bab 7 kls x

1. Perubahan musim yang terjadi dengan diikuti perubahan wujud dari cair ke padat, atau sebaliknya (musim salju), merupakan fenomena berlakunya konsep-konsep kalor yang terjadi di alam ini. A. Kalor B. Perpindahan Kalor Kalor Hasil yang harus Anda capai: menerapkan konsep kalor dan prinsip konservasi energi pada berbagai perubahan energi. Setelah mempelajari bab ini, Anda harus mampu: • menganalisis pengaruh kalor terhadap suatu zat; • menganalisis cara perpindahan kalor; • menerapkan asas Black dalam pemecahan masalah. 4@30A0@:0= EC9C3=G0 0B4@8 381430:0= 4=9038 H0B ?030B 208@ 30= 60A !4B860 H0B B4@A41CB 48;8:8 :0@0:B4@8AB8: 0A8=60A8=6 :0= B4B0?80301414@0?0H0BG0=630?0B14@C107EC9C3:0@4=0?4=60@C7AC7C 30= :0;@ 0B8;07 :4B8:0 =30 A430=6 8=C 8=C0= G0=6 38 30;0=G0B4@30?0B4A10BC(40:8=;004A10BC30;08=C0=0:0= 4=208@ 30= 14@C107 EC9C3 #4=60?0 B4@9038 348:80= 4=0@:07 4A B4@A41CB B830: 78;0=6 4;08=:0= 14@C107 EC9C3 #4=60?0 38 AC0BC B4?0B 38 C8 B4@9038 7C90= A0;9C #4=60?0 38 =3=4A80B830:B4@9038CA8A0;9C0301018=8=300:0=4?4;090@8 B4=B0=6 :0;@ 8:0 =30 B4;07 40708 :=A4? :=A4? :0;@ ?4@B0=G00=B4@A41CB0:0=C307=3090E01%;47:0@4=08BC?4;090@8;07 101 8=8 34=60= AC=66C7AC=66C7 147 Bab 7 Sumber: CD Image

2. Tes Kompetensi Awal #-6...1-+3*,0/41-03,3+,/-)40-40-#3*,65%-.#6,6-5*)/ ?0G0=638A41CB34=60=?4C080= (41CB:0=B86020@0?4@?8=3070=:0;@ Joseph Black (1728–1799) 148 Mudah dan Aktif Belajar Fisika untuk Kelas X

3. ?0:07 A00 B8B8: 38387 08@ :4B8:0 3838387:0= 38 ?46C=C=60= 3810=38=6:0= 34=60= B8B8: 38387 08@ 38 304@07?0=B08 4;0A:0=90E010==30 A. Kalor 4@=07:07 =30 4;0:C:0= ?4=6C:C@0= AC7C #8A0;=G0 ?030 AC0BC B4@ 4B4@ :; @0:A0 =30 4=600B8 ?4@C1070= ?0=90=6 :; A410608 8=38:0B@ 030=G0 ?4@C1070= AC7C (4AC=66C7=G0 ?4@C1070= AC7C 8BC B4@9038 0:810B14=30G0=6A430=638C:C@4;4?0A:0=:0;@0B0C4=4@80:0;@ 1. Pengertian Kalor !0;@ 4@C?0:0= A0;07 A0BC 14=BC: 4=4@68 G0=6 30?0B 14@?8=307 30@8 14=30 G0=6 A0BC :4 14=30 G0=6 ;08= 8:0 3C0 1C07 14=30 G0=6 AC7C=G0 14@1430 38A4=BC7:0= AC0BC A00B 0:0= B4@9038 :4A4B810=60= B4@0; AC7C=G0 A00 0; 8=8 B4@9038 :0@4=0 030=G0 ?4@?8=3070= :0;@ 30@8 14=30 G0=6 14@AC7C B8=668 :4 14=30 G0=6 14@AC7C @4=307 !0;@14@143034=60=AC7CE0;0C?C=:43C0=G048;8:87C1C=60=4@0B (C7C030;0734@090B?0=0A0B0C38=68=AC0BC14=30A430=6:0=!%'030;07 $'.$$!$'(*)*$!$.$$(C7C30=:0;@ 30?0B 381430:0= 34=60= 94;0A ?030 ?4@8AB8E0 ?4@C1070= EC9C3 AC0BC H0B *=BC: 4=6C107 4A 4=9038 08@ 38?4@;C:0= :0;@ 030 ?4@8AB8E0 ?4@C1070= EC9C3 8=8 4A 14@AC7C I 14@C107 4=9038 08@ 14@AC7C I 038 B830: 030 ?4@C1070= AC7C ?030 A00B 4A 4=208@ B4B0?8 381CBC7:0= :0;@ C=BC: 4=6C107 EC9C3 4A B4@A41CB (0BC0= ( C=BC: :0;@ 030;07 9C;4 38A8=6:0B (0BC0= :0;@ G0=6 ;08=030;07:0;@80B0C:8;:0;@8)*!%'383458=8A8:0=A410608$.! $. !%' .$ '*!$ *$)*!#$!!$ (** ()* '# ' ((' 0 2. Kalor Jenis #8A0;:0=?030 08@30= 0;:7;3814@8:0=:0;@G0=6 A00 )4@=G0B0 :4=08:0= AC7C ?030 0;:7; ;4187 14A0@ 30@8?030 08@ 48:80= 70;=G0 98:0 ?030 08@ 30= 08@ 3814@8:0= :0;@ G0=6 A00 10=G0:=G0 :4=08:0= AC7C ?030 08@ ;4187 14A0@ 30@8 ?030 08@4@8AB8E0B4@A41CB4=C=9C::0=70;70;A41060814@8:CB 0 !0;@ G0=6 3814@8:0= ?030 H0B A410=38=6 34=60= :4=08:0= AC7C 1 !0;@G0=6381CBC7:0=C=BC:4=08::0=AC7CH0BA410=38=60AA0H0B 2 !0;@G0=6381CBC7:0=C=BC:4=08::0=AC7CH0B14@60=BC=694=8AH0B (420@0 0B40B8A 38BC;8A A410608 14@8:CB # J !4B4@0=60= 10=G0:=G0 :0;@ G0=6 3814@8:0= :0;@8 0B0C 9C;4 # 0AA0 H0B 6 0B0C :6 :0;@ 94=8A :0;6I 0B0C :6I ?4@C1070= AC7C I 34./ 30?0B 38BC;8A A410608 14@8:CB # J Tokoh Joseph Black, adalah seorang kimiawan Skotlandia yang mendukung teori tentang panas, yaitu bahwa suhu merupakan konsentrasi kalori dalam suatu benda. Ia kemudian menemukan ilmu baru yang disebut kalorimetri. Ketika menyelidiki tentang panas (kalori), ia mengira bahwa kapasitas panas merupakan jumlah panas yang dapat ditampung oleh suatu benda. Padahal, ini sebenarnya merupakan ukuran tentang jumlah energi yang diperlukan untuk menaikkan suhu suatu benda dalam jumlah tertentu. Sumber: Jendela Iptek, 1996

4. Tugas Anda 7.1 Anda telah mengenal besaran kalor jenis suatu benda. Diskusikan dengan teman sekelas Anda jika kalor jenis suatu zat besar, apakah benda akan cepat panas atau lambat panas? Ingatlah Kalor 149 038:0;@94=8A30?0B383458=8A8:0=A41060810=G0:=G0:0;@G0=638?4@ ;C:0=AC0BCH0BC=BC:4=08::0=AC7C :6H0BB4@A41CBA414A0@ I#- 4=C=9C::0= :0;@ 94=8A 1414@0?0 H0B ?030 AC7C I 30= B4:0=0= B4B0? 0BA54@ #- !0;@ 4=8A414@0?0-0B?030(C7C I30=)4:0=0= 0B -03+/*4 ,(; !5%5 !C=8=60= ;C8=8C )41060 4A80B0C1090 )807 #0@4@ 4@0: !0GC (4=6

5. Mari Mencari Tahu -03+/*4 ,(;

6. Sumber: Physics, 1980 !5*3 ;:7;4B8; '0:A0 8@ 4AJI 208@ I C0? I 030=0=CA80 @B48= !4B8:0=3014@90;0=380B0A?0A8@38?0=B083810E07B4@8:#0B070@8=3030?0B 4@0A0:0= ?0A8@ :4@8=6 G0=6 B4@8=90: B4@0A0 ?0=0A !4C380= =30 14@0;87 4=68=90:08@;0CB8@;0CB@4;0B85;418738=68=1C:0=03070;?0A8@30=08@;0CB A00A00B4@:4=0A8=0@#0B070@8#8A0;=G0=30B8=90C :608@;0CB30= :6 ?0A8@?0=B08380=B0@0=G0#4=C@CB=300=0:07G0=648;8:8:0;@94=8A;4187 B8=66808@0B0C?0A8@ 3. Kapasitas Kalor ()(!%'030;07$.!$.!%'.$'*!$*$)*!#$!!$ (** (*)* $ ((' 0 (420@0 0B40B8A ?4@=G0B00= B4@A41CB 38BC;8A A410608 14@8:CB J 030 34./ 030;07 :0?0A8B0A :0;@ :0;I 0B0C ! )4@30?0B ?4@14300= ?4=64@B80= 0=B0@0 :0?0A8B0A :0;@ 30= :0;@ 94=8A B4B0?8 A420@0 0B40B8A :43C0=G0 48;8:8 7C1C=60= A410608 14@8:CB 0@8 34./ 30= 34./ 38?4@;47 # # J !4B4@0=60= #0AA0H0B:6 :0;@ 94=8A :6I 4. Hukum Kekekalan Energi untuk Kalor =30 B4;07 4?4;090@8 107E0 4=4@68 B830: 30?0B 3828?B0:0= 30= B830:?4@=07CA=07B4B0?84=4@6830?0B14@C10714=BC:30@8A0BC14=BC: 4=4@68 :4 14=BC: 4=4@68 ;08= (410608 2=B7 4=4@68 4:0=8: 30?0B 14@C107 4=9038 4=4@68 ;8AB@8: !4C380= 4=4@68 ;8AB@8: 30?0B 14@C107 ;0684=90384=4@682070G030=A4B4@CA=G048:80=9C604=4@6830;0 14=BC: :0;@ 14@0A0; 30@8 14=BC: 4=4@68 ;08= Kesetaraan kalori dan joule adalah 1 kalori = 4,2 joule 1 joule = 0,24 kalori 1 kkal = 1.000 kalori

7. C:C !4:4:0;0= =4@68 C=BC: :0;@ 30?0B 3800B8 34=60= 4=6 6C=0:0= !%'#)' A4?4@B8 B0?0: ?030 .#3 (49C;07 0AA0 H0B 208@ 38?0=0A:0= !4C380= H0B 208@ B4@A41CB 380AC::0= :4 30;0 :0;@84B4@(C7CH0B208@B4@A41CB38C:C@34=60=B4@4B4@30=3820B0B (4;0=9CB=G0 A49C;07 0AA0 H0B 208@ 34=60= AC7C G0=6 ;4187 @4=307 380AC::0= :4 30;0 :0;@84B4@ 30= 3803C: 34=60= ?4=603C: 78=660 :43C0 H0B 208@ 14@20?C@ 30=4@0B0 !4C380= AC7C H0B 208@ G0=6 B4@ 20?C@ 38C:C@ 30= 3820B0B (4;00 ?4=603C:0= 14@;0=6AC=6 B4@9038 ?4@?8=3070=:0;@-0B208@G0=614@AC7C;4187B8=6684;4?0A:0=:0;@=G0 A430=6:0= H0B 208@ G0=6 14@AC7C ;4187 @4=307 4=G4@0? :0;@ 30@8 H0B 208@ G0=6 14@AC7C B8=668 A478=660 ?030 0:78@=G0 4=20?08 AC7C :4 A4B810=60= G0=6 38A41CB 34=60= ()#$ )'# #4=C@CB 041) -$, J 9C;07 :0;@ G0=6 38;4?0A ;4?0A A00 34=60= 9C;07 :0;@ G0=6 38B4@80 B4@80 (420@0 0B40B8A ?4@=G0B00= B4@A41CB 38BC;8A A410608 14@8:CB Contoh 7.1 Contoh 7.2 (410=G0:

8. 6@008@38?0=0A:0=30@8

9. I4=9038 I 8:00AA094=8A08@030;07 :0;6I0B0C :6!B4=BC:0= 0 10=G0:=G0:0;@G0=638B4@8008@B4@A41CB30;0:0;@8 1 10=G0:=G0:0;@G0=638B4@8008@B4@A41CB30;09C;4 8# 8:4B07C8#

10. 6 :0;6I IJ

11. I I 0 #

12. 6 :0;6I I :0;@8 03810=G0:=G0:0;@G0=638B4@8008@B4@A41CB030;07 :0;@8 1 0@8:4A4B0@00=:0;@830=9C;438:4B07C8107E0 :0;@8 9C;4A478=660 K 9C;4 9C;4 air 150 Mudah dan Aktif Belajar Fisika untuk Kelas X ;4?0A B4@80 # ;4?0A # B4@80 (4;0=9CB=G0 34./ 38:4=0; A410608 ((

13. ! J termometer pengaduk kalorimeter isolator (4?B=60;C8=8C14@0AA0 630=14@AC7C I380AC::0=:430;0 608@ G0=614@AC7C I4=60=4=60108:0=?4@BC:0@0=:0;@34=60=;8=6:C=60=78BC=6 AC7C0:78@20?C@0=98:0:0;@94=8A0;C8=8C :6!30=:0;@94=8A08@ :6! 8# 8:4B07C8 #08@ 6 08@ I #; 6 ; I 08@ :6! ; :6! ;4?0A B4@80 #08@08@ # 6 :6! IJ 6 :6!J I J)

14. J I I I 038AC7C0:78@A4B4;07B4@9038:4A4B810=60=B4@0;030;07I Gambar 7.1 Kalorimeter

15. Gas menyublim menyublim menguap mengembun Padat Cair mencair membeku Kalor 151 5. Perubahan Wujud Zat )4;07 =30 ?4;090@8 107E0 H0B 30?0B 14@14=BC: B860 EC9C3 G08BC ?030B 208@ 30= 60A :810B ?4=60@C7 AC7C EC9C3 H0B 30?0B 14@C107 30@8EC9C3?030B4=9038208@30=30@8208@4=903860A0B0CA410;8:=G0 -0B A4;0;C 4=4@80 0B0C 4;4?0A:0= :0;@ A4;00 ?4@C1070= EC9C3 14@;0=6AC=6B4B0?8B830:38A4@B0834=60=:4=08:0=0B0C?4=C@C=0=AC7C (410608 2=B7 :4B8:0 4A A430=6 4=208@ B4@9038 ?4@C1070= EC9C3 30@8 ?030B 4=9038 208@ 030 ?4@8AB8E0 B4@A41CB 381CBC7:0= A49C;07 :0;@ B4B0?8 B830: 38A4@B08 34=60= :4=08:0= AC7C 4;090@8 3806@0 ?030 .#3 4@C1070= EC9C3 30@8 208@ :4 60A 38A41CB #$* @A4A A410;8:=G0 30@8 60A :4 208@ 38A41CB #$#*$ @A4A ?4@C1070= EC9C3 30@8 ?030B :4 60A 38A41CB #$.*# @A4A A410;8:=G0 30@8 60A :4 ?030B 38A41CB #$.*# a. Kalor Uap dan Kalor Embun 8:0 =30 4=4B4A:0= A?8@8BCA ?030 :C;8B B0=60= 10680= :C;8B G0=6 38B4B4A8 A?8@8BCA 0:0= B4@0A0 38=68= 0; B4@A41CB 38A4101:0= :0;@ ?030 10680= :C;8B B0=60= G0=6 38B4B4A8 A?8@8BCA 38A4@0? ;47 A?8@8BCA C=BC: 4=6C0? 0=G0:=G0 :0;@ G0=6 38?4@;C:0= C=BC: 4=6C0?:0= A?8@8BCA A410=38=6 34=60= 0AA0 A?8@8BCA 30= 14@60=BC=6 ?030 :0;@ ;0B4= C0? :0;@ C0? A?8@8BCA B4@A41CB %' * 030;07 !%' .$ '*!$ *$)*! #$* ,* * ! /) '#$ *))!$.30?C=!%'#*$030;07:%'.$ ( *$)*! #$* ,* * ! * #$ ' ))! *'$. !0;@C0?AC0BCH0BA0014A0@34=60=:0;@41C=H0BB4A41CB#8A0;=G0 :0;@C0?@0:A0 :6:0;@41C=@0:A0?C= :6 8:0:0;@C0?:0;@41C=AC0BCH0B030;07C=BC:4=6C0?:0= 4=641C=:0=H0BG0=60AA0=G0#0:0=44@;C:0=:0;@A410=G0: Gambar 7.2 Diagram perubahan wujud zat #* 34=60= * 030;07 :0;@ C0? :6 J #- 4?4@;870B:0= B8B8: 38387 30= :0;@ C0? 14@10608 H0B ?030 B4:0=0= 0BA54@ 4;090@8 B014; B4@A41CB H0B 0=0:07 G0=6 4 8;8:8 :0;@ C0? B4@14A0@ Tugas Anda 7.2 Seperempat kilogram es, dengan suhu –10°C, dicampur dengan 2 kg air yang suhunya 20°C. Diskusikan dengan teman sebangku Anda, bagaimana fase akhir campuran tersebut dan berapakah suhu akhir campuran tersebut? #- )8B8:838730=!0;@*0?4@10608-0B030)4:0=0= 0B !5 -03%*%*) ,( *5*,%*%*) ; J

16. J J J

33. 4;8C 83@64= $8B@64= %:A864= ;:7; '0:A0 8@ (C;5C@ )80778B0 =B8= 4@0: 0A )41060 Sumber: Physics, 1980

34. b. Kalor Lebur dan Kalor Beku (41=6:074A38?0=0A:0=?030AC7C I30=B4:0=0=C30@0 0BA54@ 4AC;084=208@030A00B4A4=208@381CBC7:0=:0;@(C7C:4B8:04A4= 208@ 38A41CB B8B8: ;41C@ (410;8:=G0 98:0 08@ B4@A41CB =30 38=68=:0= 08@ 8BC 0:0= 414:C 4@8AB8E0 ?414:C0= 8=8 38A4@B08 34=60= ?4;4?0A0= :0;@ (C7C :4B8:0 08@ 414:C 38A41CB B8B8: 14:C 0=G0:=G0 :0;@ G0=6 381CBC7:0= 0B0C 38;4?0A:0= ;47 AC0BC H0B C=BC: 4=208@ ?030 B8B8: ;41C@=G0 0B0C 414:C ?030 B8B8: ;41C@=G0 A410=38=6 34=60= 0AA0 H0B 8BC 30= 14@60=BC=6 ?030 :0;@ ;41C@ 14:C H0B B4@A41CB %' *' 030;07 !%' .$ '*!$ *$)*! #$* ,* * ! /) ) #$ ' ))! *'$. 30?C= !%' !* 030;07 :%' .$ (!$ *$)*! #$* ,* * ! /) ' #$ ) !0;@ ;41C@ AC0BC H0B A00 14A0@ 34=60= :0;@ 14:C H0B B4@A41CB 8:0 :0;@ ;41C@ :0;@ 14:C AC0BC H0B 030;07 C=BC: 4;41C@ 414:C H0B G0=6 0AA0=G0 # ?030 B8B8: ;41C@ B8B8: 14:C 0:0= 4 4@;C:0= :0;@ A410=G0: #- )8B8:41C@30=!0;@41C@4@10608-0B?030)4:0=0= 0B c. Perubahan Wujud Es Menjadi Uap 8:0 A49C;07 0AA0 4A G0=6 AC7C=G0 38 10E07 I 38?0=0A:0= 3814@8:0=:0;@78=660AC7C=G0380B0A IA49C;070AA04AB4@A41CB 14@C107 EC9C3 A4;C@C7=G0 4=9038 C0? 4@C1070= EC9C3 4A 4=9038 C0? 30?0B 3800B8 ?030 .#3 !4B4@0=60= J (49C;07 :0;@ 3814@8:0= :4?030 4A A478=660 AC7C 4A =08: 30@8 J) I 4=9038 I(4;C@C70AA00A8714@14=BC:H0B?030BG08BC4A J (49C;07 :0;@ 3814@8:0= C=BC: 4=6C107 EC9C3 H0B A478=660 4A 4=9038 08@ 34=60= AC7C B4B0? G08BC I J (49C;07:0;@3814@8:0=C=BC:4=08::0=AC7C30@8 I78=66008@ 4=38387?030AC7C I(4;C@C70AA00A8714@14=BC:EC9C3 H0B 208@ G08BC 08@ Pembahasan Soal D E Energi Gambar 7.3 Suhu (°C) A B C Pemanasan es pada tekanan 1 atmosfer. 100 0 t1 152 Mudah dan Aktif Belajar Fisika untuk Kelas X # J #-4=C=9C::0=B8B8:;41C@30=:0;@;41C@14@10608H0B?030 B4:0=0= 0BA54@ 4;090@8 B014; B4@A41CB H0B 0=0:07 G0=6 48;8:8 B8B8: ;41C@ B4@4=307 !5 *5*,#63 ; 4;8C 83@64= $8B@64= %:A864= ;:7; '0:A0 8@ (C;5C@ )80778B0 =B8= 4@0: 0A )41060 J J

35. J J J

36. -03#63,(

55. Sumber: Physics for Scientist Engineer, (2000). Sebanyak 320 gram campuran es dan air pada suhu 0°C berada dalam bejana yang kapasitas kalornya dapat diabaikan. Kemudian, dimasukkan 79g uap air yang bersuhu 100°C ke dalam bejana tersebut. Suhu akhir menjadi 79°C. Jika kalor lebur es 79,0 kal/g dan kalor penguapan air 540 kal/g, maka banyaknya air mula-mula adalah ... gram. a. 4 d. 65 b. 10 e. 79 c. 35 SPMB, 2002 Pembahasan Diketahui: mes + air = 320 gram, T = 0°C muap = 79 gram, T = 100°C Kalor lebur es, L = 79,0 kal/gram Kalor penguapan air , L = 540 kal/gram Takhir = 79°C Kalor yang di lepas uap air (100°C) Qlepas = muap L + muap c T = (79) (540) + (79) (1) (100 – 79) = 79 (561) kalori Qterima= mes L + mes c T + mair c T = (320 – mair) 79 + (320 – mair) (1) (79 – 0) + mair (1) (79 – 0) = (640 – mair) 79 Asas black : Qlepas = Qterima 79 (561) = (640 – mair) 79 mair = 79 gram Jawaban : E

56. Kalor 153 J (49C;07 :0;@ 3814@8:0= C=BC: 4=6C107 EC9C3 H0B A478=660 H0B 208@G0=64=3838714@C1074=9038C0??030AC7CB4B0?G08BC I Contoh 7.3 830;0A41C071490=0B4@30?0B :608@14@AC7C

57. I!430;01490=0B4@A41CB 380AC::0=A4?B=64A14@AC7CJ IA410=G0: :68:4B07C8:0;@;41C@4A

58. K :6 8:0?4@BC:0@0=:0;@70=G0B4@90380=B0@008@30=4AB4=BC:0=AC7C 0:78@:4A4B810=60=B4@0; 8# 8:4B07C8 #08@ :6

59. K :6 08@

60. I 08@ :6I 4AJ I 4A :6I #8A0;:0=

61. 030;07:0;@G0=638;4?0A:0=08@30@8

62. IA0?08 I

63. #08@08@ :6 :6I

65. 9C;4 030;07:0;@G0=6381CBC7:0=C=BC:4=6C1074A14@AC7CJ I4=90384A 14@AC7C I #4A4A :6 :6I JJ 030;07:0;@G0=638A4@0?4AC=BC:4;41C@ #4A4A :6

66. K :6

67. !0;@G0=638A4@0?4A

68. )4@=G0B0:0;@G0=638A4@0?4A;418714A0@30@8?030:0;@G0=638;4?0A:0=08@4@0@B8 :0;@G0=6381CBC7:0=4AC=BC:4=208@;418714A0@30@8?030:0;@G0=638;4?0A:0=08@ A0?08 I 03870=G0A410680=4AG0=64=208@30=AC7C:4A4B810=60=B4@0; 030;07 I 4. Pemuaian Zat (C0BC14=30108:?030B208@0C?C=60AB4@38@80B0A?0@B8:4;?0@B8:4; A0=60B :428; G0=6 A4;0;C 14@64B0@ 38A41CB ;4:C; 0@0: 0=B0@;4:C; H0B ?030B A0=60B 14@34:0B0= 030 H0B 208@ 90@0: 0=B0@;4:C;=G0 060: @4=660=6 A430=6:0= ?030 60A 90@0: 0=B0@;4:C;=G0 A0=60B @4=660=6 4@70B8:0= .#3

69. 8:0 AC0BC 14=30 38?0=0A:0= ;4:C;;4:C; 8BC 14@64B0@ A40:8= 24?0B 4B0@0= 0=B0@;4:C; B4@A41CB 4=G4101:0= ;4:C;;4:C; A0;8=6 3@=6 a. Pemuaian Zat Padat 4C080= ?030 H0B ?030B 30?0B 3800B8 4;0;C8 ?4@C1070= ?0=90=6 ;C0A 30= D;C4 .6*/ 1/+/( 30=14@AC7C 38?0=0A:0= (4CB0A:0E0B;60G0=6?0=90=6=G0 A0?08AC7C0:0:0E0B;608BC0:0=4C08A478=660?0=90=6=G0 4=9038 Gambar 7.4 Partikel-partikel pada (a) zat padat (b) cair (c) gas. (a) (b) (c)

70. 4@70B8:0= .#3 0B0C amperemeter mistar termometer digital 4@30A0@:0= ,5*7*54 *4*, 38 0B0A 14@0?0 :458A84= C08 ?0=90=6 :0E0B B41060 0=38=6:0= 70A8;=G0 34=60= #-

71. 14@8:CB 8=8G0=64C0B:458A84=C08?0=90=61414@0?01070=?030AC7C I untuk Anda Information for You 154 Mudah dan Aktif Belajar Fisika untuk Kelas X J 4@B01070= ?0=90=6 :0E0B 30?0B 38B4=BC:0= 34=60= ?4@A000= A410608 14@8:CB J 030 34./ B4@A41CB 030;07 :458A84= C08 ?0=90=6 0@8 34./ 30= 34./ 38?4@;47 J J . J / 60@=30;418740708?4C080=?0=90=6;0:C:0=;07:4680B0= 14@8:CB J Gambar 7.5 Pemuaian panjang pada kawat logam. Aktivitas Fisika 7.1 Pemuaian Panjang Tujuan Percobaan Menentukan koefisien muai panjang suatu kabel konduktor Alat-Alat Percobaan 1. Kabel (bahan uji: kawat tembaga dan kawat besi) diameter 1,5 mm 2. Statif 3. Beban (massa 5–50 gram) 4. Mistar tegak (100 cm) 5. Amperemeter digital 6. Termometer digital Langkah-Langkah Percobaan 1. Susun alat seperti pada gambar berikut. 2. Ikatkan kedua ujung kabel (kawat) pada batang peyangga bagian atas. 3. Gantungkan beban bermassa 50 gram pada kawat, tepat di tengah kawat. 4. Tempatkan mistar sejajar dengan be-ban untuk mengetahui besar perubah-an yang dapat terjadi. 5. Hubungkan amperemeter, sumber tegangan dan termometer seperti pada gambar. 6. Catat awal dan T0 kabel dan posisi beban x0 dan y0. 7. Catat perubahan yang terjadi pada kawat x, y, dan T setiap perubahan tegangan yang dinaikkan. 8. Gunakan perhitungan metode grafik untuk menentukan koefisien muai panjangnya. 9. Bandingkan koefisien muai panjang antara kedua kawat tersebut, apakah sama untuk kedua bahan kawat tersebut? Menurut Anda, mengapa koefisiennya berbeda/sama? 10. Apa kesimpulan yang Anda peroleh dari percobaan tersebut? ) 20 40 60 80 0 100 sumber tegangan x 0 y 0 + – + – kabel statif beban statif Informasi Di toko peralatan elektronik yang menjual kabel (penghubung), terdapat berbagai jenis merek dagang. Untuk memperoleh kabel yang baik dapat dilakukan percobaan seperti pada Aktivitas 7.1 untuk membandingkan sifat termal antara merek yang satu dan yang lainnya. Kabel penghubung yang baik sebagai konduktor adalah kabel yang menghantarkan listrik secara baik sekalipun pada suhu yang tinggi. Ini artinya bahwa koefisien muai panjangnya relatif kecil nilainya. Dengan demikian, Anda dapat memilih untuk menggunakan kabel berdasarkan kualitasnya. In wire selling market, there is a various kind of trade mark. To obtain wire we can do an activities as in Physics Activity 7.1 to compare thermal properties between one trademark to others. Good wire as a conductor is a wire that conducting electricity good enough even in high temperature. Its mean that the linear expansion coefficient relatively small. By that, you can choose to use wire according to its quality.

72. baja roda sungai Kalor 155 !41CBC70= ?4=64B07C0= 4=64=08 :458A84= C08 ?0=90=6 AC0BC 1070= 030;07 C=BC: 4?4@78BC=6:0= ?4=66C=00= 1070= B4@A41CB #8A0;=G0?48;870=1070=30=C:C@0=G0=6386C=0:0=C=BC::=AB@C:A8 9410B0= 030A0;07A0BCC9C=6:=AB@C:A89410B0=34@=3814@8:0=@301090 G0=6 30?0B 14@?CB0@ 1410A !4B8:0 9410B0= 4C08 0:810B ?0=0A 30A0@ 9410B0= 30?0B 4=664@0::0= @30 1090 B4@A41CB 030 C9C=6 G0=6 ;08= 9C603814@8:0=24;07G0=64C=6:8=:0=30A0@9410B0=30?0B14@64@0: Gambar 7.6 Pada salah satu ujung jembatan ini, dipasang roda dan diberi celah untuk memberi ruangan ketika jembatan memuai. #-

73. !458A84=#C080=90=64@10608-0B?030(C7C I !5 0'*4*/6*/+/( ; ;C8=8C !C=8=60=30=?4@C=66C 4B=30=10BC !020180A0 !020?G@4F )807 4A8 !E0@A0 090 #0@4@ )41060 K J K J K J K J

74. K J K J K J K J K J K JJ

75. K J K J Sumber: Physics, 1980 Contoh 7.4 030AC7C I?0=90=6:0E0B14A8030;07 4@0?0:07?0=90=6:0E0B14A8 B4@A41CB?030AC7C I98:0:458A84=C08?0=90=614A8 K JI 8# 8:4B07C8 I I K JIJ . J / . K JI IJ I/ 038?0=90=6:0E0B14A8B4@A41CB?030AC7C I030;07 .6*/ 64 (4:4?8=6 ;60 G0=6 ?0=90=6=G0 - 30= ;410@=G0 . 0:0= 4=60;08 C08 ;C0A 98:0 38?0=0A:0= 4C080= ;C0A AC0BC H0B 14@60=BC=6 :4?030 :458A84= C08 ;C0A G0=6 3814@8 ;010=6 #C08 ;C0A B4@14=BC: 30@8 3C0 ?4C080= G08BC ?4@B01070= ?0=90=6 30= ?4@B01070= ;410@ :810B=G014A0@:458A84=C08;C0A A0034=60=3C0:0;8:458A84= C08 ?0=90=6 G08BC 8:0 A4:4?8=6 ;60 G0=6 ;C0A=G0 30= AC7C=G0 38?0=0A:0= A0?08 AC7C ;60 B4@A41CB 0:0= 4C08 A478=660 ;C0A=G0 4=9038 4A0@=G0?4@B01070=;C0A:4?8=6;60 B4@A41CB30?0B38BC;8A:0= 30;0 ?4@A000= 14@8:CB #8A0;=G0 ;C0A ?4@A468 J

76. J 4=60=40AC::0=70@60 AJ0:0 34./ 4=9038 J 0B0C

77. J J

78. Contoh 7.5 (4:4?8=60;C8=8C34=60=?0=90=6 230=;410@

79. 238?0=0A:0=30@8 I A0?08 I 8:0:458A84=C08?0=90=60;C8=8CB4@A41CB 030;07K I B4=BC:0=;C0A:4?8=60;C8=8CA4B4;0738?0=0A:0= 8# 8:4B07C8 2K

80. 2 2 2 K JIK JI IJ I I0:0 2. K JIK I/ 2 038;C0A:4?8=60;C8=8CA4B4;0738?0=0A:0=030;07 2 .6*/ 0-6. 4C080= D;C4 14=30 14@60=BC=6 :4?030 :458A84= C08 D;C4 G0=6 3814@8 ;010=6 4C080= D;C4 B4@14=BC: 30@8 B860 ?4C080= G08BC ?4@B01070= ?0=90=6 ?4@B01070= ;410@ 30= ?4@B01070= B8=668 :810B=G0 14A0@ :458A84= C08 D;C4 A00 34=60= B860 :0;8 :4 58A84= C08 ?0=90=6 G08BC 0 2 0 0 J 156 Mudah dan Aktif Belajar Fisika untuk Kelas X

81. J 8:0A41C0714=3014@14=BC:10;:G0=6D;C4=G030=AC7C=G0 38?0=0A:0= A0?08 AC7C ) 14=30 B4@A41CB 0:0= 4C08 A478=660 D;C4 =G04=9038 4A0@=G0 ?4@B01070= D;C4 14=30 14@14=BC: @C0=6 30?0B 38BC;8A 30;0 14=BC: ?4@A000= 14@8:CB

82. #8A0;=G0 D;C4 :C1CA

83. J 4=60=40AC::0=70@60 0 V V V 0:034./ 4=9038 Gambar 7.7 Pemuaian luas pada keping logam. Gambar 7.8 Pemuaian volume pada benda berbentuk balok. 2 0 2 2 0 0 3 2 0 2 0 2 0 0 0B0C

84. J J

85. Kata Kunci • kalor • energi • kalor jenis • kapasitas kalor • hukum kekekalan energi untuk kalor • titik didih • kalor uap • kalor embun • kalor lebur • kalor beku • koefisien muai panjang • koefisien muai volume • koefisien muai luas V olume (cm3) 0 4 8 12 16 18 Suhu (°C) Sumber: Physics, 1980 Kalor 157 Contoh 7.6 (41C0714A814@D;C4

86. 38?0=0A:0=30@8 IA0?08 I 8:00AA094=8A14A8 ?030AC7C I030;07 :6

87. 30=:458A84=C08?0=90=6=G0 K JI 78BC=6;070AA094=8A14A8?030AC7C I 8# 8:4B07C8

88. :6

90. K J

91. K JI I +;C414A8A4B4;0738?0=0A:0=030;07

93. K JI I/

95. (4B4;0738?0=0A:0=D;C414=3014@C107B4B0?80AA0=G0B4B0? 7.200 kg 1,033 m 3 m V :6

96. 0380AA094=8A14A84=9038:6

97. b. Pemuaian Zat Cair 030CC=G0?4C080=H0B208@70=G030?0B3800B84;0;C8?4@C107 0= D;C4=G0 8:0 A41C07 1490=0 64;0A G0=6 14@8A8 08@ 70?8@ ?4=C7 38?0=0A:0= A4B4;07 :4=08:0= AC7C 08@ 0:0= BC?07 4@8AB8E0 B4@A41CB 30?0B 38B4@0=6:0= A410608 14@8:CB 'C0=6 64@0: ?0@B8:4; ?030 H0B 208@ ;4187 14A0@ 30@8?030 @C0=6 64@0: ?0@B8:4; ?030 H0B ?030B 8:0 :43C0 H0B 8BC 4=60;08 ?40=0A0= A420@0 14@A000= ?0@B8:4; ?030 H0B 208@ ;4187 ;4;C0A0 14@64@0: 3810=38=6:0= 34=60= ?0@B8:4; H0B ?030B %;47 :0@4=0 8BC D;C4 08@ ;4187 24?0B 14@ B010730@8?030D;C41490=064;0AA478=66008@0:0=BC?074@8AB8E0 B4@A41CB 4=C=9C::0= 107E0 :458A84= C08 D;C4 H0B 208@ ;4187 14A0@ 30@8?030 :458A84= C08 D;C4 H0B ?030B c. Anomali Air )4;07 =30 :4B07C8 107E0 98:0 H0B ?030B H0B 208@ 30= 60A 38?0=0A:0= H0BH0B B4@A41CB 0:0= 4C08 A48@8=6 34=60= :4=08:0= AC7C 030 ?40=0A0= 08@ 030 70; 4=0@8: G0=6 30?0B 3800B8 #8A0;:0= =30 40=0A:0= 08@ 14@AC7C I A410=G0: 2

98. 8 0=B0@0 AC7C I 30= ID;C408@0:0=4=GCACBB4B0?80AA0=G0B4B0?A478=6600AA0 94=8A=G0 =08: 8:0 ?40=0A0= 38B4@CA:0= A0?08 38 0B0A I D;C4 08@ 0:0= 4C08 A4?4@B8 H0BH0B G0=6 ;08= (850B ?4C080= 08@ G0=6 B830: B4@0BC@ 8=8 38A41CB $%# ' A4?4@B8 38?4@;870B:0= 6@058: ?030 .#3 0@8 6@058: B4@A41CB B0?0: 107E0 30@8 I A0?08 I D;C4 08@ B4@CA4=GCACBA0?08:C@0=630@8 2

99. !4C380=380B0AAC7C I D;C4 08@ 4C08 *=BC: 0AA0 G0=6 A00 ?030 AC7C I D;C4 4A ;4187 14A0@ 30@8?030D;C408@4@0@B80AA094=8A4A;4187:428;30@8?0300AA0 94=8A 08@ :0@4=0 0AA0 94=8A 14@10=38=6 B4@10;8: 34=60= D;C4 # 1.0016 1.0014 1.0012 1.0010 1.0008 1.0006 1.0004 1.0002 1.0000 BC;07 A4101=G0 1=6:070= 4A 30?0B 4=60?C=6 38 0B0A 08@ Gambar 7.9 Grafik peristiwa anomali air

100. d. Pemuaian Gas 030 $264 )3-4 4;0:C:0= ?4@2100= G0=6 4=C=9C::0= 107E0 A4C0 60A 4C08 34=60= :458A84= C08 G0=6 A00 G08BC A4 14A0@

101. #4=C@CB=G0 98:0 A4C0 60A 38?0=0A:0= D;C4 30= B4:0=0==G0 14@C107 .6*/ (4 1% 5,// 551 4@70B8:0= .#3 (41C07 B01C=6 :0;4=6 BCBC?=G0 3814@8 ?8?0 :428; G0=6 C9C=6=G0 38?0A0=68 10;= :4?4A 8:0 B01C=6 B4@A41CB 38?0=0A:0= AC7C 60A G0=6 14@030 38 30;0 B01C=6 0:0= =08: 030 :4 0300= 8BC ?0@B8:4; 60A 14@64@0: A0;8=6 14@34A0:0= :4 A460;0 0@07 30= 0AC: :4 30;0 10;= 4;0;C8 ?8?0 A478=660 10;= 414A0@ 4@8AB8E0 B4@A41CB 4=C=9C::0= B4;07 B4@9038 ?4C080= 60A 38 30;0 B01C=6 ?030 B4:0=0= B4B0? A00 34=60= B4:0=0= C30@0 38 ;C0@ B01C=6 @058: AC7C B4@7030? D;C4 ?030 ?@A4A ?40=0A0= 60A 34=60= B4:0=0= B4B0? B0?0: ?030 .#3 0@86010@B4@A41CBB0?0:107E098:0B4:0=0=60AB4B0? D;C4 60A 14@10=38=6 ;C@CA 34=60= AC7C=G0 (4;0=9CB=G0 ?4@=G0B00= 8=8 38A4 1CB C:C 70@;4A (420@0 0B40B8A 7C:C 8=8 38BC;8A pipa balon tabung kaleng balon sebelum memuai bunsen 0 100 200 300 400 500 pipa sumbat tabung 158 Mudah dan Aktif Belajar Fisika untuk Kelas X 0B0C J !4B4@0=60= D;C4 0E0;

102. D;C4 60A A4B4;07 38?0=0A:0=

103. AC7C 0E0; ! AC7C 60A A4B4;07 38?0=0A:0= ! 4A0@=G0 ?4@B01070= D;C4 60A A00B 38?0=0A:0= 44=C78 ?4@ A000=G0=6A0034=60=14A0@=G0?4@B01070=D;C4H0B?030BG08BC ) J %;47:0@4=0 30= 1 273 0:0 )

104. ./4/ (4 1% 70-6. 551 J 4@70B8:0= .#3 030 10680= C9C=6 ?8?0 38BCBC? @0?0B 34=60= AC10B ?0@0;= 8:0 B01C=6 ?030 .#3 38?0=0A:0= D;C4 60A B4B0? :810B=G0 B4:0=0= 60A 14@B0107 14A0@ 038 ?030 ?40=0A0= 8=8 D;C4 60A B4B0? A430=6:0= B4:0=0==G0 14@C107 =30 30?0B 4=64B07C8 7C1C=60= 0=B0@0 :4=08:0= AC7C 30= B4 :0=0= ?030 ?40=0A0= 60A ?030 D;C4 B4B0? 34=60= 4;0:C:0= ?4@2100= 14@8:CB Gambar 7.10 Gas dipanaskan pada tekanan tetap Gambar 7.11 Volume gas sebagai fungsi dari suhu pada pemanasan gas dengan tekanan tetap. V T(K) Gambar 7.12 Gas dipanaskan pada volume tetap

105. termometer Kalor 159 Aktivitas Fisika 7.2 Hubungan Suhu dan Tekanan Tujuan Percobaan Mengetahui hubungan antara kenaikan suhu dan tekanan. Alat-Alat Percobaan Alat ukur Bourdon, termometer raksa, tabung gas, pembakar bunsen, dan kaki tiga. Langkah-Langkah Percobaan 1. Rancanglah peralatan tersebut seperti pada Gambar 7.13. Pastikan tidak ada kebocoran gas dalam tabung. 2. Setelah terjadi kesetimbangan termal antara termometer dan gas dalam tabung, catat suhu awal gas (terlihat pada termometer) dan tekanan udara (terlihat pada alat Bourdon). 3. Panaskan tabung dengan pembakar bunsen. Catat kenaikan suhu udara dan tekanan udara dalam tabung pada suatu tabel. 4. Dari tabel tersebut, buatlah grafik suhu terhadap tekanan gas. 5. Buatlah kesimpulan dari kegiatan ini. Bourdon Gambar 7.13 Perangkat percobaan untuk mengetahui hubungan antara kenaikan suhu dan tekanan. raksa bunsen kaki tiga 4@2100= A4@C?0 34=60= :4680B0= B4@A41CB ?4@=07 38;0:C:0= ?0@0 07;8 8A8:0 0@8 70A8; ?4@2100= G0=6 4@4:0 ;0:C:0= 38?4@;47 6@058: B4:0=0=60AB4@7030?AC7C?030D;C4B4B0?14@14=BC:60@8A;C@CA;870B :410;8 .#3

106. (420@0 0B40B8:0 60@8A 8=8 44=C78 ?4@A000= 0B0C !4B4@0=60= B4:0=0= 60A 0B AC7C 60A ! :=AB0=B0 J Gambar 7.14 0@8 ?4@A000= 38:4B07C8 107E0 B4:0=0= 60A 14@10=38=6 ;C@CA 34=60= AC7C=G0 ?030 D;C4 B4B0? (4;0=9CB=G0 ?4@=G0B00= B4@A41CB 38A41CB C:C 0GCAA02 Grafik tekanan terhadap suhu T(K) p(Pa) Tes Kompetensi Subbab A 3+,/-)%-.#6,6-5*)/ 8BC=610=G0::0;@G0=638?4@;C:0=C=BC:4=6C107 0 :64A30@8 I4=903808@ I 1 64A30@8 I4=9038C0? I (41C0764;0A14@8A8 608@?030 I4@0?06@04A G0=6 AC7C=G0 J I 70@CA 3820?C@ :4 30;0 08@ A478=660AC7C0:78@20?C@0=4A30=08@4=9038

107. I 38:4B07C8 :0;@ 94=8A 08@ :6 ! 108:0= ?4@BC:0@0=:0;@34=60=64;0A

108. 0301490=0G0=614@8A8 608@14@AC7C I380AC: :0= 4A 10;: G0=6 0AA0=G0 6 14@AC7C J I 8BC=6;07AC7C0:78@20?C@0=30=0AA04AG0=64;41C@ !4B8=6680=08@B4@9C=38:4B07C8 8:0A4;C@C74=4@68 4:0=8:38C1074=90384=4@68:0;@:0;@94=8A08@ :6!30= A78BC=6;07:4=08:0=AC7C 08@B4@9C=A4B4;0790BC7 4BC=9C: 6C=0:0= ?4@A000= 4=4@68 ?B4=A80; 4=4@68:0;@C=BC:4=08::0=AC7C 4@0?0 10=G0:=G0 :0;@ G0=6 38?4@;C:0= C=BC: 4=6C1074AJG0=60AA0=G0 6@04=9038 08@ 98:0:0;@94=8A4A :0;6I:0;@94=8A08@ :0;6@I30=:0;@;41C@4A :0;64@70B8:0= 6010@14@8:CB –5°C 0°C 0°C 20°C

109. B. Perpindahan Kalor 8:08=68=40A0:08@38?4@;C:0=:0;@G0=614@0A0;30@80?8:?@ 030?40=0A0=A4?0=2808@C;0C;0:0;@30@80?8:?@14@?8=307 :4?0=28B4?0B08@:4C380=:0;@30@8?0=28B4@A41CB14@?8=307:408@ 4@?8=3070=:0;@B4@903830@814=3014@AC7CB8=668?0=0A:414=30 14@AC7C ;4187 @4=307 38=68= 4=30 G0=6 ?0=0A 414@8:0= :0;@ :4 ?030 14=30 G0=6 38=68= (40=308=G0 :0;@ B830: 30?0B 14@?8=307 0:0= AC;8BC=BC:40A0:08@A4101:0;@30@80?8:?@B830:30?0B14@?8=307 :4 08@ 30 B860 14=BC: ?4@?8=3070= :0;@ G08BC !%$*!( 70=B0@0= !%$+!( 0;8@0= 30= '( ?0=20@0= 1. Perpindahan Kalor Secara Konduksi #8A0;:0= =30 40=0A:0= A0;07 A0BC C9C=6 10B0=6 ;60 A4?4@B8 .#3 )4=BC=G0?0@B8:4;?0@B8:4;?030C9C=6;60G0=638?0=0A8 14@64B0@ ;4187 24?0B (40:8= 14A0@ 9C;07 :0;@ G0=6 3814@8:0= :4?030 C9C=6 ;60 8BC A40:8= 24?0B 64B0@0= ?0@B8:4;=G0 (410680= 4=4@68 :8=4B8: G0=6 388;8:8 ?0@B8:4; G0=6 14@64B0@ B4@A41CB 3814@8:0= :4?030 ?0@B8:4;?0@B8:4; 38 34:0B=G0 4;0;C8 BC1C:0= :810B=G0 ?0@B8:4; ?0@B8:4; G0=6 38BC1C: 8BC 8:CB 14@64B0@ 348:80= A4B4@CA=G0 78=660 64B0@0= ?0@B8:4; A0?08 :4 C9C=6 ;60 G0=6 B830: 38?0=0A8 4@010B0= 64B0@0= ?0@B8:4; B4@A41CB 38A4@B08 34=60= ?4@010B0= :0;@ 30@8 C9C=6 ;60 G0=6 38?0=0A8 A0?08 :4 C9C=6 ;60 G0=6 B830: 38?0=0A8 :810B=G0 C9C=6 ;60 G0=6 B830: 38?0=0A8 4=9038 ?0=0A '010B0= :0;@ 8=8 B830: 38A4@B08 34=60= ?4@?8=3070= ?0@B8:4;?0@B8:4; ;60'$$!%'#*/)'$)'%#$$)!(') '$$ ')!')! /) )'(*) (' '#$$ 38A41CB $)'$ 0B0C !%$*!( .#3 4?4@;870B:0= A410B0=6 ;60 34=60= ;C0A ?4=0?0=6 30= ?0=90=6 10B0=6 38?0=0A:0= A0;07 A0BC C9C=6=G0 030 ?4@8AB8E0 B4@A41CB :0;@ 0:0= 4@010B :4 C9C=6 A414;07 :0=0= G0=6 AC7C=G0 ;4187 @4=307 4@30A0@:0= ?4=4;8B80= G0=6 B4;07 38;0:C:0= 38?4@;47 107E0 ?4@?8=3070= :0;@ A420@0 :=3C:A8 14@60=BC=6 ?030 94=8A ;60 ;C0A ?4=0?0=6 ?4=670=B0@ :0;@ ?4@14300= AC7C 30@8 C9C=6C9C=6 ;60B4?0B:0;@4@010BA4@B0?0=90=690;0=G0=638;0;C8;47:0;@ 4@?8=3070= :0;@ A4B80? A0BC0= E0:BC 38@CCA:0= A410608 14@8:CB Gambar 7.15 Contoh perpindahan kalor secara konduksi. T2 A Gambar 7.16 Hantaran kalor T1 160 Mudah dan Aktif Belajar Fisika untuk Kelas X ! J J

110. ! J

111. !4B4@0=60= :0;@ G0=6 4@010B ?4@A0BC0= E0:BC A 0B0C E0BB ! :458A84= :=3C:A8 B4@0; H0B A! 0B0C,! ;C0A ?4=0?0=6 10B0=6 ?0=90=6 10B0=6

112. Tantangan untuk Anda Kalor 161 #- 30= #- 4?4@;870B:0= :458A84= :=3C:A8 B4@0; H0B 1C:0= ;60 30= ;60 #- !458A84=!=3C:A8)4@0;-0B C:0=60'0B0'0B0 .)/ .4; A 0B04@07 4B= !020 !0GC:4@8=6 01CA (BG@50 !08=B410; *30@0 #- !458A84=!=3C:A8)4@0;60 '0B0'0B0 .)/ .4; 4@0: )41060

113. ;C8=8C !C=8=60= )810;

114. 090 '0:A0

115. 2. Perpindahan Kalor Secara Konveksi 030 A00B =30 40=0A:0= 08@ 38 30;0 ?0=28 :0;@ 0:0= 14@?8=307 30@8 30A0@ ?0=28 :4 ?4@C:00= 08@ A420@0 :=D4:A8 4@?8=3070= :0;@ A420@0 :=D4:A8 38A4@B08 64@0:0= 0AA0 0B0C 64@0:0= ?0@B8:4;?0@B8:4; H0B ?4@0=B0@0=G0 !=D4:A8 70=G0 B4@9038 ?030 H0B G0=6 30?0B 4=60;8@ 5;C830 30 3C0 20@0 ?4@?8=3070= :0;@ 4;0;C8 70=B0@0= :=D4:A8 G08BC :=D4:A8 A420@0 0;0807 30= :=D4:A8 ?0:A0 a. Konveksi Alamiah #8A0;:0==3040=0A:0=A4?0=2808@A4?4@B8?030.#3 (4B4;07 08@ 38 10680= 10E07 ?0=28 4=4@80 :0;@ 08@ B4@A41CB 0:0= 4C08 A478=660 0AA0 94=8A=G0 ;4187 :428; 30@8?030 0AA0 94=8A 08@ 38 10680= 0B0A 4@14300= 0AA0 94=8A B4@A41CB 4=60:810B:0= ?0@B8:4; ?0@B8:4; 08@ G0=6 14@0AA0 94=8A ;4187 :428; 0:0= 14@64@0: :4 0B0A )4?0BG0=638B8=660;:0=?0@B8:4;08@G0=614@0AA094=8A;4187:428; 0:0= B4@8A8 ;47 ?0@B8:4; 08@ G0=6 14@0AA0 94=8A ;4187 14A0@ 4@8AB8E0 B4@A41CB 14@;0=6AC=6 B4@CA4=4@CA A478=660 ?0@B8:4;?0@B8:4; 08@ 30;0 ?0=28 14@?CB0@ =08: 30= BC@C= ;8@0=0;8@0= ?0@B8:4; G0=6 14@64@0: B4@A41CB 38A4@B08 34=60= ?4@?8=3070= :0;@ 4@?8=3070= :0;@ 34=60= 4=60;8@:0= ?0@B8:4;?0@B8:4; 08@ A4?4@B8 8=8 38A41CB !%$+!( # !=D4:A8 0;0807 10=G0: 389C?08 38 ?01@8:?01@8: G0=6 4=6 6C=0:0=24@1=60A0?0A70A8;?410:0@0=48;8:80AA094=8A;4187 :428; 30@8?030 0AA0 94=8A C30@0 38 A4:8B0@=G0 :810B=G0 60A 70A8; ?410:0@0= 0:0= 4=60;8@ :4 0B0A )4?0B G0=6 38B8=660;:0= ;47 60A 70A8;?410:0@0=0:0=388A8;47C30@0A4:8B0@G0=648;8:80AA094=8A ;4187 14A0@ 30@8?030 0AA0 94=8A 60A 70A8; ?410:0@0= =68= ;0CB 30= 0=68= 30@0B B4@9038 14@30A0@:0= :=D4:A8 0;0807 C30@0 =68= ;0CB 30= 0=68= 30@0B G0=6 380=500B:0= =4;0G0= C=BC: 14@;0G0@B4@90384;0;C8:=D4:A80;0807C30@0380=0?0=0A38?8=307 :0= 30@8 AC0BC B4?0B :4 B4?0B ;08= 34=60= ?4@64@0:0= ?0@B8:4; G0=6 Berapa kalor yang diperlukan untuk mengubah 500 g es dari –10°C menjadi uap bersuhu 120°C (kalor lebur, dan kalor uap dapat dilihat pada Tabel 7.2 dan Tabel 7.3). Gambar 7.17 Peristiwa konveksi alamiah Sumber: Fisika Universitas, 2002 Sumber: Fisika Universitas, 2002

116. 38?8=307:0= 030 A80=6 70@8 30@0B0= ;4187 24?0B ?0=0A 30@8?030 ;0CB A478=660 C30@0 ?0=0A 38 0B0A 30@0B0= =08: 30= B4?0B=G0 3860=B8:0= C30@038=68=30@80B0A;0CB8=8G0=638A41CB0=68=;0CB0300;070@8 B4@9038 A410;8:=G0 0@0B0= ;4187 24?0B 38=68= 30@8?030 ;0CB A478=660 C30@0 38 0B0A ;0CB =08: 30= B4?0B=G0 3860=B8:0= ;47 C30@0 30@8 0B0A 30@0B0= G0=6 38A41CB 0=68= 30@0B b. Konveksi Paksa !=D4:A8 ?0:A0 10=G0: 386C=0:0= ?030 A8AB4 ?4=38=68= 4A8= 8A0;=G0 ?030 4A8= 18;4A8= :0?0; ;0CB4A8= 384A4; AB0A8=4@ 30= :8?0A 0=68= !=D4:A8 ?0:A0 A4?4@B8 ?030 .#3 38?0:08 30;0 A8AB4 ?4=38=68=4A8=18;8@4=60;8@38A4:8B0@@C0=64A8=4;0;C8?8?0 ?8?0 3810=BC ;47 A41C07 ??0 08@ ,)' *# !0;@ G0=6 38B4@80 4A8= 18; 30@8 70A8; ?@A4A ?410:0@0= 4=20?08 AC7C I 030 AC7C 8=8 4C=6:8=:0= 4A8= 18; 4C08 4;41878 10B0A :400=0= 30=4=60:810B:0=10680=10680=4A8=18;4=9038;407!4@CA0:0= ?4@B00 G0=6 A4@8=6 389C?08 030;07 ?030 :? A8;8=34@ 4A8= 4=9038 4;4=6:C=6 4=60@C7 14@8:CB=G0 D8A:A8B0A 8=G0: ?4;C0A 4=9038 @4=307 4=24@ 0=0A ?030 4A8= 18; 14@?8=307 ;47 A8@:C;0A8 08@ 4=C9C :4 @0 380B@ *30@0 38=68= 30@8 ;C0@ 4A8= 38B0@8: ;47 A41C07 :8?0A C=BC: 4=38=68=:0= 08@ ?030 @0380B@ A478=660 08@ G0=6 38=68= 8=8 :410;8 4=60;8@ 30= 14@A4=BC70= 34=60= 1;:1;: 4A8= C=BC: 4=6C;0=6 A8@:C;0A8 14@8:CB=G0 038 5C=6A8 @0380B@ 030;07 4=9060 AC7C 4A8= 060@ B830: 4;0?0C8 10B0A ?0=0A G0=6 388H8=:0= C;07 4=4@68 :0;@ ?4@ A0BC0= E0:BC G0=6 38B4@80 ;47 5;C830 A4:8B0@=G0 A420@0 :=D4:A8 030;07 A410=38=6 34=60= ;C0A ?4@C:00= 14=30 G0=6 14@A4=BC70= 34=60= 5;C830 34=60= 1430 AC7C t (420@0 0B40B8A 38BC;8A Contoh 7.7 Tantangan untuk Anda 162 Mudah dan Aktif Belajar Fisika untuk Kelas X Q t 0B0C J

117. !4B4@0=60= 9C;07 :0;@ ?4@ A0BC0= E0:BC A :458A84= :=D4:A8 B4@0; AI ;C0A ?4=0?0=6 ?4@?8=3070= :0;@ t ?4@14300= AC7C G0=638?0=0A834=60= AC7C 5;C830 I J $8;08 :458A84= :=D4:A8 B4@0; 14@60=BC=6 ?030 14=BC: 30= :43C3C: 0= ?4@C:00= 08@ 94=8A 5;C830 G0=6 14@A4=BC70= 34=60= ?4@C:00= (C7CC30@030;0A41C07@C0=60=A414A0@ IA430=6:0=AC7C?4@C:00=94=34;0 ?030@C0=60=B4@A41CB

118. I4@0?0;09C:0;@G0=638B4@80;4794=34;0:020A4;C0A 98:0:458A84=:=D4:A8C30@0A00B8BCK J :0;AI pipa-pipa kecil kipas pompa air saluran air dalam mesin Gambar 7.18 Sistem peredaran pendingin air pada mobil. Misalkan Anda menyeduh secangkir kopi dengan air panas, lalu Anda mengaduknya dengan sendok yang terbuat dari logam. Mengapa sendok itu terasa lebih panas dibandingkan jika Anda mengaduknya menggunakan sendok yang terbuat dari plastik?

119. untuk Anda Information for You Kalor 163 8# )J)

120. IJ I I K J :0;AI K J :0;AI I :0; 038;09C:0;@G0=638B4@80;4794=34;0:020 :0; 3. Perpindahan Kalor secara Radiasi 0=0A0B070@84@C?0:0=AC14@4=4@68B4@14A0@1068:4;0=6AC=60= 783C?0:7;C:381C80=0A30@80B070@8A0?08:4C8B830:4;0;C8 70=B0@0= :=3C:A8 0B0C?C= 0;8@0= :=D4:A8 A4101 :=3C:A8 30= :=D4:A8 44@;C:0= H0B ?4@0=B0@0 A430=6:0= 0=B0@0 #0B070@8 30= C8B4@30?0B@C0=670?00B0C34=60=:0B0;08=B830:030H0B?4@0=B0@0 0=0A :0;@ 30@8 #0B070@8 A0?08 :4 C8 B4@9038 A420@0 @0380A8 ?0=20@0=!0;@30@8#0B070@8A0?08:4C830;014=BC:64;10=6 4;4:B@06=4B8: 038 '( 030;07 '$$ !%' # $)*! %#$ !)'%#$)! 4@C:00= 14=30 G0=6 14@E0@=0 78B0 30?0B 4=G4@0? 30= 4 0=20@:0=4=4@68:0;@@0380A834=60=108:A430=6:0=?4@C:00=14=30 14@E0@=0?CB874=G4@0?30=40=20@:0=:0;@@0380A834=60=1C@C: 0=4; AC@G0 (%' $ 386C=0:0= C=BC: 4=G4@0? 30= 40=20@:0= @0380A8A8=0@#0B070@8%;47:0@4=08BC1830=6;6014@=660=G03814@8 E0@=0 78B0 =4@68 :0;@ @0380A8 380=500B:0= C=BC: 40=0A:0= 08@ #18;18; B0=6:8 ?4=60=6:CB 8=G0: ?030 10680= 0B0A B0=6:8 3820B 34=60=E0@=0?CB870;B4@A41CB380:AC3:0=6C=04=678=30@8?4=G4 @0?0= 4=4@68 ?0=0A A420@0 :=D4:A8 ;47 8=G0: 4@C:00= 14=30 78B0 ;4187 10=G0: 4=G4@0? 30= 40=20@:0= 4=4@68 :0;@ A410;8:=G0 ?4@C:00= 14=30 14@E0@=0 ?CB87 ;4187 A438:8B 4=G4@0? 30= 40=20@:0= :0;@ @0380A8 a. Api Unggun 4@=07:07 =30 4;0:C:0= #$ 38 304@07 ?46C=C=60= (C307 4=9038 70; G0=6 180A0 38 30;0 ?4@:4070= 41C0B 0?8 C=66C= ?8 C=66C= 381C0B 30@8 @0=B8=6@0=B8=6 ?7= :4@8=6 G0=6 3810:0@ (4B4;07 38 1C0B 0?8 C=66C= B4=BC =30 4=9038 4@0A0 70=60B E0;0C?C= BC1C7 =30 B830: 14@A4=BC70= 34=60= 0?8 0=0A 0?8 4=60;8@ 4;0;C8 C30@0 ;4187 10=G0: A420@0 @0380A8 30@8 0?8 C=66C= 4=64=08 BC1C7 =30 30@8 ?030 A420@0 :=D4:A8 b. Rumah Kaca )830:A4C064;10=6@0380A8A8=0@#0B070@830?0B4@010B0AC: :430;0@C07:0200=G02070G0B0?0:G0=630?0B4@010B38=38=6 :020 0B0C ?;0AB8: A430=6:0= A8=0@ C;B@0D8;4B 30= A8=0@ 8=5@0 4@07 38 ?0=BC;:0=:410;8;4738=38=6:020=4@68:0;@@0380A830@82070G0B0?0: 38A4@0? ;47 B0=07 30= B0=00= 38 30;0 @C0=6 @C07 :020 (410;8:=G0 B0=07 30= B0=00= 0:0= 40=20@:0= :410;8 64;10=6 @0380A8 14@C?0 A8=0@ 8=5@04@07 0=90=6 64;10=6 G0=6 ;4187 14A0@ 4=G4101:0= 64;10=68=5@04@07B4@?4@0=6:0?;4738=38=6:020A478=660AC7C@C0=60= 4=9038 ;4187 70=60B 30= B0=00= 30?0B 783C? 34=60= A460@ Informasi Alat Penukar Kalor Seperti namanya, alat penukar kalor adalah seperangkat instrumen di mana terjadi pertukaran kalor antara dua aliran fluida bergerak tanpa pencampur-an. Alat penukar kalor banyak digunakan diberbagai industri dengan berbagai model. Bentuk paling sederhana dari alat penukar kalor adalah penukar kalor pipa ganda, yakni tersusun oleh dua pipa konsentris dengan diameter berbeda. Satu fluida mengalir di dalam pipa, dan fluida lainnya mengalir pada pipa yang menembus ruang antara pipa. Kalor dipindahkan dari fluida yang panas ke fluida yang dingin melalui dinding pemisahnya. Terkadang pipa yang berada di dalam dibuat dua putaran di dalam selongsong untuk menambah pertukaran kalor. As the name implies, heat exchangers are devices where two moving fluid streams exchange heat without mixing. Heat exchanger are widely used in various industries, and they come in numerous designs. The simplest form of a heat exchanger is a double tube heat exchanger. It is composed of two concentric pipes of different diameters. One fluids flows in the inner pipes, and the other in the annular space between the two pipes. Heat is transferred from the hot fluid to the cold one through the wall separating them. Sometimes the inner tube makes a couple of turn inside the shell to increase the heat transfer area. Sumber: Thermodynamics, 1998

121. *0? 08@ G0=6 38:4;C0@:0= ;47 B0=07 30= B0=00= 4=G4101:0= ?4=6C0?0= B4@?4@0=6:0? ;47 38=38=6 :020 ;8@0= C0? 08@ :4 0B0A A420@0 :=D4:A8 B830: 14@90;0= A478=660 C0? 08@ :410;8 90BC7 :4 30;0 B0=07 30= B0=00= 30;0 14=BC: H0B 208@ 030 0:78@=G0 :4;41010= C30@0 30;0 @C07 :020 30?0B B4@9060 A4?0=90=6 @0380A8 :0;@ #0B070@8 14@ ;0=6AC=6 c. Radiasi Benda Hitam 041) 5'/

122. J

123. 30= 6%8*( 0-5:.// J 4=G0B0:0= 107E0 14A0@=G0 4=4@68 G0=6 38?0=20@:0= ;47 AC0BC ?4@C:00= ?4@ A0BC0= E0:BC ?4@ A0BC0= ;C0A A410=38=6 34=60= ?0=6:0B 4?0B AC7C ?4@C:00= 8BC 30?0B 38BC;8A 34=60= ?4@A000= 4=4@0?0= :=A4? @0380A8 30;0 :4783C?0= A470@870@8 10=G0: 38 9C?082=B7=G0?410:0@0=?0300;0B?40=660=6%+$?4=64@8=60= :?8 30= B410:0C 30= A8AB4 ?4=38=68= 0B0C ?40=0A0= @C07 Contoh 7.8 Tantangan untuk Anda 164 Mudah dan Aktif Belajar Fisika untuk Kelas X J !4B4@0=60= 4=4@68 G0=6 38?0=20@:0= 0B0C 38A4@0? ?4@ A0BC0= E0:BC ?4@ A0BC0= ;C0A A 0B0C E0BB :=AB0=B0CC(B450=;BH0==K JE0BB! AC7C CB;0: ! 48A8D8B0A ?4@C:00= B830: 14@A0BC0= *=BC: 14=30 78B0 A4?C@=0 70@60 14=3014=30 ;08= 70@60 :458A84=48A8D8B0A=G0;4187:428;30@8?030A0BCA430=6:0=C=BC:14=30 14@E0@=0 ?CB87 A4?C@=0 0@60 14@60=BC=6 ?030 :40300= ?4@ C:00= 14=30 G08BC :4:0A0@0==G0 A4@B0 E0@=0 30@8 14=30 =4@68 G0=6 38?0=20@:0= ;47 A41C07 14=30 30;0 A0BC0= 9C;4 38B4=BC:0= 34=60= ?4@A000= ) J !4B4@0=60= 4=4@68 G0=6 38?0=20@:0= ;47 ?4@C:00= 14=30 ;C0A ?4@C:000= 14=30 ) ;00 E0:BC 48A8 4=4@68 A4:= 8:03810=38=6:0=B4@7030?;8=6:C=60=G0=614@AC7C ?4@A000= =G0 0:0= 4=9038 J J ) (41C0714=3048;8:8?4@C:00=78B0A4?C@=014@AC7C IC0A?4@C:00=

124. 240=20@:0=4=4@68:4;8=6:C=60=G0=614@AC7CI)4=BC:0=4=4@68?4@ A0BC0=E0:BCG0=638?0=20@:0=14=30B4@A41CB 8# 14=3078B0

125. !

126. !

127. 2

128. K J Seorang petinju profesional setelah selesai bertanding berada di ruangan yang bersuhu 15°C. Berapa laju energi yang dikeluarkan tubuhnya jika suhu tubuhnya saat itu 34°C, emisivitasnya 0,7, dan luas tubuhnya yang berhubungan langsung dengan udara adalah 1,5 m2?

129. Kalor 165 K JE0BB! ) K J. J

130. /

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136. A Tes Kompetensi Subbab B 3+,/-)%-.#6,6-5*)/ )47?0=0A14@AC7CI38BC0=6:0=:430;03C0 1C07B4:G0=6A0BCB4::4@08:E0@=078B0 30=A0BC=G0;068B4:4=68;0B

137. 8:0AC7C@C0=6I14@0?0:424?0B0=78;0=6=G0 :0;@ 30@8 A4B80? B4: :4 ;8=6:C=60= 98:0 ;C0A ?4@C:00=B4:030;07K J 4=603C: B4@1C0B 30@8 0;C8=8C 34=60= ;C0A ?4=0?0=630=?0=90=6 2 8:0AC7C08@ 30;01490=0 IAC7C38C9C=6?4=603C:I 30=:=3C:B8D8B0AB4@0; ,IB4=BC:0= 0 :424?0B0=0;8@0=:0;@?0300;8@0=:0;@?030 ?4=603C: 1 :0;@G0=64=60;8@4;0;C8?4=603C:A4;00 90 !?8 ?0=0A 14@AC7C I 38BC0=6:0= :4 30;0 20=6:8@:4@08:E0@=02:4;0B 8:0AC7C @C0=6IB4=BC:0=:424?0B0=78;0=6=G0:0;@30@8 20=6:8@:4;8=6:C=60=;C0A20=6:8@ 2 A0A;02:4=G0B0:0=107E09C;07:0;@ G0=6 38;4?0A ?030 AC0BC A8AB4 0:0= A00 34=60= 9C;07 :0;@ G0=6 38B4@80 4C080=?030H0B?030B30?0B3800B84 ;0;C8 ?4@C1070= ?0=90=6 ;C0A 30= D;C4 030 CC=G0 ?4C080= H0B 208@ 70=G0 30?0B 3800B8 4;0;C8 ?4@C1070= D;C4 A090 Rangkuman !0;@ 030;07 4=4@68 G0=6 38?8=307:0= 30@8 AC0BC 14=30 :4 14=30 G0=6 ;08= !0;@ 94=8A 30?0B 383458=8A8:0= A410608 10=G0:=G0 :0;@ G0=6 38?4@;C:0= AC0BC H0B C=BC: 4=08::0= AC7C :6 H0B B4@A41CB A414A0@ I

138. !0?0A8B0A :0;@ 030;07 10=G0:=G0 G0=6 38?4@;C:0= C=BC: 4=08::0= AC7C 14=30 A414A0@ I Kata Kunci • konduksi • konveksi • radiasi • rumah kaca • radiasi benda hitam

139. !0;@ 030;07 4=4@68 G0=6 38?8=307:0= 30@8 AC0BC 14=30 :4 14=30 G0=6 ;08= 30 B860 14=BC: ?4@?8=3070= :0;@ G08BC :=3C:A8 :=D4:A8 30= @0380A8 4@?8=3070= :0;@ 4;0;C8 H0B ?4@0=B0@0 34=60=B830:38A4@B08?4@?8=3070=?0@B8:4; Peta Konsep #4;41C@ #414:C #4=6C0? #4=641C= #4=GC1;8 2=B7 ?4@8AB8E0=G0 4@C1070= ,C9C3 Refleksi 166 Mudah dan Aktif Belajar Fisika untuk Kelas X ?0@B8:4;H0BB4@A41CBA020@0?4@0=4=38A41CB 70=B0@0= 0B0C :=3C:A8 4@?8=3070= :0;@ A420@0 :=D4:A8 0;8@0= 38A4@B08 64@0:0= 0AA0 0B0C 64@0:0= ?0@B8:4; ?4@0=B0@0=G0 5;C830 '0380A8 030;07 ?4@?8=3070= :0;@ 30;0 14=BC: 64;10=6 4;4:B@06=4B8: 30?0B 14@?8=307 A420@0 !=D4:A8;8@0= !=3C:A80=B0@0= '0380A80=20@0= 30?0B 4=G4101:0= -03 4C080= 4=30 A478=660 B4@9038 4@C1070=(C7C 94=8A=G0 4C080=0=90=6 4C080=C0A 4C080=+;C4 Setelah Anda mempelajari bab ini, tentunya Anda telah memahami tentang perubahan wujud zat akibat perubahan kalor. Dapatkah Anda menerangkan mengenai cara perpindahan kalor pada suatu zat? Bagaimana dari cara perpindahan tersebut yang belum Anda pahami? Diskusikan dengan teman Anda tentang materi yang belum Anda pahami tersebut. Jika masih menemui kesulitan, bertanyalah kepada guru Fisika Anda.

140. Kalor 167 *-*)-)4-)456+8#/9/(1-*/(515%/,3+,/-)1%#6,6-5*)/ 0;0 A8AB4 A0BC0= 8=B4@=0A8=0; 14A0@0= AC7C 48;8:8A0BC0= 0 =4EB= 3 :4;D8= 1 E0BB 4 :0;@8 2 9C;4 0;01B;B4@AB4@30?0B

141. 6@0:?8?030AC7C I8B0107:0=ACACA410=G0: 6@014@AC7C I 8:0B830:030:0;@20?C@0=0C?C=:0;@G0=6 B4@A4@0?1B;B4@A08@:?8ACAC :0;6I (C7C20?C@0==G0030;07 0 I 3

142. I 1 I 4 I 2 I

143. (C7CA410B0=61090G0=6?0=90=6=G0 38B8=6:0B:0= 30@8 I4=9038 IA478=660?0=90=6=G014@B0107 8;84B4@4@B01070=?0=90=6AC0BC10B0=61090 G0=6?0=90=6=G0 298:038?0=0A:0=30@8 IA0?08 I030;07 0 3 1 4 2 (41C071;014@=660B4@1C0B30@8?4@C=66C34=60= :458A84=C08;8=40@ K JI030 I 90@890@8=G0 8:01;0B4@A41CB38?0=0A:0=A0?08 I?4@B01070=;C0A?4@C:00=1;0030;07A414A0@ 30;0 0

144. K J 3 K J 1 K J 4 K J 2 K J 30=14@BC@CBBC@CB030;07:458A84=C08?0=90=6 :458A84=C08;C0A30=:458A84=C08D;C44@8:CB 8=8 G0=6 4=C=9C::0= 7C1C=60= G0=6 14=0@ 30@8 :4B860=G0030;07 0 3

145. 1

146. 4

147. 2

148. 38?0=0A:0= (410B0=614A8C;0C;0?0=90=6=G0 0 30@8AC7C) 4=9038) 8:0:458A84=C08?0=90=6 10B0=6 B4@A41CB ?4@B01070= ?0=90=6 10B0=6 030;07 0 ) 1 ) J ) 0 2 ) J ) 0 3 0 ) J ) 4 ) 4;0BB8?8A30@8;60?0=90=6=G0 230=;410@=G0 2?030AC7C I 8:0?4;0BB4@A41CB38?0=0A:0= A0?0834=60=AC7C IB4@=G0B0;C0A=G0 2!458A84=C08?0=90=6;60B4@A41CB030;07 0 K JI 3 K JI 1 K JI 4

149. K JI 2 K JI 030AC7CI0AA094=8A40A

150. :6

151. :458A84=C08?0=90=640A

152. K JI#0AA0 94=8A40AB4@A41CB?030AC7C I030;07 0

153. :6

154. 3

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161. 0A30;0@C0=6B4@BCBC?A410=G0:B4:0=0==G0 38C1074=9038:0;8A4C;0?030AC7CG0=6A00 +;C460A0:0=14@C1074=9038 0 3 1 4 2

162. (C0BC60A8340;A410=G0: ?030B4:0=0= 0B 30=AC7CI!4C380=60AB4@A41CB38?0=0A:0= 78=660AC7C=G0 I30=D;C4=G0

163. )4:0=0= 60AA4:0@0=6030;07 0 0B 3 0B 1 0B 4 0B 2 0B (0BC0=:0;@94=8A30;0(030;07 0 :0;I 3 9C;4! 1 :0;6I 4 9C;4:6! 2 ::0;:6I C;07?0=0AG0=638?4@;C:0=C=BC:4=08::0=AC7C I?030 6@008@030;07 0 :0; 3 :0; 1

164. :0; 4 :0; 2 :0;

165. 0=0A94=8AAC0BC14=3038B4=BC:0=;47 0 94=8A14=30=G0 1 :4=08:0=AC7C14=30 2 D;C414=30 3 0AA014=30 4 ;C0A?4@C:00=14=30 8 30;0 64;0A 14@8A8 6 08@ 14@AC7C I 38 0AC::0= 64A14@AC7CJI:0;@94=8A4AB4@A41CB :0;6I 8:070=G0B4@9038?4@BC:0@0=:0;@0=B0@0 08@30=4AA4B4;07B4@9038:4A4B810=60=0:0=38 ?4@;47 0 A4;C@C74A4=208@30=AC7C=G0380B0A I 1 A4;C@C74A4=208@30=AC7C=G0 I 2 B830:A4;C@C74A4=208@30=AC7C=G0 I 3 AC7CA4;C@C7A8AB43810E07 I 4 A410680=08@414:C30=AC7CA4;C@C7A8AB4 I Tes Kompetensi Bab 7

166. C0 1C07 1;0 ;60 G0=6 A494=8A 0A8=60A8=6 14@AC7CI380AC::0=:430;008@G0=6AC7C=G0 I!43C0=G04=20?08:4A4B810=60=B4@0;?030 AC7C I 8:00AA0A4B80?1;0 :630=

167. :6:0;@ 94=8A;60030;07

168. K

169. :6!30=:0;@94=8A08@ 030;07 K

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171. K 3 K 1

172. K 4 K 2 K )4@9038=G0?4@?8=3070=:0;@?030AC0BCH0B:0@4=0 64B0@0=0B0B=G038A41CB 0 @0380A8 3 :=3C:A8 1 :=D4:A8 4 @B0A8 2 B@0=A;0A8 4?4=6B41060B410;=G0 230=;C0A?4=0 ?0=6=G0 4@C:00=G0=6A0BC48;8:8AC7C

173. ! A430=6:0= G0=6 ;08= 14@AC7C

174. ! 8:0 :=3C:B8D8B0AB4@0;B41060

175. A!14A0@ =G0:0;@G0=64@010BB80?A4:=030;07 0 K 3

176. K 1 K 4

177. K 2

178. K 8#-)135/9/#3*,65*/*%/(/515 168 Mudah dan Aktif Belajar Fisika untuk Kelas X 8:0AC7C14=30G0=614@?890@4=9038:0;8A4C;0 4=4@68G0=638?0=20@:0=B80?34B8:B80?A0BC0=;C0A 4=9038 0 :0;8A4C;0 3 :0;8A4C;0 1 :0;8A4C;0 4 :0;8A4C;0 2 :0;8A4C;0 ;0G0=6?4@C:00==G078B0A4?C@=048;8:8 AC7C :=AB0= ! 4A0@=G0 4=4@68 G0=6 38?0=20@:0=B80?34B8:;47?4@C:00=1;0B4@A41CB 98:0:=AB0=B0(B450=;BH0==K J, ! 030;07 0 K , 1 K , 2

179. K , 3

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182. I 30= I 8:0 :43C0 14=30 B4@A41CB 40=20@:0=4=4@6830;014=BC:@0380A8?4@10=38=6 0=9C;074=4@68?4@34B8:G0=638?0=20@:0=030;07 0 3 1 4

183. 2 030AC7CIA41C07:C1CA764=?0=90=6@CAC: @CAC:=G0 2 8:0:458A84=C08?0=90=6:C1CA8BC I14@0?0:07D;C4:C1CA?030AC7C I C0@4;10900A8=60A8=6?0=90=6=G0

184. ?030AC7C I0:0=38A01C=6C=BC:414=BC:AC0BC;8=B0A0= :4@4B0 8:0AC7C0:A8C=G0 I14@0?0:07;410@ 24;07 8=8C G0=6 70@CA 38A4380:0= 38 0=B0@0 A01C=60=@4;

185. 030AC7C IA41C071B;14@8A8?4=C7208@0=G0=6 D;C4=G0 22 8:0:458A84=C08D;C41B; 30=208@0=14@BC@CBBC@CB I30= I 14@0?0:07208@0=G0=6BC?0798:038?0=0A:0=A0?08 I 8:4B07C8?0=0A94=8A4A :0;6I?0=0A94=8A 08@ :0;6I30=:0;@;41C@4A :0;68BC=6;07 AC7C0:78@20?C@0= 64AG0=6AC7C=G0JI14@030 30;0:0;@84B4@G0=6:0?0A8B0A?0=0A=G0 :0;I 38BC0=68 608@G0=614@AC7C I C0?C;C76@0AC0BCH0B3814@8:0;@ :0;A478=660 AC7C14=30B4@A41CB14@C10730@8 I4=9038 I 4@0?0:07 :0;@ G0=6 38;4?0A:0= 060@ AC7C 14=30 BC@C=30@8 I4=9038 I (C7C ?030 :43C0 ?4@C:00= :4?8=6 14A8 G0=6 48;8:8;C0A 230=B410;2030;07 I30= I 8:0 :=3C:B8D8B0A 14A8 :0;A2I B4=BC:0=?0=0AG0=63870=B0@:0=B80?4=8B (41C07;C10=6:428;?03038=38=6BC=6:C4=G4@C?08 14=3078B0 8:0;C0A;C10=6 230=AC7C=G0 A0034=60=AC7CBC=6:CG08BC I14@0?0:0;@ G0=638@0380A8:0=;C10=6A4B80?A4:= ;0G0=6@038CA=G0248;8:848A8D8B0A ;0B4@A41CB4=G4@0?4=4@68:0;@30@8;8=6:C=60= =G0 34=60= ;09C

186. J/s 4@0?0 AC7C 1;0 8BC A4:0@0=6

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