{"id":589,"date":"2018-02-12T16:47:51","date_gmt":"2018-02-12T08:47:51","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=589"},"modified":"2018-02-12T16:47:51","modified_gmt":"2018-02-12T08:47:51","slug":"linear-expansion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/linear-expansion-problems-and-solutions.htm","title":{"rendered":"Linear expansion \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A steel is 40 cm long at 20 <\/span><\/span>o<\/span><\/span><\/sup>C. The coefficient of linear expansion<\/a> for steel is 12 x 10<\/span><\/span>-6<\/span><\/span><\/sup> (C<\/span><\/span>o<\/span><\/span><\/sup>)<\/span><\/span>-1<\/span><\/span><\/sup>. The increase in length and the final length when it is at 70 <\/span><\/span>o<\/span><\/span><\/sup>C will be…<\/span><\/span><\/p>\n

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

The change in temperature (<\/span><\/span>\u0394T) = 70<\/span><\/span>o<\/span><\/span><\/sup>C \u2013 20<\/span><\/span>o<\/span><\/span><\/sup>C = 50<\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The original length (L<\/span><\/span>1<\/span><\/sub>) = 40 cm<\/span><\/span><\/p>\n

Coefficient of linear expansion for steel (\u03b1<\/span><\/span>) <\/span><\/span>= 12 x 10<\/span><\/span>-6<\/span><\/span><\/sup> (C<\/span><\/span>o<\/span><\/span><\/sup>)<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> The change in length (<\/span><\/span>\u0394L) and the final length (L<\/span><\/span>2<\/span><\/sub>) <\/span><\/span><\/p>\n

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

a) The change in length (<\/b><\/span><\/span>\u0394L)<\/b><\/span><\/span><\/p>\n

\u0394L = \u03b1<\/span><\/span> L<\/span><\/span>1<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

\u0394L <\/span><\/span>= (<\/span><\/span>12×10<\/span><\/span>-6 o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>)(40 cm)(50<\/span><\/span>o<\/span><\/span><\/sup>C) <\/span><\/span><\/p>\n

\u0394L = (10<\/span><\/span>-6<\/span><\/span><\/sup>)(24 x 10<\/span><\/span>3<\/span><\/span><\/sup>) cm <\/span><\/span><\/p>\n

\u0394L = 24 x 10<\/span><\/span>-3<\/span><\/span><\/sup> cm<\/span><\/span><\/p>\n

\u0394L = 24 \/ 10<\/span><\/span>3<\/span><\/span><\/sup> cm<\/span><\/span><\/p>\n

\u0394L = 24 \/ 1000 cm<\/span><\/span><\/p>\n

\u0394L = 0.024 cm<\/span><\/span><\/p>\n

b) The final length (L<\/b><\/span><\/span>2<\/b><\/span><\/sub>) <\/b><\/span><\/span><\/p>\n

L<\/span><\/span>2<\/span><\/sub> = L<\/span><\/span>1<\/span><\/sub> + <\/span><\/span>\u0394L <\/span><\/span><\/p>\n

L<\/span><\/span>2<\/span><\/sub> = 40 cm + 0.024 cm <\/span><\/span><\/p>\n

L<\/span><\/span>2<\/span><\/sub> = <\/span><\/span>40.024 cm <\/span><\/span><\/p>\n

<\/p>\n

2. An iron <\/span><\/span>rod heated from 30 <\/span><\/span>o<\/span><\/span><\/sup>C to 80 <\/span><\/span>o<\/span><\/span><\/sup>C. The final length of iron is 115 cm and the coefficient of linear expansion is 3×10<\/span><\/span>-3<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. What is the original length and the change in length of the iron ? <\/span><\/span><\/p>\n

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

The change in temperature <\/a><\/span><\/span>(<\/span><\/span>\u0394T) = 80 <\/span><\/span>o<\/span><\/span><\/sup>C \u2013 30 <\/span><\/span>o<\/span><\/span><\/sup>C = 50 <\/span><\/span>o<\/span><\/span><\/sup>C <\/span><\/span><\/p>\n

The final length <\/span><\/span>(L<\/span><\/span>2<\/span><\/sub>) = 115 cm<\/span><\/span><\/p>\n

The coefficient of linear expansion <\/span><\/span>(<\/span><\/span>\u03b1) <\/span><\/span>= 3×10<\/span><\/span>-3 o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> the original length <\/span><\/span>(L<\/span><\/span>1<\/span><\/sub>) <\/span><\/span>and the change in length <\/span><\/span>(<\/span><\/span>\u0394L)<\/span><\/span><\/p>\n

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

a) <\/b><\/span><\/span>The original length <\/b><\/span><\/span>(L<\/b><\/span><\/span>1<\/b><\/span><\/sub>)<\/b><\/span><\/span><\/p>\n

Formula of the change in length for the linear expansion :<\/span><\/span><\/p>\n

\u0394L =\u00a0<\/span><\/span>\u03b1 L<\/span><\/span>1<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

Formula of the final length <\/span><\/span>:<\/span><\/span><\/p>\n

L<\/span><\/span>2<\/span><\/sub> = <\/span><\/span>L<\/span><\/span>1 <\/span><\/sub>+ <\/span><\/span>\u0394L <\/span><\/span><\/p>\n

L<\/span><\/span>2<\/span><\/sub> = L<\/span><\/span>1<\/span><\/sub> + \u03b1<\/span><\/span> L<\/span><\/span>1<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

L<\/span><\/span>2 <\/span><\/sub>= <\/span><\/span>L<\/span><\/span>1<\/span><\/sub> (1 +\u00a0<\/span><\/span>\u03b1 \u0394T) <\/span><\/span><\/p>\n

115 cm = <\/span><\/span>L<\/span><\/span>1 <\/span><\/sub>(1 + (3.10<\/span><\/span>-3 o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>)(50<\/span><\/span>o<\/span><\/span><\/sup>C)<\/span><\/span><\/p>\n

115 cm = L<\/span><\/span>1 <\/span><\/sub>(1 + 150.10<\/span><\/span>-3<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

115 cm = L<\/span><\/span>1 <\/span><\/sub>(1 + 0.15)<\/span><\/span><\/p>\n

115 cm = L<\/span><\/span>1 <\/span><\/sub>(1.15)<\/span><\/span><\/p>\n\n\n\n<\/colgroup>\n\n\n
\n

L<\/span><\/span>1 <\/span><\/sub>= 115 cm \/ 1.15 <\/span><\/span><\/p>\n

L<\/span><\/span>1 <\/span><\/sub>= 100 cm<\/span><\/span><\/p>\n

b) <\/b><\/span><\/span>the change in length <\/b><\/span><\/span>(<\/b><\/span><\/span>\u0394L)<\/b><\/span><\/span><\/p>\n

\u0394L = L<\/span><\/span>2 <\/span><\/sub>\u2013 L<\/span><\/span>1<\/sub> <\/span><\/span><\/p>\n

\u0394L = 115 cm \u2013 100 cm<\/span><\/span><\/p>\n

\u0394L = 15 cm <\/span><\/span><\/p>\n<\/td>\n

<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

<\/p>\n

3. At 25 <\/span><\/span>o<\/span><\/span><\/sup>C, the length of the glass is 50 cm. After heated, the final length of the glass is 50.9 cm. The coefficient of linear expansion is <\/span><\/span>\u03b1 = 9 x 10<\/span><\/span>-6 <\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup>. <\/span><\/span>Determine the final temperature of the glass…<\/span><\/span><\/p>\n

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

The original length <\/span><\/span>(L<\/span><\/span>1<\/span><\/sub>) = 50 cm<\/span><\/span><\/p>\n

The final length <\/span><\/span>(L<\/span><\/span>2<\/span><\/sub>) = 50.09 cm<\/span><\/span><\/p>\n

The change in length <\/span><\/span>(<\/span><\/span>\u0394L) = 50.2 cm \u2013 50 cm = 0.09 cm<\/span><\/span><\/p>\n

The coefficient of linear expansion <\/span><\/span>(\u03b1) = 9 x 10<\/span><\/span>-6 o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

The original 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> The final temperature <\/span><\/span>(T<\/span><\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

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

\u0394L = \u03b1<\/span><\/span> L<\/span><\/span>1<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

\u0394L =\u00a0<\/span><\/span>\u03b1 L<\/span><\/span>1<\/span><\/sub> (T<\/span><\/span>2 <\/span><\/span><\/sub>– T<\/span><\/span>1<\/span><\/span><\/sub>)<\/span><\/span><\/p>\n

0.09 <\/span><\/span>cm <\/span><\/span>= (<\/span><\/span>9 x 10<\/span><\/span>-6 o<\/span><\/span><\/sup>C)<\/span><\/span>(50 <\/span><\/span>cm<\/span><\/span>)(T<\/span><\/span>2 <\/span><\/sub>\u2013 25 <\/span><\/span>o<\/span><\/span><\/sup>C)<\/span><\/span><\/p>\n

0.09 = (45 <\/span><\/span>x 10<\/span><\/span>-5<\/span><\/span><\/sup>)(T<\/span><\/span>2 <\/span><\/sub>– 25)<\/span><\/span><\/p>\n

0.09 \/ (45 <\/span><\/span>x 10<\/span><\/span>-5<\/span><\/span><\/sup>) = T<\/span><\/span>2 <\/span><\/sub>– 25<\/span><\/span><\/p>\n

0.002 <\/span><\/span>x 10<\/span><\/span>5 <\/span><\/span><\/sup>= T<\/span><\/span>2<\/span><\/sub> \u2013 25<\/span><\/span><\/p>\n

2 <\/span><\/span>x 10<\/span><\/span>2 <\/span><\/span><\/sup>= T<\/span><\/span>2<\/span><\/sub> \u2013 25<\/span><\/span><\/p>\n

200<\/span><\/span> = <\/span><\/span>T<\/span><\/span>2<\/span><\/sub> \u2013 25<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= 200 + 25<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= 225<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

The final temperature is <\/span><\/span>225 <\/span><\/span>o<\/span><\/span><\/sup>C.<\/span><\/span><\/p>\n

<\/p>\n

4. <\/span><\/span>The original length of metal is 1 meter and the final length is 1.02 m. The change in temperature is <\/span><\/span>50 Kelvin. <\/span><\/span>Determine the coefficient of linear expansion!<\/span><\/span><\/p>\n

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

The initial length <\/span><\/span>(L<\/span><\/span>1<\/span><\/sub>) = 1 meter<\/span><\/span><\/p>\n

The final length <\/span><\/span>(L<\/span><\/span>2<\/span><\/sub>) = 1.02 meter<\/span><\/span><\/p>\n

The change in length <\/span><\/span>(<\/span><\/span>\u0394<\/span><\/span>L) = L<\/span><\/span>2 <\/span><\/sub>\u2013 L<\/span><\/span>1<\/span><\/sub> = 1.02 meter – 1 meter = 0.02 meter <\/span><\/span><\/p>\n

The change in temperature <\/span><\/span>(\u0394<\/span><\/span>T<\/span><\/span>) = 50 Kelvin = 50<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

Wanted : <\/u><\/span><\/span>The coefficient of linear expansion<\/span><\/span><\/p>\n

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

\u0394<\/span><\/span>L =\u00a0<\/span><\/span>\u03b1 L<\/span><\/span>1<\/span><\/sub> \u0394T <\/span><\/span><\/p>\n

0.02 m =\u00a0<\/span><\/span>\u03b1 (1 <\/span><\/span>m<\/span><\/span>)<\/span><\/span>(50<\/span><\/span>o<\/span><\/span><\/sup>C) <\/span><\/span><\/p>\n

0.02 =\u00a0<\/span><\/span>\u03b1 (<\/span><\/span>50<\/span><\/span>o<\/span><\/span><\/sup>C) <\/span><\/span><\/p>\n

\u03b1\u00a0= <\/span><\/span>0.02 \/ 50<\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span><\/p>\n

\u03b1 = <\/span><\/span>0.0004 <\/span><\/span>o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

\u03b1\u00a0= <\/span><\/span>4 x 10<\/span><\/span>-4<\/span><\/span><\/sup> o<\/span><\/span><\/sup>C<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

[wpdm_package id=’694′]<\/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. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be… Known : The change in temperature (\u0394T) = 70oC \u2013 20oC = 50oC The original length (L1) … 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":"Linear expansion \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-589","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\/589","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=589"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/589\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}