Halaman
A. KalorB. PerpindahanKalorKalor4@30A0@:0= EC9C3=G0 <0B4@8 381430:0= <4=9038 H0B ?030B 208@30= 60A !4B860 H0B B4@A41CB <4<8;8:8 :0@0:B4@8AB8: <0A8=6<0A8=6 :0=B4B0?80301414@0?0H0BG0=630?0B14@C107EC9C3:0@4=0?4=60@C7AC7C30= :0;>@ <0B8;07 :4B8:0 =30 A430=6 <8=C< <8=C<0= G0=6 3830;0<=G0B4@30?0B4A10BC(4<0:8=;0<04A10BC30;0<<8=C<0=0:0=<4=208@ 30= 14@C107 EC9C3 #4=60?0 B4@9038 34<8:80= 4=0@:07 4AB4@A41CB B830: 78;0=6 <4;08=:0= 14@C107 EC9C3#4=60?0 38 AC0BC B4<?0B 38 C<8 B4@9038 7C90= A0;9C #4=60?0 38 =3>=4A80B830:B4@9038<CA8<A0;9C&0301018=8=300:0=<4<?4;090@8B4=B0=6 :0;>@ 8:0 =30 B4;07 <4<070<8 :>=A4?:>=A4? :0;>@?4@B0=G00=B4@A41CB0:0=<C307=3090E01%;47:0@4=08BC?4;090@8;07101 8=8 34=60= AC=66C7AC=66C7•menganalisis pengaruh kalor terhadap suatu zat;•menganalisis cara perpindahan kalor;•menerapkan asas Black dalam pemecahan masalah.Setelah mempelajari bab ini, Anda harus mampu:menerapkan konsep kalor dan prinsip konservasi energi pada berbagai perubahanenergi.Hasil yang harus Anda capai:147Perubahan musim yang terjadi dengan diikuti perubahan wujud daricair ke padat, atau sebaliknya (musim salju), merupakan fenomenaberlakunya konsep-konsep kalor yang terjadi di alam ini.Bab7Sumber: CD Image
148Mudah dan Aktif Belajar Fisika untuk Kelas X &#&-6..&.1&-"+"3*,0/4&1"-03,&3+","/-")40"-40"-#&3*,65%"-".#6,6-"5*)"/Tes Kompetensi Awal
?0G0=638A41CB34=60=?4<C080= (41CB:0=B86020@0?4@?8=3070=:0;>@ ?0:07 A0<0 B8B8: 38387 08@ :4B8:0 3838387:0= 38?46C=C=60= 3810=38=6:0= 34=60= B8B8: 38387 08@ 38304@07?0=B08 4;0A:0=90E010==30A. Kalor&4@=07:07 =30 <4;0:C:0= ?4=6C:C@0= AC7C #8A0;=G0 ?030 AC0BC B4@<><4B4@ :>;>< @0:A0 =30 <4=60<0B8 ?4@C1070= ?0=90=6 :>;>< A4106088=38:0B>@ 030=G0 ?4@C1070= AC7C (4AC=66C7=G0 ?4@C1070= AC7C 8BC B4@90380:810B14=30G0=6A430=638C:C@<4;4?0A:0=:0;>@0B0C<4=4@8<0:0;>@1.Pengertian Kalor!0;>@ <4@C?0:0= A0;07 A0BC 14=BC: 4=4@68 G0=6 30?0B 14@?8=307 30@814=30 G0=6 A0BC :4 14=30 G0=6 ;08= 8:0 3C0 1C07 14=30 G0=6 AC7C=G014@1430 38A4=BC7:0= AC0BC A00B 0:0= B4@9038 :4A4B8<10=60= B4@<0;AC7C=G0 A0<0 0; 8=8 B4@9038 :0@4=0 030=G0 ?4@?8=3070= :0;>@30@814=30 G0=6 14@AC7C B8=668 :4 14=30 G0=6 14@AC7C @4=307!0;>@14@143034=60=AC7CE0;0C?C=:43C0=G0<4<8;8:87C1C=60=4@0B(C7C030;0734@090B?0=0A0B0C38=68=AC0BC14=30A430=6:0=!"%'030;07 $ ' .$ & $ !$ ' (*)* $ ! $ .$ " $(C7C30=:0;>@30?0B 381430:0= 34=60= 94;0A ?030 ?4@8AB8E0 ?4@C1070= EC9C3 AC0BC H0B*=BC: <4=6C107 4A <4=9038 08@ 38?4@;C:0= :0;>@ &030 ?4@8AB8E0?4@C1070= EC9C3 8=8 4A 14@AC7C I 14@C107 <4=9038 08@ 14@AC7C I 038 B830: 030 ?4@C1070= AC7C ?030 A00B 4A <4=208@ B4B0?8 381CBC7:0=:0;>@ C=BC: <4=6C107 EC9C3 4A B4@A41CB(0BC0= ( C=BC: :0;>@ 030;07 9>C;4 38A8=6:0B (0BC0= :0;>@ G0=6;08=030;07:0;>@80B0C:8;>:0;>@8)*!"%' 383458=8A8:0=A410608$.!$. !"%' .$ & '"*!$ *$)*! # $ !!$ (* * ()* '# ' ( ('0
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#J To k o hJoseph Black, adalah seorangkimiawan Skotlandia yangmendukung teori tentang panas,yaitu bahwa suhu merupakankonsentrasi kalori dalam suatubenda. Ia kemudian menemukanilmu baru yang disebutkalorimetri. Ketika menyelidikitentang panas (kalori), ia mengirabahwa kapasitas panas merupakanjumlah panas yang dapatditampung oleh suatu benda.Padahal, ini sebenarnyamerupakan ukuran tentang jumlahenergi yang diperlukan untukmenaikkan suhu suatu bendadalam jumlah tertentu.Sumber:Jendela Iptek, 1996Joseph Black(1728–1799)
149Kalor3.Kapasitas Kalor&( )(!"%'030;07$.!$.!"%'.$ & '"*!$*$)*!# $ !!$(* * (*)* $ ( (' 0
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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 <0B4<0B8A :43C0=G0 <4<8;8:8 7C1C=60= A410608 14@8:CB0@8 &34".""/<
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!4B4@0=60=#<0AA0H0B:6 :0;>@ 94=8A :6I4.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@=07<CA=07B4B0?84=4@6830?0B14@C10714=BC:30@8A0BC14=BC:4=4@68 :4 14=BC: 4=4@68 ;08= (410608 2>=B>7 4=4@68 <4:0=8: 30?0B14@C107 <4=9038 4=4@68 ;8AB@8: !4<C380= 4=4@68 ;8AB@8: 30?0B 14@C107;068<4=90384=4@682070G030=A4B4@CA=G04<8:80=9C604=4@6830;0<14=BC: :0;>@ 14@0A0; 30@8 14=BC: 4=4@68 ;08=Sumber: Physics, 1980 "#&- !0;>@ 4=8A414@0?0-0B?030(C7C I30=)4:0=0=
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Kesetaraan kalori dan joule adalah1 kalori= 4,2 joule1 joule= 0,24 kalori1 kkal= 1.000 kaloriIngatlahTugas Anda 7.1Anda telah mengenal besarankalor jenis suatu benda. Diskusikandengan teman sekelas Anda jikakalor jenis suatu zat besar, apakahbenda akan cepat panas ataulambat panas?
150Mudah dan Aktif Belajar Fisika untuk Kelas X(410=G0: 6@0<08@38?0=0A:0=30@8 I<4=9038 I 8:0<0AA094=8A08@030;07
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151Kalor5.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@C10730@8EC9C3?030B<4=9038208@30=30@8208@<4=903860A0B0CA410;8:=G0-0B A4;0;C <4=4@8<0 0B0C <4;4?0A:0= :0;>@ A4;0<0 ?4@C1070= EC9C314@;0=6AC=6B4B0?8B830:38A4@B0834=60=:4=08:0=0B0C?4=C@C=0=AC7C(410608 2>=B>7 :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 60A38A41CB# $ *& &@>A4A A410;8:=G0 30@8 60A :4 208@ 38A41CB# $ #*$&@>A4A ?4@C1070= EC9C3 30@8 ?030B :4 60A 38A41CB# $.*" # &@>A4AA410;8:=G0 30@8 60A :4 ?030B 38A41CB# $.*" #a.Kalor Uap dan Kalor Embun 8:0 =30 <4=4B4A:0= A?8@8BCA ?030 :C;8B B0=60= 10680= :C;8B G0=638B4B4A8 A?8@8BCA 0:0= B4@0A0 38=68= 0; B4@A41CB 38A4101:0= :0;>@ ?03010680= :C;8B B0=60= G0=6 38B4B4A8 A?8@8BCA 38A4@0? >;47 A?8@8BCA C=BC:<4=6C0? 0=G0:=G0 :0;>@ G0=6 38?4@;C:0= C=BC: <4=6C0?:0= A?8@8BCAA410=38=6 34=60= <0AA0 A?8@8BCA 30= 14@60=BC=6 ?030 :0;>@ ;0B4=C0?:0;>@ C0? A?8@8BCA B4@A41CB"%' *&030;07!"%' .$ & '"*!$ *$)*! # $ * ,* * ! /) '# $ *&&) ) ! $.30?C=!"%' #*$030;07:"%'.$ " &( *$)*! # $ * ,* * ! *& # $ ' & ) ) ! " *'$.!0;>@C0?AC0BCH0BA0<014A0@34=60=:0;>@4<1C=H0BB4A41CB#8A0;=G0:0;>@C0?@0:A0 :6:0;>@4<1C=@0:A0?C= :6 8:0:0;>@C0?:0;>@4<1C=AC0BCH0B030;07C=BC:<4=6C0?:0=<4=64<1C=:0=H0BG0=6<0AA0=G0#0:0=<4<4@;C:0=:0;>@A410=G0:Gambar 7.2Diagram perubahan wujud zatGasCairPadatmencairmembekumenyublimmenyublimmenguapmengembun
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0B<>A54@ &4;090@8 B014; B4@A41CB H0B <0=0:07 G0=6 <4<8;8:8 :0;>@ C0? B4@14A0@JTugas Anda 7.2Seperempat kilogram es, dengansuhu –10°C, dicampur dengan 2 kgair yang suhunya 20°C. Diskusikandengan teman sebangku Anda,bagaimana fase akhir campurantersebut dan berapakah suhu akhircampuran tersebut? "#&-
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4;8C< 83@>64=$8B@>64=%:A864=;:>7>;'0:A08@(C;5C@)8<0778B0<=B8<>=&4@0:<0A)4 < 1 0 6 0Sumber: Physics, 1980
152Mudah dan Aktif Belajar Fisika untuk Kelas Xc.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
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I(4;C@C7<0AA0<0A8714@14=BC:EC9C3H0B208@ G08BC08@Gambar 7.3Pemanasan es pada tekanan 1atmosfer.EnergiABCDESuhu (°C)1000t1
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Sumber: Physics for Scientist & Engineer, (2000).Sebanyak 320 gram campuran es danair pada suhu 0°C berada dalam bejanayang kapasitas kalornya dapatdiabaikan. Kemudian, dimasukkan 79guap air yang bersuhu 100°C ke dalambejana tersebut. Suhu akhir menjadi79°C. Jika kalor lebur es 79,0 kal/g dankalor penguapan air 540 kal/g, makabanyaknya air mula-mula adalah ...gram.a.4d. 65b.1 0e. 79c.3 5SPMB, 2002PembahasanDiketahui:mes + air = 320 gram, T = 0°Cmuap = 79 gram, T = 100°CKalor lebur es, L = 79,0 kal/gramKalor penguapan air , L = 540 kal/gramTakhir = 79°CKalor yang di lepas uap air (100°C)Qlepas= muap L + muap cT= (79) (540) + (79) (1) (100 – 79)= 79 (561) kaloriQterima= mes L + mes cT + mair cT= (320 – mair) 79 + (320 – mair) (1) (79 – 0) + mair (1) (79 – 0)= (640 – mair) 79Asas black :Qlepas = Qterima 79 (561) = (640 – mair) 79 mair = 79 gramJawaban : EPembahasan Soal
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IGambar 7.4Partikel-partikel pada(a) zat padat(b) cair(c) gas.(a)(b)(c)
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Gambar 7.5Pemuaian panjang pada kawat logam.Aktivitas Fisika 7.1Pemuaian PanjangTujuan PercobaanMenentukan koefisien muai panjang suatu kabel konduktorAlat-Alat Percobaan1.Kabel (bahan uji: kawat tembaga dan kawat besi) diameter 1,5 mm2.Statif3.Beban (massa 5–50 gram)4.Mistar tegak (100 cm)5.Amperemeter digital6.Termometer digitalLangkah-Langkah Percobaan1.Susun alat seperti pada gambar berikut.2.Ikatkan kedua ujung kabel (kawat)pada batang peyangga bagian atas.3.Gantungkan beban bermassa 50 grampada kawat, tepat di tengah kawat.4.Tempatkan mistar sejajar dengan be-ban untuk mengetahui besar perubah-an yang dapat terjadi.5.Hubungkan amperemeter, sumbertegangan dan termometer sepertipada gambar.6.Catat awal 0 dan T0 kabel danposisi beban x0 dan y0.7.Catat perubahan yang terjadi pada kawat x, y, dan T setiap perubahantegangan yang dinaikkan.8.Gunakan perhitungan metode grafik untuk menentukan koefisien muaipanjangnya.9.Bandingkan koefisien muai panjang antara kedua kawat tersebut, apakahsama untuk kedua bahan kawat tersebut? Menurut Anda, mengapakoefisiennya berbeda/sama?10.Apa kesimpulan yang Anda peroleh dari percobaan tersebut?)2040 60801000amperemetersumber teganganx0y0+ –+ –kabelmistarstatifbebanstatiftermometerdigitalInformasiDi toko peralatan elektronik yangmenjual kabel (penghubung),terdapat berbagai jenis merekdagang. Untuk memperoleh kabelyang baik dapat dilakukan percobaanseperti pada Aktivitas 7.1 untukmembandingkan sifat termalantara merek yang satu dan yanglainnya.Kabel penghubung yang baiksebagai konduktor adalah kabelyang menghantarkan listrik secarabaik sekalipun pada suhu yangtinggi. Ini artinya bahwa koefisienmuai panjangnya relatif kecilnilainya. Dengan demikian, Andadapat memilih untukmenggunakan kabel berdasarkankualitasnya.In wire selling market, there is avarious kind of trade mark. Toobtain wire we can do an activitiesas in Physics Activity 7.1 tocompare thermal propertiesbetween one trademark to others.Good wire as a conductor is awire that conducting electricitygood enough even in hightemperature. Its mean that the linearexpansion coefficient relativelysmall. By that, you can choose touse wire according to its quality.untuk AndaInformation for You
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158Mudah dan Aktif Belajar Fisika untuk Kelas Xd.Pemuaian Gas&030
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Gambar 7.10Gas dipanaskan pada tekanan tetappipabalontabungkalengbalonsebelummemuaibunsenGambar 7.11Volume gas sebagai fungsi dari suhupada pemanasan gas dengan tekanantetap.0100200300400500VT(K)Gambar 7.12Gas dipanaskan pada volume tetappipasumbattabung
159KalorAktivitas Fisika 7.2Hubungan Suhu dan TekananTujuan PercobaanMengetahui hubungan antara kenaikan suhu dan tekanan.Alat-Alat PercobaanAlat ukur Bourdon, termometer raksa, tabung gas, pembakar bunsen, dan kakitiga.Langkah-Langkah Percobaan1.Rancanglah peralatan tersebut seperti pada Gambar 7.13. Pastikan tidak adakebocoran gas dalam tabung.2.Setelah terjadi kesetimbangan termal antara termometer dan gas dalamtabung, catat suhu awal gas (terlihat pada termometer) dan tekanan udara(terlihat pada alat Bourdon).3.Panaskan tabung dengan pembakar bunsen. Catat kenaikan suhu udara dantekanan udara dalam tabung pada suatu tabel.4.Dari tabel tersebut, buatlah grafik suhu terhadap tekanan gas.5.Buatlah kesimpulan dari kegiatan ini.Gambar 7.13Perangkat percobaan untukmengetahui hubungan antarakenaikan suhu dan tekanan.Bourdontermometerraksabunsenkakitiga&4@2>100= A4@C?0 34=60= :4680B0= B4@A41CB ?4@=07 38;0:C:0= ?0@007;8 8A8:0 0@8 70A8; ?4@2>100= G0=6 <4@4:0 ;0:C:0= 38?4@>;47 6@058:B4:0=0=60AB4@7030?AC7C?030D>;C<4B4B0?14@14=BC:60@8A;C@CA;870B:4<10;8".#"3 (420@0 <0B4<0B8:0 60@8A 8=8 <4<4=C78 ?4@A0<00=&
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160Mudah dan Aktif Belajar Fisika untuk Kelas XB. Perpindahan Kalor 8:08=68=<4<0A0:08@38?4@;C:0=:0;>@G0=614@0A0;30@80?8:><?>@&030?4<0=0A0=A4?0=2808@<C;0<C;0:0;>@30@80?8:><?>@14@?8=307:4?0=28B4<?0B08@:4<C380=:0;>@30@8?0=28B4@A41CB14@?8=307:408@&4@?8=3070=:0;>@B4@903830@814=3014@AC7CB8=668?0=0A:414=3014@AC7C ;4187 @4=307 38=68= 4=30 G0=6 ?0=0A <4<14@8:0= :0;>@ :4?030 14=30 G0=6 38=68= (40=308=G0 :0;>@ B830: 30?0B 14@?8=3070:0=AC;8BC=BC:<4<0A0:08@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 <4<0=0A:0= A0;07 A0BC C9C=6 10B0=6 ;>60< A4?4@B8".#"3
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162Mudah dan Aktif Belajar Fisika untuk Kelas X38?8=307:0= &030 A80=6 70@8 30@0B0= ;4187 24?0B ?0=0A 30@8?030 ;0CBA478=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=;0CB&030<0;0<70@8B4@9038 A410;8:=G0 0@0B0= ;4187 24?0B 38=68= 30@8?030 ;0CB A478=660C30@0 38 0B0A ;0CB =08: 30= B4<?0B=G0 3860=B8:0= >;47 C30@0 30@8 0B0A30@0B0= G0=6 38A41CB 0=68= 30@0Bb.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;;0CB<4A8=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=6<4A8=<4;0;C8?8?0?8?0 3810=BC >;47 A41C07 ?><?0 08@ ,) ' &*#& !0;>@ G0=6 38B4@8<0<4A8= <>18; 30@8 70A8; ?@>A4A ?4<10:0@0= <4=20?08 AC7C
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:0;InformasiAlat Penukar KalorSeperti namanya, alat penukarkalor adalah seperangkatinstrumen di mana terjadipertukaran kalor antara dua aliranfluida bergerak tanpa pencampur-an. Alat penukar kalor banyakdigunakan diberbagai industridengan berbagai model.Bentuk paling sederhana darialat penukar kalor adalah penukarkalor pipa ganda, yakni tersusunoleh dua pipa konsentris dengandiameter berbeda. Satu fluidamengalir di dalam pipa, dan fluidalainnya mengalir pada pipa yangmenembus ruang antara pipa.Kalor dipindahkan dari fluida yangpanas ke fluida yang dinginmelalui dinding pemisahnya.Terkadang pipa yang berada didalam dibuat dua putaran di dalamselongsong untuk menambahpertukaran kalor.As the name implies, heatexchangers are devices where twomoving fluid streams exchangeheat without mixing. Heatexchanger are widely used invarious industries, and they come innumerous designs.The simplest form of a heatexchanger is a double tube heatexchanger. It is composed of twoconcentric pipes of differentdiameters. One fluids flows in theinner pipes, and the other in theannular space between the twopipes. Heat is transferred from thehot fluid to the cold one throughthe wall separating them.Sometimes the inner tube makes acouple of turn inside the shell toincrease the heat transfer area.Sumber: Thermodynamics, 1998untuk AndaInformation for You
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