{"id":1583,"date":"2018-03-11T02:35:43","date_gmt":"2018-03-10T18:35:43","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1583"},"modified":"2023-08-19T01:17:25","modified_gmt":"2023-08-19T01:17:25","slug":"carnot-engine-application-of-the-second-law-of-thermodynamics-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/carnot-engine-application-of-the-second-law-of-thermodynamics-problems-and-solutions.htm","title":{"rendered":"Carnot engine (application of the second law of thermodynamics) – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

25 Carnot engine (application of the second law of thermodynamics) – problems and solutions<\/span><\/p>\n

1. An engine operates between 1200 Kelvin and 300 Kelvin. What is the efficiency of this engine?<\/span><\/p>\n

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

High temperature (TH<\/sub>) = 1200 K<\/span><\/p>\n

Low temperature (TL<\/sub>) = 300 K<\/span><\/p>\n

Wanted: Efficiency (e)<\/span><\/p>\n

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

\"Carnot<\/span><\/p>\n

2. An engine operates between 727o<\/sup>C and 127o<\/sup>C. The engine’s heat input is 6000 Joule. What is the efficiency of the engine and work done by the engine each cycle?<\/span><\/p>\n

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

High temperature<\/a> (TH<\/sub>) = 727o<\/sup>C + 273 = 1000 K<\/span><\/p>\n

Low temperature (TL<\/sub>) = 127o<\/sup>C + 273 = 400 K<\/span><\/p>\n

Heat input (QH<\/sub>) = 6000 Joule<\/span><\/p>\n

Wanted:<\/u> Efficiency (e) and work done by the engine (W)<\/span><\/p>\n

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

The efficiency of the Carnot engine<\/a> :<\/b><\/span><\/p>\n

\"Carnot<\/span><\/p>\n

Work is done by the Carnot engine :<\/u><\/span><\/p>\n

W = e QH<\/sub><\/span><\/p>\n

W = (0.6)(6000) <\/span><\/p>\n

W = 3600 Joule<\/span><\/p>\n

3. An engine operates between 527o<\/sup>C and 127o<\/sup>C. The engine’s heat input is 10,000 Joule. How much heat is discharged as waste heat from this engine?<\/span><\/p>\n

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

High temperature (TH<\/sub>) = 527o<\/sup>C + 273 = 800 K<\/span><\/p>\n

Low temperature (TL<\/sub>) = 127o<\/sup>C + 273 = 400 K<\/span><\/p>\n

Heat<\/a> input (QH<\/sub>) = 10,000 Joule<\/span><\/p>\n

Wanted :<\/u> Heat output (QL<\/sub>)<\/span><\/p>\n

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

Efficiency of the Carnot engine :<\/span><\/p>\n

\"Carnot<\/span><\/p>\n

Work is done by Carnot engine :<\/span><\/p>\n

W = e Q1<\/sub><\/span><\/p>\n

W = (0.5)(10.000) <\/span><\/p>\n

W = 5000 Joule<\/span><\/p>\n

Heat output (QL<\/sub>) = Heat input (QH<\/sub>) \u2013 Work (W)<\/span><\/p>\n

QL<\/sub> = QH<\/sub> \u2013 W<\/span><\/p>\n

QL<\/sub> = 10,000 \u2013 5,000<\/span><\/p>\n

QL<\/sub> = 5000 Joule<\/span><\/p>\n

4. What is the efficiency of the engine and work done by the engine according to diagram below.<\/span><\/p>\n

\"CarnotKnown :<\/u><\/span><\/p>\n

High temperature (TH<\/sub>) = 800 K<\/span><\/p>\n

Low temperature (TL<\/sub>) = 300 K<\/span><\/p>\n

Heat input (QH<\/sub>) = 1000 Joule<\/span><\/p>\n

Wanted: Efficiency (e) and work<\/a> (W) done by the engine<\/span><\/p>\n

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

The efficiency of the Carnot engine :<\/span><\/p>\n

\"Carnot<\/span><\/p>\n

Work is done by the Carnot engine :<\/u><\/span><\/p>\n

W = e QH<\/sub><\/span><\/p>\n

W = (0.625)(1000) <\/span><\/p>\n

W = 625 Joule<\/span><\/p>\n

5. If a Carnot engine operates between \\( 500\\,K \\) and \\( 300\\,K \\), calculate its efficiency.<\/span>
\nSolution: \\( \\eta = 1 – \\frac{T_C}{T_H} = 1 – \\frac{300}{500} = 0.4 = 40\\% \\)<\/span><\/p>\n

6. Find the efficiency of a Carnot refrigerator operating between \\( 300\\,K \\) and \\( 250\\,K \\).<\/span>
\nSolution: \\( \\eta = \\frac{T_C}{T_H – T_C} = \\frac{250}{50} = 5 = 500\\% \\)<\/span><\/p>\n

7. A Carnot engine’s efficiency is \\( 30\\% \\). If \\( T_C = 300\\,K \\), find \\( T_H \\).<\/span>
\nSolution: \\( T_H = \\frac{T_C}{1 – \\eta} = \\frac{300}{0.7} \\approx 428.57\\,K \\)<\/span><\/p>\n

8. If a Carnot heat pump operates between \\( 270\\,K \\) and \\( 330\\,K \\), find the coefficient of performance (COP).<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_H}{T_H – T_C} = \\frac{330}{60} = 5.5 \\)<\/span><\/p>\n

9. Determine \\( T_C \\) for a Carnot engine with \\( 50\\% \\) efficiency and \\( T_H = 400\\,K \\).<\/span>
\nSolution: \\( T_C = T_H \\times (1 – \\eta) = 400 \\times 0.5 = 200\\,K \\)<\/span><\/p>\n

10. Calculate the efficiency of a Carnot engine operating between \\( 373\\,K \\) and \\( 273\\,K \\).<\/span>
\nSolution: \\( \\eta = 1 – \\frac{273}{373} \\approx 0.268 = 26.8\\% \\)<\/span><\/p>\n

11. Find the COP of a Carnot refrigerator with \\( T_C = 260\\,K \\) and \\( T_H = 300\\,K \\).<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_C}{T_H – T_C} = \\frac{260}{40} = 6.5 \\)<\/span><\/p>\n

12. A Carnot engine has \\( 25\\% \\) efficiency and \\( T_C = 200\\,K \\). Find \\( T_H \\).<\/span>
\nSolution: \\( T_H = \\frac{T_C}{1 – \\eta} = \\frac{200}{0.75} \\approx 266.67\\,K \\)<\/span><\/p>\n

13. If a Carnot heat pump operates between \\( 220\\,K \\) and \\( 320\\,K \\), find the COP.<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_H}{T_H – T_C} = \\frac{320}{100} = 3.2 \\)<\/span><\/p>\n

14. Determine \\( T_C \\) for a Carnot engine with \\( 60\\% \\) efficiency and \\( T_H = 500\\,K \\).<\/span>
\nSolution: \\( T_C = T_H \\times (1 – \\eta) = 500 \\times 0.4 = 200\\,K \\)<\/span><\/p>\n

15. Calculate the efficiency of a Carnot engine operating between \\( 600\\,K \\) and \\( 400\\,K \\).<\/span>
\nSolution: \\( \\eta = 1 – \\frac{400}{600} \\approx 0.333 = 33.3\\% \\)<\/span><\/p>\n

16. Find the COP of a Carnot refrigerator with \\( T_C = 280\\,K \\) and \\( T_H = 330\\,K \\).<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_C}{T_H – T_C} = \\frac{280}{50} = 5.6 \\)<\/span><\/p>\n

17. A Carnot engine has \\( 40\\% \\) efficiency and \\( T_C = 250\\,K \\). Find \\( T_H \\).<\/span>
\nSolution: \\( T_H = \\frac{T_C}{1 – \\eta} = \\frac{250}{0.6} \\approx 416.67\\,K \\)<\/span><\/p>\n

18. If a Carnot heat pump operates between \\( 270\\,K \\) and \\( 370\\,K \\), find the COP.<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_H}{T_H – T_C} = \\frac{370}{100} = 3.7 \\)<\/span><\/p>\n

15. Determine \\( T_C \\) for a Carnot engine with \\( 75\\% \\) efficiency and \\( T_H = 800\\,K \\).<\/span>
\nSolution: \\( T_C = T_H \\times (1 – \\eta) = 800 \\times 0.25 = 200\\,K \\)<\/span><\/p>\n

16. Calculate the efficiency of a Carnot engine operating between \\( 700\\,K \\) and \\( 300\\,K \\).<\/span>
\nSolution: \\( \\eta = 1 – \\frac{300}{700} \\approx 0.571 = 57.1\\% \\)<\/span><\/p>\n

17. Find the COP of a Carnot refrigerator with \\( T_C = 250\\,K \\) and \\( T_H = 350\\,K \\).<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_C}{T_H – T_C} = \\frac{250}{100} = 2.5 \\)<\/span><\/p>\n

18. A Carnot engine has \\( 10\\% \\) efficiency and \\( T_C = 150\\,K \\). Find \\( T_H \\).<\/span>
\nSolution: \\( T_H = \\frac{T_C}{1 – \\eta} = \\frac{150}{0.9} \\approx 166.67\\,K \\)<\/span><\/p>\n

19. If a Carnot heat pump operates between \\( 200\\,K \\) and \\( 400\\,K \\), find the COP.<\/span>
\nSolution: \\( \\text{COP} = \\frac{T_H}{T_H – T_C} = \\frac{400}{200} = 2 \\)<\/span><\/p>\n

20. Determine \\( T_C \\) for a Carnot engine with \\( 80\\% \\) efficiency and \\( T_H = 1000\\,K \\).<\/span>
\nSolution: \\( T_C = T_H \\times (1 – \\eta) = 1000 \\times 0.2 = 200\\,K \\)<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

25 Carnot engine (application of the second law of thermodynamics) – problems and solutions 1. An engine operates between 1200 Kelvin and 300 Kelvin. What is the efficiency of this engine? Known : High temperature (TH) = 1200 K Low temperature (TL) = 300 K Wanted: Efficiency (e) Solution : 2. An engine operates between … 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":"Carnot engine (application of the second 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-1583","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\/1583","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=1583"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1583\/revisions"}],"predecessor-version":[{"id":9019,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1583\/revisions\/9019"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1583"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1583"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1583"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}