{"id":1509,"date":"2018-03-09T12:50:43","date_gmt":"2018-03-09T04:50:43","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1509"},"modified":"2023-08-19T03:04:06","modified_gmt":"2023-08-19T03:04:06","slug":"the-first-law-of-thermodynamics-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/the-first-law-of-thermodynamics-problems-and-solutions.htm","title":{"rendered":"The first law of thermodynamics – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

30 The first law of thermodynamics – problems and solutions<\/span><\/p>\n

1. 3000 J of heat<\/a> is added to a system and 2500 J of work<\/a> is done by the system. What is the change in internal energy of the system?<\/span><\/p>\n

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

Heat (Q) = +3000 Joule<\/span><\/p>\n

Work (W) = +2500 Joule <\/span><\/p>\n

Wanted: the change in internal energy of the system<\/span><\/p>\n

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

The equation of the first law of thermodynamics<\/a><\/b><\/span><\/p>\n

\u0394U = Q-W<\/span><\/p>\n

The sign conventions :<\/span><\/p>\n

Q is positive if the heat added to the system<\/span><\/p>\n

W is positive if work is done by the system<\/span><\/p>\n

Q is negative if heat leaves the system<\/span><\/p>\n

W is negative if work is done on the system<\/span><\/p>\n

The change in internal energy of the system :<\/span><\/p>\n

\u0394U = 3000-2500<\/span><\/p>\n

\u0394U = 500 Joule<\/span><\/p>\n

Internal energy increases by 500 Joule. <\/span><\/p>\n

2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?<\/span><\/p>\n

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

Heat (Q) = +2000 Joule<\/span><\/p>\n

Work (W) = -2500 Joule <\/span><\/p>\n

Wanted: The change in internal energy of the system <\/span><\/p>\n

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

\u0394U = Q-W<\/span><\/p>\n

\u0394U = 2000-(-2500)<\/span><\/p>\n

\u0394U = 2000+2500<\/span><\/p>\n

\u0394U = 4500 Joule<\/span><\/p>\n

Internal energy increases by 4500 Joule. <\/span><\/p>\n

3. 2000 J of heat leaves the system and 2500 J of work is done on the system. What is the change in internal energy of the system?<\/span><\/p>\n

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

Heat (Q) = -2000 Joule<\/span><\/p>\n

Work (W) = -3000 Joule <\/span><\/p>\n

Wanted: The change in internal energy of the system <\/span><\/p>\n

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

\u0394U = Q-W<\/span><\/p>\n

\u0394U = -2000-(-3000)<\/span><\/p>\n

\u0394U = -2000+3000<\/span><\/p>\n

\u0394U = 1000 Joule<\/span><\/p>\n

Internal energy increases by 4500 Joule. <\/span><\/p>\n

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

– If heat is added to the system, then the internal energy of the system increases<\/span><\/p>\n

– If heat leaves the system, then the internal energy of the system decreases<\/span><\/p>\n

– If the work is done by the system, then the internal energy of the system decreases<\/span><\/p>\n

– If the work is done on the system, then the internal energy of the system increases<\/span><\/p>\n

4. Calculate the change in internal energy of 2 moles of an ideal gas when 400 J of heat is added, and the gas expands, doing 300 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics: \\( \\Delta U = Q – W \\):<\/span>
\n\\[ <\/span>\\Delta U = 400 – 300 = 100\\, \\text{J} <\/span>\\]<\/span><\/p>\n

5. Determine the heat transfer for a system that performs 200 J of work on its surroundings and has a change in internal energy of 50 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[ <\/span>Q = \\Delta U + W = 50 + 200 = 250\\, \\text{J} <\/span>\\]<\/span><\/p>\n

6.\u00a0 Calculate the work done by a system when it absorbs 600 J of heat and its internal energy increases by 150 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = 600 – 150 = 450\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

7. Determine the heat transfer for a system that performs 500 J of work and its internal energy decreases by 100 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = (-100) + 500 = 400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

8. Calculate the change in internal energy when 300 J of heat is lost and 200 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = -300 + 200 = -100\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

9. Determine the work done on a system when it loses 400 J of heat and its internal energy decreases by 200 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = \\Delta U – Q = (-200) – (-400) = 200\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

10. Calculate the heat transfer when a system\u2019s internal energy increases by 100 J and 50 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = 100 + 50 = 150\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

11. Determine the change in internal energy when a system absorbs 250 J of heat and performs 150 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = Q – W = 250 – 150 = 100\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

12. Calculate the work done by the system when it loses 300 J of heat, and its internal energy decreases by 100 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = -300 – (-100) = -200\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

13. Determine the heat transfer for a system that does 400 J of work on its surroundings and its internal energy increases by 150 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = 150 + 400 = 550\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

14. Calculate the change in internal energy when 500 J of heat is added, and the system does 300 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = Q – W = 500 – 300 = 200\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

15. Determine the work done by the system when it absorbs 600 J of heat, and its internal energy increases by 200 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = 600 – 200 = 400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

16. Calculate the heat transfer when a system performs 700 J of work and its internal energy decreases by 300 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = (-300) + 700 = 400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

17. Determine the change in internal energy when 800 J of heat is lost and 400 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = -800 + 400 = -400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

18. Calculate the work done on the system when it loses 900 J of heat, and its internal energy decreases by 500 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = \\Delta U – Q = (-500) – (-900) = 400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

19. Determine the heat transfer when a system’s internal energy increases by 600 J, and 300 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = 600 + 300 = 900\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

20. Calculate the change in internal energy when a system absorbs 700 J of heat and performs 350 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = Q – W = 700 – 350 = 350\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

21. Determine the work done by the system when it loses 800 J of heat, and its internal energy decreases by 400 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = -800 – (-400) = -400\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

22. Calculate the heat transfer when a system does 900 J of work on its surroundings, and its internal energy increases by 450 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = 450 + 900 = 1350\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

23. Determine the change in internal energy when 1000 J of heat is added, and the system does 500 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = Q – W = 1000 – 500 = 500\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

24. Calculate the work done by the system when it absorbs 1100 J of heat, and its internal energy increases by\u00a0550 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = 1100 – 550 = 550\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

25. Determine the heat transfer when a system performs 1200 J of work, and its internal energy decreases by 600 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = (-600) + 1200 = 600\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

26. Calculate the change in internal energy when 1300 J of heat is lost, and 650 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = -1300 + 650 = -650\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

27. Determine the work done on the system when it loses 1400 J of heat, and its internal energy decreases by 700 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = \\Delta U – Q = (-700) – (-1400) = 700\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

28. Calculate the heat transfer when a system’s internal energy increases by 800 J, and 400 J of work is done on the system.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nQ = \\Delta U + W = 800 + 400 = 1200\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

29. Determine the change in internal energy when a system absorbs 1500 J of heat and performs 750 J of work on its surroundings.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\n\\Delta U = Q – W = 1500 – 750 = 750\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

30. Calculate the work done by the system when it loses 1600 J of heat, and its internal energy decreases by 800 J.<\/span>
\nSolution:<\/span>
\nUsing the first law of thermodynamics:<\/span>
\n\\[<\/span>
\nW = Q – \\Delta U = -1600 – (-800) = -800\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

These problems and solutions are designed to provide a comprehensive understanding of the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 The first law of thermodynamics – problems and solutions 1. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system? Known : Heat (Q) = +3000 Joule Work (W) = +2500 Joule Wanted: the change … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"The first law of thermodynamics - 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-1509","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\/1509","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=1509"}],"version-history":[{"count":4,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1509\/revisions"}],"predecessor-version":[{"id":9029,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1509\/revisions\/9029"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1509"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1509"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1509"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}