{"id":1578,"date":"2018-03-11T02:00:26","date_gmt":"2018-03-10T18:00:26","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1578"},"modified":"2023-08-19T01:25:44","modified_gmt":"2023-08-19T01:25:44","slug":"heat-engine-application-of-the-second-law-of-thermodynamics-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/heat-engine-application-of-the-second-law-of-thermodynamics-problems-and-solutions.htm","title":{"rendered":"Heat engine (application of the second law of thermodynamics) – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

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

1. Heat input QH<\/sub> = 3000 Joule and heat output QL<\/sub> = 1000 Joule. What is the efficiency of the heat engine<\/a>?<\/span><\/p>\n

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

Heat<\/a> input (QH<\/sub>) = 3000 Joule<\/span><\/p>\n

Heat output (QL<\/sub>) = 1000 Joule<\/span><\/p>\n

Work<\/a> done by the engine (W) = 3000 \u2013 1000 = 2000 Joule<\/span><\/p>\n

Wanted:<\/u> Efficiency of the heat engine (e)<\/span><\/p>\n

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

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

2. A heat engine produces 2000 Joule of mechanical work and the engine discharges heat to the environment at a rate of 500 Joule. What is the efficiency of the heat engine?<\/span><\/p>\n

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

The work is done by the engine (W) = 2000 Joule<\/span><\/p>\n

Heat output (QL<\/sub>) = 5000 Joule<\/span><\/p>\n

Heat input (QH<\/sub>) = 2000 + 5000 = 7000 Joule<\/span><\/p>\n

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

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

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

3. A heat engine has an efficiency of 30%. If the engine produces 10,000 Joule of mechanical work, how much heat is discharged as waste heat from this engine?<\/span><\/p>\n

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

Efficiency (e) = 30 % = 30\/100 = 0.3<\/span><\/p>\n

The work is done by engine (W) = 3,000 Joule <\/span><\/p>\n

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

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

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

Heat output (QL<\/sub>) = heat input (QH<\/sub>) \u2013 work is done by engine (W)<\/span><\/p>\n

QL <\/sub>= 10,000 Joule \u2013 3,000 Joule<\/span><\/p>\n

QL<\/sub> = 7,000 Joule<\/span><\/p>\n

4. Determine the efficiency of a Carnot engine operating between two reservoirs at temperatures of 500 K and 300 K.<\/span>
\nSolution: Using the Carnot efficiency formula, \\(\\eta = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}}\\), we get \\(\\eta = 1 – \\frac{300}{500} = 0.4\\) or \\(40\\%\\).<\/span><\/p>\n

5. A heat engine absorbs 2000 J of heat from a hot reservoir and expels 1200 J to a cold reservoir. Calculate the work done by the engine.<\/span>
\nSolution: Using the first law of thermodynamics, \\(Q_{\\text{hot}} = W + Q_{\\text{cold}}\\), the work done is \\(W = 2000 – 1200 = 800\\ \\text{J}\\).<\/span><\/p>\n

6. Calculate the efficiency of a heat engine that absorbs 1000 J from a hot reservoir and does 600 J of work.<\/span>
\nSolution: \\(\\eta = \\frac{W}{Q_{\\text{hot}}} = \\frac{600}{1000} = 0.6\\) or \\(60\\%\\).<\/span><\/p>\n

7. A heat engine operates between 400 K and 200 K. Determine the maximum possible efficiency.<\/span>
\nSolution: \\(\\eta_{\\text{max}} = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}} = 1 – \\frac{200}{400} = 0.5\\) or \\(50\\%\\).<\/span><\/p>\n

8. A heat engine rejects 300 J to the cold reservoir while doing 200 J of work. Calculate the heat absorbed from the hot reservoir.<\/span>
\nSolution: \\(Q_{\\text{hot}} = W + Q_{\\text{cold}} = 200 + 300 = 500\\ \\text{J}\\).<\/span><\/p>\n

9. Determine the efficiency of an Otto cycle with a compression ratio of 8.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{1}{r^{(\\gamma – 1)}} = 1 – \\frac{1}{8^{(1.4 – 1)}} \\approx 0.564\\) or \\(56.4\\%\\), where \\(\\gamma\\) is the heat capacity ratio.<\/span><\/p>\n

10. A heat engine performs 150 J of work and rejects 100 J to the cold reservoir. Find the heat absorbed from the hot reservoir.<\/span>
\nSolution: \\(Q_{\\text{hot}} = W + Q_{\\text{cold}} = 150 + 100 = 250\\ \\text{J}\\).<\/span><\/p>\n

11. Determine the efficiency of a Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{1}{r^{(\\gamma – 1)}} \\left( \\frac{r_c – 1}{r_c} \\right) \\approx 0.627\\) or \\(62.7\\%\\), where \\(r\\) is the compression ratio, \\(r_c\\) is the cutoff ratio, and \\(\\gamma\\) is the heat capacity ratio.<\/span><\/p>\n

12. Calculate the work done by a heat engine that absorbs 400 J and has an efficiency of 50%.<\/span>
\nSolution: \\(W = \\eta \\cdot Q_{\\text{hot}} = 0.5 \\cdot 400 = 200\\ \\text{J}\\).<\/span><\/p>\n

13. A heat engine operates between 600 K and 300 K. Find the Carnot efficiency.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}} = 1 – \\frac{300}{600} = 0.5\\) or \\(50\\%\\).<\/span><\/p>\n

14. A heat engine absorbs 3000 J and rejects 1800 J to the cold reservoir. Calculate the work done.<\/span>
\nSolution: \\(W = Q_{\\text{hot}} – Q_{\\text{cold}} = 3000 – 1800 = 1200\\ \\text{J}\\).<\/span><\/p>\n

15. Determine the efficiency of a Stirling engine operating between two reservoirs at 800 K and 400 K.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}} = 1 – \\frac{400}{800} = 0.5\\) or \\(50\\%\\).<\/span><\/p>\n

16. A heat engine rejects 500 J to the cold reservoir while doing 300 J of work. Calculate the heat absorbed from the hot reservoir.<\/span>
\nSolution: \\(Q_{\\text{hot}} = W + Q_{\\text{cold}} = 300 + 500 = 800\\ \\text{J}\\).<\/span><\/p>\n

17. Determine the efficiency of an Ericsson cycle with hot and cold reservoirs at 1000 K and 250 K.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}} = 1 – \\frac{250}{1000} = 0.75\\) or \\(75\\%\\).<\/span><\/p>\n

18. A heat engine does 700 J of work and has an efficiency of 70%. Calculate the heat rejected to the cold reservoir.<\/span>
\nSolution: \\(Q_{\\text{cold}} = \\frac{W}{\\eta} – W = \\frac{700}{0.7} – 700 \\approx 1000 – 700 = 300\\ \\text{J}\\).<\/span><\/p>\n

19. Determine the efficiency of a Brayton cycle with pressure ratios of 6.5 and \\(\\gamma = 1.4\\).<\/span>
\nSolution: \\(\\eta = 1 – \\frac{1}{r_p^{(\\gamma – 1)\/\\gamma}} \\approx 0.433\\) or \\(43.3\\%\\), where \\(r_p\\) is the pressure ratio.<\/span><\/p>\n

20. Calculate the work done by a heat engine that absorbs 250 J and rejects 100 J to the cold reservoir.<\/span>
\nSolution: \\(W = Q_{\\text{hot}} – Q_{\\text{cold}} = 250 – 100 = 150\\ \\text{J}\\).<\/span><\/p>\n

21. A heat engine operates between 450 K and 150 K. Find the Carnot efficiency.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{T_{\\text{cold}}}{T_{\\text{hot}}} = 1 – \\frac{150}{450} \\approx 0.667\\) or \\(66.7\\%\\).<\/span><\/p>\n

22. Determine the efficiency of an Atkinson cycle with a compression ratio of 10.<\/span>
\nSolution: \\(\\eta = 1 – \\frac{1}{r^{(\\gamma – 1)}} \\approx 0.593\\) or \\(59.3\\%\\),<\/span><\/p>\n

where \\(r\\) is the compression ratio and \\(\\gamma\\) is the heat capacity ratio.<\/span><\/p>\n

20. A heat engine absorbs 180 J and has an efficiency of 40%. Calculate the work done.<\/span>
\nSolution: \\(W = \\eta \\cdot Q_{\\text{hot}} = 0.4 \\cdot 180 = 72\\ \\text{J}\\).<\/span><\/p>\n

Note: For problems involving specific heat cycles (like Otto, Diesel, etc.), certain values like heat capacity ratio \\(\\gamma\\) were used, which are typical for certain gases (e.g., \\(\\gamma = 1.4\\) for air). These values might differ depending on the specific substances used in the engine.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 Heat engine (application of the second law of thermodynamics) – problems and solutions 1. Heat input QH = 3000 Joule and heat output QL = 1000 Joule. What is the efficiency of the heat engine? Known : Heat input (QH) = 3000 Joule Heat output (QL) = 1000 Joule Work done by the engine … 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":"Heat 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-1578","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\/1578","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=1578"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1578\/revisions"}],"predecessor-version":[{"id":9020,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1578\/revisions\/9020"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1578"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1578"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1578"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}