><\/span>T<\/span><\/span>2 <\/span><\/sub>\u2013 600<\/span><\/span><\/p>\n

300 + 600 = 20T<\/span><\/span>2<\/span><\/sub><\/p>\n

900 = 20T2<\/sub><\/p>\n

T2<\/sub> = 900 \/ 20<\/span><\/span><\/p>\n

T2<\/sub> = 45<\/span><\/span><\/p>\n

The change in temperature is 45<\/span><\/span>o<\/span><\/span><\/sup>C \u2013 30<\/span><\/span>o<\/span><\/span><\/sup>C = 15<\/span><\/span>o<\/span><\/span><\/sup>C.<\/span><\/span><\/p>\n\n

4<\/span><\/span>. <\/span><\/span>T<\/span><\/span>he change in temperature <\/span><\/span>of the sea water <\/span><\/span>is <\/span><\/span>1<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>when water absorbs 3900 Joule of heat. The specific heat of the sea water is <\/span><\/span>3.9 \u00d7 10<\/span><\/span>3<\/span><\/span><\/sup> J\/<\/span><\/span>k<\/span><\/span>g\u00b0C, <\/span><\/span>what is the mass of the sea water.<\/span><\/span><\/p>\n

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

The change in temperature <\/span><\/span>(<\/span><\/span>\u0394<\/span><\/span>T) = 1<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Heat <\/span><\/span>(Q) = 3900 Joule<\/span><\/span><\/p>\n

The specific heat of the sea water <\/span><\/span>(c) = 3.9 x <\/span><\/span>10<\/span><\/span>3<\/span><\/span><\/sup> J\/<\/span><\/span>k<\/span><\/span>g\u00b0C = 3900 J\/Kg\u00b0C <\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Mass <\/span><\/span>(m)<\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394<\/span><\/span>T<\/span><\/span><\/p>\n

Q = <\/span><\/span>heat<\/span><\/span>, m = mass, c = <\/span><\/span>specific heat<\/span><\/span>, <\/span><\/span>\u0394<\/span><\/span>T = <\/span><\/span>the change in temperature<\/span><\/span><\/i><\/p>\n

m = Q \/ c <\/span><\/span>\u0394<\/span><\/span>T = 3900 \/ (3900)(1) = 3900 \/ 3900 = 1 <\/span><\/span>kg<\/span><\/span><\/p>\n

5<\/span><\/span>. <\/span><\/span>A 2-kg copper <\/span><\/span>absorbs 39,000 J of heat at 30\u00b0C. If the specific heat of copper is 390 J\/kg \u00b0C, what is the final temperature of the copper…<\/span><\/span><\/p>\n

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

Mass (m) = 2 kg<\/span><\/span><\/p>\n

Initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/span><\/sub>) = 30<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Heat<\/span><\/span> (Q) = 39,000 Joule<\/span><\/span><\/p>\n

Specific heat <\/span><\/span>(c) <\/span><\/span>of copper <\/span><\/span>= 390 J\/kg <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>:<\/u><\/span><\/span> The final temperature<\/span><\/span> (T<\/span><\/span>2<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394T<\/span><\/span><\/p>\n

Q = <\/span><\/span>heat<\/span><\/span>, m = mass, c = <\/span><\/span>specific heat<\/span><\/span>, \u0394T = <\/span><\/span>the change in temperature<\/span><\/span><\/i><\/p>\n

Q = m c \u0394T = m c (T<\/span><\/span>2 <\/span><\/span><\/sub>– T<\/span><\/span>1<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

39,000 <\/span><\/span>= (2)(<\/span><\/span>390)(T<\/span><\/span>2 <\/span><\/span><\/sub>– 30)<\/span><\/span><\/p>\n

100 = (2)(1)(T<\/span><\/span>2 <\/span><\/span><\/sub>– 30)<\/span><\/span><\/p>\n

100 = (2)(T<\/span><\/span>2 <\/span><\/span><\/sub>– 30)<\/span><\/span><\/p>\n

50 = T<\/span><\/span>2<\/span><\/span><\/sub> – <\/span><\/span>30<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/span><\/sub>= 50 + 30<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/span><\/sub>= 80<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

<\/p>\n

6<\/span><\/span>. <\/span><\/span>A 5-kg water is heated from 15\u00b0C to 40\u00b0C. What is the heat is absorbed by water. Specific heat of water is <\/span><\/span>4.2 \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/Kg\u00b0 C.<\/span><\/span><\/p>\n

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

Mass (m) = 5 kg<\/span><\/span><\/p>\n

Initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/span><\/sub>) = 15\u00b0C <\/span><\/span><\/p>\n

Final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/span><\/sub>) = 40\u00b0C <\/span><\/span><\/p>\n

Specific heat of water <\/span><\/span>(c) = 4.2 \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg\u00b0 C <\/span><\/span><\/p>\n

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

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

Q = m c <\/span><\/span>\u0394T <\/span><\/span><\/p>\n

Q = (5 kg)(<\/span><\/span>4.2 \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/kg\u00b0C)(40\u00b0C \u2013 15\u00b0C) <\/span><\/span><\/p>\n

Q = <\/span><\/span>(5)(<\/span><\/span>4.2 \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>J)(25) <\/span><\/span><\/p>\n

Q = 525 x 10<\/span><\/span>3 <\/span><\/span><\/sup>J <\/span><\/span><\/p>\n

Q = 525,000 Joule<\/span><\/span><\/p>\n

7<\/span><\/span>. <\/span><\/span>A 2-kg water is heated from 24\u00b0C to 90\u00b0C. What is the heat is absorbed by water. Specific heat of water is <\/span><\/span>4.2 \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>J\/Kg\u00b0 C.<\/span><\/span><\/p>\n

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

Mass (m) = 2 kg<\/span><\/span><\/p>\n

Initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/span><\/sub>) = 24\u00b0C <\/span><\/span><\/p>\n

Final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/span><\/sub>) = 90\u00b0C <\/span><\/span><\/p>\n

Specific heat of water <\/span><\/span>(c) = 4,200 Joule\/kg\u00b0C<\/span><\/span><\/p>\n

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

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

Q = m <\/span><\/span>c <\/span><\/span>\u0394T <\/span><\/span><\/p>\n

Q = (2 kg)(<\/span><\/span>4,200 Joule\/kg\u00b0C)(90\u00b0C – 24\u00b0C)<\/span><\/span><\/p>\n

Q = (2 kg)(<\/span><\/span>4,200 Joule\/kg\u00b0C)(66\u00b0C)<\/span><\/span><\/p>\n

Q = (132)(<\/span><\/span>4,200 Joule)<\/span><\/span><\/p>\n

Q = 554,400 Joule<\/span><\/span><\/p>\n

<\/p>\n

8<\/span><\/span>. <\/span><\/span>A 5-gram water is heated from 10\u00b0C to 40\u00b0C. What is the heat is absorbed by water. Specific heat of water is 1<\/span><\/span> \u00d7 10<\/span><\/span>3 <\/span><\/span><\/sup>cal<\/span><\/span>\/g<\/span><\/span>r<\/span><\/span>\u00b0 C.<\/span><\/span><\/p>\n

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

Mass (m) = 5 gram<\/span><\/span><\/p>\n

Initial temperature<\/span><\/span> (T<\/span><\/span>1<\/span><\/span><\/sub>) = 10<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/span><\/sub>) = 40<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Specific heat of water<\/span><\/span> (c) = 1 <\/span><\/span>c<\/span><\/span>al\/ gr\u00b0C<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : Heat<\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394T<\/span><\/span><\/p>\n

Q = (5 gram)(<\/span><\/span>1 cal\/ gr\u00b0C)(40<\/span><\/span>o<\/span><\/span><\/sup>C \u2013 10<\/span><\/span>o<\/span><\/span><\/sup>C) <\/span><\/span><\/p>\n

Q = <\/span><\/span>(5)(<\/span><\/span>1 <\/span><\/span>c<\/span><\/span>al)(30) <\/span><\/span><\/p>\n

Q = 150 <\/span><\/span>calories<\/span><\/span><\/p>\n

9<\/span><\/span>. <\/span><\/span>A 0.2-kg water absorbs 42,000 Joule of heat at <\/span><\/span>25<\/span><\/span>o<\/span><\/span><\/sup>C. The specific heat of water is 4200 J\/kg <\/span><\/span>o<\/span><\/span><\/sup>C, what is the final temperature of water.<\/span><\/span><\/p>\n

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

Mass <\/span><\/span>of water <\/span><\/span>(m) = 0.2 kg<\/span><\/span><\/p>\n

Heat <\/span><\/span>(Q) = 42,000 Joule<\/span><\/span><\/p>\n

Specific heat of water <\/span><\/span>(c) = 4200 J\/kg <\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

Initial temperature <\/span><\/span>(T<\/span><\/span>1<\/span><\/sub>) = 25<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

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

Q = m c <\/span><\/span>\u0394T = m c (T<\/span><\/span>2<\/span><\/sub> \u2013 T<\/span><\/span>1<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

Q = <\/i><\/span><\/span>heat<\/i><\/span><\/span>, m = mass, c = <\/i><\/span><\/span>specific heat<\/i><\/span><\/span>, <\/i><\/span><\/span>\u0394T = <\/i><\/span><\/span>the change in temperature<\/i><\/span><\/span>, T<\/i><\/span><\/span>1 <\/i><\/span><\/sub>= <\/i><\/span><\/span>the initial temperature<\/i><\/span><\/span>, T<\/i><\/span><\/span>2 <\/i><\/span><\/sub>= <\/i><\/span><\/span>the final temperature<\/i><\/span><\/span><\/p>\n

Q = m c (T<\/span><\/span>2<\/span><\/sub> \u2013 T<\/span><\/span>1<\/span><\/sub>) <\/span><\/span><\/p>\n

42,000 = (0.2)(4200)(T<\/span><\/span>2<\/span><\/sub> \u2013 25)<\/span><\/span><\/p>\n

42,000 = 840 (T<\/span><\/span>2<\/span><\/sub> \u2013 25)<\/span><\/span><\/p>\n

42,000 = 840 T<\/span><\/span>2<\/span><\/sub> \u2013 21,000<\/span><\/span><\/p>\n

42,000 + 21,000 = 840 T<\/span><\/span>2<\/span><\/sub><\/p>\n

63,000 = 840 T<\/span><\/span>2<\/span><\/sub><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = 63,000 \/ 840<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = 75<\/span><\/span>o<\/span><\/span><\/sup>C<\/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 Heat Mass Specific heat The change in temperature \u2013 Problems and Solutions 1. A 2 kg lead is heated from 50oC to 100oC. The specific heat of lead is 130 J.kg-1 oC-1. How much heat is absorbed by the lead? Known : Mass (m) = 2 kg The specific heat (c) = 130 J.kg-1C-1 … 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":"Heat mass specific heat the change in temperature \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-600","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\/600","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=600"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/600\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=600"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=600"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}