<\/ins><\/p>\n\n\n\nCh\u1ea5t r\u1eafn \u0111a tinh th\u1ec3 \u0111\u01b0\u1ee3c c\u1ea5u t\u1ea1o b\u1edfi v\u00f4 s\u1ed1 c\u00e1c tinh th\u1ec3 nh\u1ecf s\u1eafp x\u1ebfp\n h\u1ed7n \u0111\u1ed9n, do \u0111\u00f3 t\u00ednh d\u1ecb h\u01b0\u1edbng c\u1ee7a m\u1ed7i tinh th\u1ec3 nh\u1ecf \u0111\u01b0\u1ee3c b\u00f9 tr\u1eeb trong \nto\u00e0n kh\u1ed1i ch\u1ea5t , d\u1eabn \u0111\u1ebfn ch\u1ea5t r\u1eafn \u0111a tinh th\u1ec3 kh\u00f4ng c\u00f3 tinh d\u1ecb h\u01b0\u1edbng, \nch\u1ec9 c\u00f3 t\u00ednh \u0111\u1eb3ng h\u01b0\u1edbng.<\/p>\n\n\n\n
C3.<\/strong> ( trang 186 sgk V\u1eadt L\u00fd 10): Ch\u1ea5t r\u1eafn v\u00f4 \u0111\u1ecbnh h\u00ecnh c\u00f3 t\u00ednh d\u1ecb h\u01b0\u1edbng kh\u00f4ng? C\u00f3 nhi\u1ec7t \u0111\u1ed9 n\u00f3ng ch\u1ea3y x\u00e1c \u0111\u1ecbnh kh\u00f4ng? T\u1ea1i sao?<\/p>\n\n\n\nTr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\nV\u00ec ch\u1ea5t r\u1eafn v\u00f4 \u0111\u1ecbnh h\u00ecnh kh\u00f4ng c\u00f3 c\u1ea5u tr\u00fac tinh th\u1ec3, do \u0111\u00f3 c\u00f3 t\u00ednh \u0111\u1eb3ng h\u01b0\u1edbng v\u00e0 kh\u00f4ng c\u00f3 nhi\u1ec7t \u0111\u1ed9 n\u00f3ng ch\u1ea3y x\u00e1c \u0111\u1ecbnh.<\/p>\n\n\n\n
B\u00e0i 3 (trang 179 SGK V\u1eadt L\u00fd 10) :<\/strong> Trong c\u00e1c h\u1ec7 th\u1ee9c sau, h\u1ec7 th\u1ee9c n\u00e0o di\u1ec5n t\u1ea3 qu\u00e1 tr\u00ecnh nung n\u00f3ng khi trong m\u1ed9t b\u00ecnh k\u00edn khi b\u1ecf qua s\u1ef1 n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a b\u00ecnh?<\/ins><\/p>\n\n\n\nA. \u0394U = A ; B. \u0394U = Q + A<\/p>\n\n\n\n
C. \u0394U = 0 ; D. \u0394U = Q.<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n– Ch\u1ecdn D<\/strong>. <\/p>\n\n\n\n– Qu\u00e1 tr\u00ecnh nung n\u00f3ng kh\u00ed trong b\u00ecnh k\u00edn l\u00e0 qu\u00e1 tr\u00ecnh \u0111\u1eb3ng t\u00edch => A = 0<\/p>\n\n\n\n
=> \u0394U = Q. Hay nhi\u1ec7t l\u01b0\u1ee3ng m\u00e0 ch\u1ea5t kh\u00ed nh\u1eadn \u0111\u01b0\u1ee3c ch\u1ec9 d\u00f9ng l\u00e0m t\u0103ng n\u1ed9i n\u0103ng.<\/p>\n\n\n\n
B\u00e0i 4 (trang 180 SGK V\u1eadt L\u00fd 10) :<\/strong> Trong qu\u00e1 tr\u00ecnh ch\u1ea5t kh\u00ed nh\u1eadn nhi\u1ec7t v\u00e0 sinh c\u00f4ng th\u00ec Q v\u00e0 A trong h\u1ec7 th\u1ee9c \u03b4U = A + Q ph\u1ea3i c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o sau \u0111\u00e2y?<\/p>\n\n\n\nA. Q < 0 v\u00e0 A > 0<\/ins><\/p>\n\n\n\nB. Q > 0 v\u00e0 A > 0<\/p>\n\n\n\n
C. Q > 0 v\u00e0 A < 0<\/p>\n\n\n\n
D. Q < 0 v\u00e0 A < 0<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n– Ch\u1ecdn C<\/strong>.<\/p>\n\n\n\n– V\u00ec v\u1eadt nh\u1eadn nhi\u1ec7t th\u00ec Q > 0, v\u1eadt sinh c\u00f4ng th\u00ec A < 0.<\/p>\n\n\n\n
B\u00e0i 5 (trang 180 SGK V\u1eadt L\u00fd 10) :<\/strong> Tr\u01b0\u1eddng h\u1ee3p n\u00e0o sau \u0111\u00e2y \u1ee9ng v\u1edbi qu\u00e1 tr\u00ecnh \u0111\u1eb3ng t\u00edch khi nhi\u1ec7t \u0111\u1ed9 t\u0103ng?<\/p>\n\n\n\nA. \u0394U = Q v\u1edbi A > 0<\/ins><\/p>\n\n\n\nB. \u0394U = Q + A v\u1edbi A > 0<\/p>\n\n\n\n
C. \u0394U = Q + A v\u1edbi A < 0 <\/p>\n\n\n\n
D. \u0394U = Q v\u1edbi Q < 0<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n– Ch\u1ecdn A<\/strong>.<\/p>\n\n\n\n– V\u00ec trong qu\u00e1 tr\u00ecnh \u0111\u1eb3ng t\u00ednh nhi\u1ec7t l\u01b0\u1ee3ng m\u00e0 kh\u00ed nh\u1eadn \u0111\u01b0\u1ee3c ch\u1ec9 \u0111\u1ec3 l\u00e0m t\u0103ng n\u1ed9i n\u0103ng c\u1ee7a kh\u00ed.<\/p>\n\n\n\n
B\u00e0i 6 (trang 180 SGK V\u1eadt L\u00fd 10) :<\/strong> Ng\u01b0\u1eddi \nta th\u1ef1c hi\u1ec7n c\u00f4ng 100 J \u0111\u1ec3 n\u00e9n kh\u00ed trong m\u1ed9t xilanh. T\u00ednh \u0111\u1ed9 bi\u1ebfn thi\u00ean \nn\u1ed9i n\u0103ng c\u1ee7a kh\u00ed, bi\u1ebft kh\u00ed truy\u1ec1n ra m\u00f4i tr\u01b0\u1eddng xung quanh nhi\u1ec7t l\u01b0\u1ee3ng \n20 J.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c nguy\u00ean l\u00ed I N\u0110LH: \u0394U = A + Q<\/p>\n\n\n\n
V\u00ec ch\u1ea5t kh\u00ed nh\u1eadn c\u00f4ng (kh\u00ed b\u1ecb n\u00e9n) v\u00e0 truy\u1ec1n nhi\u1ec7t n\u00ean A > 0, Q < 0<\/p>\n\n\n\n
Do \u0111\u00f3 : \u0394U = A + Q = 100 \u2013 20 = 80 J.<\/p>\n\n\n\n
B\u00e0i 7 (trang 180 SGK V\u1eadt L\u00fd 10) :<\/strong> Ng\u01b0\u1eddi \nta truy\u1ec1n cho kh\u00ed trong xilanh nhi\u1ec7t l\u01b0\u1ee3ng 100 J. Kh\u00ed n\u1edf ra th\u1ef1c hi\u1ec7n \nc\u00f4ng 70 J \u0111\u1ea9y pit-t\u00f4ng l\u00ean. T\u00ednh \u0111\u1ed9 bi\u1ebfn thi\u00ean n\u1ed9i n\u0103ng c\u1ee7a kh\u00ed.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\nKh\u00ed trong xilanh nh\u1eadn \u0111\u01b0\u1ee3c nhi\u1ec7t l\u01b0\u1ee3ng => Q > 0.<\/p>\n\n\n\n
Kh\u00ed th\u1ef1c hi\u1ec7n c\u00f4ng => A < 0<\/p>\n\n\n\n
\u0110\u1ed9 bi\u1ebfn thi\u00ean n\u1ed9i n\u0103ng c\u1ee7a kh\u00ed trong xilanh l\u00e0:<\/p>\n\n\n\n
\u0394U = A + Q = 100 – 70 = 30J.<\/p>\n\n\n\n
B\u00e0i 8 (trang 180 SGK V\u1eadt L\u00fd 10) :<\/strong> Khi truy\u1ec1n nhi\u1ec7t l\u01b0\u1ee3ng 6.106<\/sup> J cho kh\u00ed trong m\u1ed9t xilanh h\u00ecnh tr\u1ee5 th\u00ec kh\u00ed n\u1edf ra \u0111\u1ea9y pit-t\u00f4ng l\u00ean l\u00e0m th\u1ec3 t\u00edch c\u1ee7a kh\u00ed t\u0103ng th\u00eam 0,50 m3<\/sup>. T\u00ednh \u0111\u1ed9 bi\u1ebfn thi\u00ean n\u1ed9i n\u0103ng c\u1ee7a kh\u00ed. Bi\u1ebft \u00e1p su\u1ea5t c\u1ee7a kh\u00ed l\u00e0 8.106<\/sup> N\/m2<\/sup> v\u00e0 coi \u00e1p su\u1ea5t n\u00e0y kh\u00f4ng \u0111\u1ed5i trong qu\u00e1 tr\u00ecnh kh\u00ed th\u1ef1c hi\u1ec7n c\u00f4ng.<\/p>\n\n\n\nL\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\nG\u1ecdi S l\u00e0 di\u1ec7n t\u00edch ti\u1ebft di\u1ec7n th\u1eb3ng c\u1ee7a xilanh, h l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng pitt\u00f4ng d\u1ecbch chuy\u1ec3n, p l\u00e0 \u00e1p su\u1ea5t kh\u00ed trong xilanh, ta c\u00f3:<\/p>\n\n\n\n
V\u00ec \u00e1p su\u1ea5t kh\u00ed kh\u00f4ng \u0111\u1ed5i trong qu\u00e1 tr\u00ecnh kh\u00ed th\u1ef1c hi\u00ean c\u00f4ng n\u00ean \u00e1p l\u1ef1c F l\u00ean pit-t\u00f4ng kh\u00f4ng \u0111\u1ed5i.<\/p>\n\n\n\n
\u2192 C\u00f4ng m\u00e0 ch\u1ea5t kh\u00ed th\u1ef1c hi\u1ec7n c\u00f3 \u0111\u1ed9 l\u1edbn l\u00e0:<\/p>\n\n\n\n
A = F.h = p.S.h = p.\u0394V = 8.106<\/sup>.0,5 = 4.106<\/sup> J.<\/p>\n\n\n\nV\u00ec ch\u1ea5t kh\u00ed th\u1ef1c hi\u1ec7n c\u00f4ng v\u00e0 nh\u1eadn nhi\u1ec7t n\u00ean theo quy d\u1ea5u ta c\u00f3: Q > 0, A < 0<\/p>\n\n\n\n
Ta c\u00f3: \u0394U = A + Q = – 4.106<\/sup> + 6.106<\/sup> = 2.106<\/sup> (J)<\/p>\n\n\n\nV\u1eady \u0111\u1ed9 bi\u1ebfn thi\u00ean n\u1ed9i n\u0103ng c\u1ee7a kh\u00ed l\u00e0 \u0394U = 2.106<\/sup> (J)<\/p>\n","protected":false},"excerpt":{"rendered":"B\u00e0i 34 : Ch\u1ea5t r\u1eafn k\u1ebft tinh. Ch\u1ea5t r\u1eafn v\u00f4 \u0111\u1ecbnh h\u00ecnh C1. ( trang 184 sgk V\u1eadt L\u00fd 10): Tinh th\u1ec3 c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c h\u00ecnh th\u00e0nh trong qu\u00e1 tr\u00ecnh n\u00f3ng ch\u1ea3y hay \u0111\u00f4ng \u0111\u1eb7c c\u1ee7a ch\u1ea5t \u0111\u00f3? Tr\u1ea3 l\u1eddi: Tinh th\u1ec3 c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c h\u00ecnh th\u00e0nh trong qu\u00e1 tr\u00ecnh \u0111\u00f4ng \u0111\u1eb7c c\u1ee7a […]<\/p>\n","protected":false},"author":12,"featured_media":44191,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\n
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Tr\u1ea3 l\u1eddi: Tinh th\u1ec3 c\u1ee7a m\u1ed9t ch\u1ea5t \u0111\u01b0\u1ee3c h\u00ecnh th\u00e0nh trong qu\u00e1 tr\u00ecnh \u0111\u00f4ng \u0111\u1eb7c c\u1ee7a […]","og_url":"https:\/\/lop12.edu.vn\/giai-vat-li-10-bai-34-chat-ran-ket-tinh-chat-ran-vo-dinh-hinh\/","og_site_name":"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam","article_published_time":"2019-10-13T07:25:35+00:00","article_modified_time":"2019-10-13T07:25:36+00:00","og_image":[{"width":696,"height":385,"url":"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2019\/10\/i34.png","type":"image\/png"}],"author":"Nguy\u1ec5n M\u01a1","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Nguy\u1ec5n M\u01a1","Est. reading time":"5 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/lop12.edu.vn\/giai-vat-li-10-bai-34-chat-ran-ket-tinh-chat-ran-vo-dinh-hinh\/","url":"https:\/\/lop12.edu.vn\/giai-vat-li-10-bai-34-chat-ran-ket-tinh-chat-ran-vo-dinh-hinh\/","name":"Gi\u1ea3i V\u1eadt L\u00ed 10 B\u00e0i 34 : Ch\u1ea5t r\u1eafn k\u1ebft tinh. 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