{"id":628,"date":"2018-02-14T07:34:54","date_gmt":"2018-02-13T23:34:54","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=628"},"modified":"2018-02-14T07:34:54","modified_gmt":"2018-02-13T23:34:54","slug":"energy-conservation-for-heat-transfer-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/energy-conservation-for-heat-transfer-problems-and-solutions.htm","title":{"rendered":"Energy conservation for heat transfer \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>1-kg water at <\/span><\/span>100<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>mixed with 1-kg water at <\/span><\/span>10<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>in an isolated system. The specific heat<\/a> of water is <\/span><\/span>4200 J\/kg <\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>Determine the final temperature<\/a> of the mixture!<\/span><\/span><\/p>\n

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

Mass<\/a> of hot water <\/span><\/span>(m<\/span><\/span>1<\/span><\/sub>) = 1 kg<\/span><\/span><\/p>\n

Temperature of hot water <\/span><\/span>(T<\/span><\/span>1<\/span><\/sub>) = 100<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Mass of cold water <\/span><\/span>(m<\/span><\/span>2<\/span><\/sub>) = 1 kg <\/span><\/span><\/p>\n

Temperature of cold water <\/span><\/span>(T<\/span><\/span>2<\/span><\/sub>) = 10<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

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

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

Heat lost = Heat gained (isolated system)<\/span><\/span><\/p>\n

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

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

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

(1)(100 – T) <\/span><\/span>= (1)(T – 10)<\/span><\/span><\/p>\n

100 \u2013 T <\/span><\/span>= T \u2013 10<\/span><\/span><\/p>\n

100 + 10 = T + T<\/span><\/span><\/p>\n

110 = 2T<\/span><\/span><\/p>\n

T = 110 \/ 2<\/span><\/span><\/p>\n

T = 55<\/span><\/span><\/p>\n

The final temperature is <\/span><\/span>55<\/span><\/span>o<\/span><\/span><\/sup>C.<\/span><\/span><\/p>\n

<\/p>\n

2. <\/span><\/span>A 3 kg block of lead at 80<\/span><\/span>o<\/span><\/span><\/sup> placed in 10 kg of water. The specific heat of lead is 1400 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup> and the specific heat of water is 4200 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. The final temperature in thermal equilibrium is <\/span><\/span>20<\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>Determine the initial temperature of water! <\/span><\/span><\/p>\n

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

Mass <\/span><\/span>(m<\/span><\/span>1<\/span><\/sub>) = 3 kg<\/span><\/span><\/p>\n

The specific heat of lead<\/span><\/span> (c<\/span><\/span>1<\/span><\/sub>) = 1400 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The temperature of lead<\/span><\/span> (T<\/span><\/span>1<\/span><\/sub>) = 80 <\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

Mass of water <\/span><\/span>(m<\/span><\/span>2<\/span><\/sub>) = 10 kg<\/span><\/span><\/p>\n

The specific heat of water <\/span><\/span>(c<\/span><\/span>2<\/span><\/sub>) = 4200 J.kg<\/span><\/span>-1<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The temperature of thermal equilibrium <\/span><\/span>(T) = 20 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

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

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

Heat lost = Heat gained <\/span><\/span><\/p>\n

Q <\/span><\/span>lead<\/span><\/span> = Q <\/span><\/span>water<\/span><\/span><\/p>\n

m<\/span><\/span>1<\/span><\/span><\/sub> c<\/span><\/span>1<\/span><\/span><\/sub> \u0394T = m<\/span><\/span>2<\/span><\/span><\/sub> c<\/span><\/span>2<\/span><\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

(3)(1400)(80-20) = (10)(4200)(20-T)<\/span><\/span><\/p>\n

(4200)(60) = (42,000)(20-T)<\/span><\/span><\/p>\n

252,000 = 840,000 \u2013 42,000 T<\/span><\/span><\/p>\n

42,000 T = 840,000 \u2013 252,000 <\/span><\/span><\/p>\n

42,000 T = 588,000 <\/span><\/span><\/p>\n

T = 588,000 \/ 42,000<\/span><\/span><\/p>\n

T = 14<\/span><\/span><\/p>\n

The initial temperature of water is <\/span><\/span>14<\/span><\/span>o<\/span><\/span><\/sup>C.<\/span><\/span><\/p>\n

<\/p>\n

3. <\/span><\/span>A block of copper at <\/span><\/span>100<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>placed in 128 gram water at <\/span><\/span>30 <\/span><\/span>o<\/span><\/span><\/sup>C. <\/span><\/span>The specific heat of water is <\/span><\/span>1 <\/span><\/span>c<\/span><\/span>al.g<\/span><\/span>-1o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup> and the specific heat of copper is <\/span><\/span>0.1 <\/span><\/span>c<\/span><\/span>al.g<\/span><\/span>-1o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. <\/span><\/span>If the temperature of the thermal equilibrium is <\/span><\/span>36 <\/span><\/span>o<\/span><\/span><\/sup>C, <\/span><\/span>determine the mass of the copper!<\/span><\/span><\/p>\n

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

The temperature of copper <\/span><\/span>(T<\/span><\/span>1<\/span><\/sub>) = 100 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The specific heat of copper <\/span><\/span>(c<\/span><\/span>1<\/span><\/sub>) = 0.1 <\/span><\/span>c<\/span><\/span>al.g<\/span><\/span>-1o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Mass of water <\/span><\/span>(m<\/span><\/span>2<\/span><\/sub>) = 128 gram<\/span><\/span><\/p>\n

Temperature of water <\/span><\/span>(T<\/span><\/span>2<\/span><\/sub>) = 30 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The specific heat of water <\/span><\/span>(c<\/span><\/span>2<\/span><\/sub>) = 1 <\/span><\/span>c<\/span><\/span>al.g<\/span><\/span>-1o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The temperature of thermal equilibrium <\/span><\/span>(T) = 36 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Wanted : <\/u><\/span><\/span>Mass of copper<\/span><\/span> (m<\/span><\/span>1<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

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

Heat lost = Heat gained<\/span><\/span><\/p>\n

Q <\/span><\/span>copper<\/span><\/span> = Q <\/span><\/span>water<\/span><\/span><\/p>\n

m<\/span><\/span>1<\/span><\/sub> c<\/span><\/span>1<\/span><\/sub> \u0394T = m<\/span><\/span>2<\/span><\/sub> c<\/span><\/span>2<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

(m<\/span><\/span>1<\/span><\/sub>)(0.1)(100-36) = (128)(1)(36-30)<\/span><\/span><\/p>\n

(m<\/span><\/span>1<\/span><\/sub>)(0.1)(64) = (128)(1)(6)<\/span><\/span><\/p>\n

(m<\/span><\/span>1<\/span><\/sub>)(6.4) = 768<\/span><\/span><\/p>\n

m<\/span><\/span>1<\/span><\/sub> = 768 \/ 6.4<\/span><\/span><\/p>\n

m<\/span><\/span>1<\/span><\/sub> = 120<\/span><\/span><\/p>\n

The mass of copper is <\/span><\/span>120 gram. <\/span><\/span><\/p>\n

<\/p>\n

4. <\/span><\/span>A M-kg block of ice at <\/span><\/span>0<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>placed in 340-gram water at <\/span><\/span>20<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span>in a vat. If the heat of fusion<\/a> for water <\/span><\/span>= 80 cal g<\/span><\/span>-1<\/span><\/span><\/sup>, <\/span><\/span>the specific heat of water is <\/span><\/span>1 cal g<\/span><\/span>-1<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. <\/span><\/span>All ice melts and the temperature of thermal equilibrium<\/a> is <\/span><\/span>5<\/span><\/span>o<\/span><\/span><\/sup>C, <\/span><\/span>determine the mass of ice!<\/span><\/span><\/p>\n

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

Mass of water <\/span><\/span>(m) = 340 gram <\/span><\/span><\/p>\n

The temperature of ice <\/span><\/span>(T <\/span><\/span>ice<\/span><\/span><\/sub>) = 0<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The temperature of water <\/span><\/span>(T <\/span><\/span>water<\/span><\/span><\/sub>) = 20<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The temperature of thermal equilibrium <\/span><\/span>(T) = 5<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The heat of fusion for water <\/span><\/span>(<\/span><\/span>L<\/span><\/span>) = 80 <\/span><\/span>c<\/span><\/span>al g<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

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

Wanted :<\/u><\/span><\/span> Mass <\/span><\/span>of ice <\/span><\/span>(M)<\/span><\/span><\/p>\n

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

Heat lost = Heat gained<\/i><\/span><\/span><\/p>\n

Q water = Q ice<\/i><\/span><\/span><\/p>\n

m<\/span><\/span> c<\/span><\/span> (<\/span><\/span>\u0394T) = m<\/span><\/span>es<\/span><\/span><\/sub> L<\/span><\/span>es<\/span><\/span><\/sub> + m<\/span><\/span> c<\/span><\/span> (<\/span><\/span>\u0394<\/span><\/span>T) <\/span><\/span><\/p>\n

(340)(1)(20-5) = M (80) + M (1)(5-0)<\/span><\/span><\/p>\n

(340)(15) = 80M + 5M<\/span><\/span><\/p>\n

5100 = 85M<\/span><\/span><\/p>\n

M = 5100\/85 <\/span><\/span><\/p>\n

M = 60 gram<\/span><\/span><\/p>\n

[wpdm_package id=’714′]<\/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":"

    1. 1-kg water at 100oC mixed with 1-kg water at 10oC in an isolated system. The specific heat of water is 4200 J\/kg oC. Determine the final temperature of the mixture! Known : Mass of hot water (m1) = 1 kg Temperature of hot water (T1) = 100oC Mass of cold water (m2) = 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":"Energy conservation for heat transfer \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-628","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\/628","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=628"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/628\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=628"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=628"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=628"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}