A. Kalor B. Perpindahan
Kalor
Kalor
430A0:0=EC9C3=G00B48381430:0=4=9038H0B?030B208 30=60A4B860H0BB4A41CB48;8:8:00:B48AB8:0A8=60A8=6:0=
B4B0?803014140?0H0BG0=630?0B14C107EC9C3:04=0?4=60C7AC7C 30= :0; 0B8;07 :4B8:0 =30 A430=6 8=C 8=C0= G0=6 38
30;0=G0B430?0B4A10BC40:8=;004A10BC30;08=C0=0:0= 4=20830=14C107EC9C34=60?0B49038348:80=4=0:074A
B4A41CBB830:78;0=64;08=:0=14C107EC9C3 4=60?038AC0BCB4?0B38C8B490387C90=A0;9C4=60?038
=3=4A80B830:B49038CA8A0;9C0301018=8=300:0=4?4;0908 B4=B0=6 :0; 8:0 =30 B4;07 40708 :=A4?:=A4? :0;
?4B0=G00=B4A41CB0:0=C307=3090E01;47:04=08BC?4;0908;07 1018=834=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 perubahan energi.
Hasil yang harus Anda capai:
147
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.
Bab
7
Sumber:
CD Image
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148
Mudah dan Aktif Belajar Fisika untuk Kelas X
-6...1-+3,041-03,3+,-40-40-3,65-.6,6-5
Tes Kompetensi Awal
?0G0=638A41CB34=60=?4C080= 41CB:0=B860200?4?8=3070=:0;
?0:07 A00 B8B8: 38387 08 :4B8:0 3838387:0= 38 ?46C=C=60=3810=38=6:0=34=60=B8B8:383870838
30407?0=B08 4;0A:0=90E010==30
A. Kalor
4=07:07=304;0:C:0=?4=6C:C0=AC7C8A0;=G0?030AC0BCB4 4B4:;0:A0=304=600B8?4C1070=?0=90=6:;A410608
8=38:0B030=G0?4C1070=AC7C4AC=66C7=G0?4C1070=AC7C8BCB49038 0:810B14=30G0=6A430=638C:C4;4?0A:0=:0;0B0C4=480:0;
1. Pengertian Kalor
0;4C?0:0=A0;07A0BC14=BC:4=468G0=630?0B14?8=307308 14=30G0=6A0BC:414=30G0=6;08= 8:03C01C0714=30G0=6AC7C=G0
141430 38A4=BC7:0= AC0BC A00B 0:0= B49038 :4A4B810=60= B40; AC7C=G0A000;8=8B49038:04=0030=G0?4?8=3070=:0;308
14=30G0=614AC7CB8=668:414=30G0=614AC7C4=307 0;14143034=60=AC7CE0;0C?C=:43C0=G048;8:87C1C=60=40B
C7C030;0734090B?0=0A0B0C38=68=AC0BC14=30A430=6:0=030;07 ..C7C30=:0;
30?0B381430:0=34=60=94;0A?030?48AB8E0?4C1070=EC9C3AC0BCH0B =BC:4=6C1074A4=90380838?4;C:0=:0;030?48AB8E0
?4C1070=EC9C38=84A14AC7C I14C1074=90380814AC7C I 038B830:030?4C1070=AC7C?030A00B4A4=208B4B0?8381CBC7:0=
:0;C=BC:4=6C107EC9C34AB4A41CB 0BC0=C=BC::0;030;079C;438A8=6:0B 0BC0=:0;G0=6
;08=030;07:0;80B0C:8;:0;8383458=8A8:0=A410608. ..
0
2. Kalor Jenis
8A0;:0=?030 0830= 0;:7;38148:0=:0;G0=6 A004=G0B0:4=08:0=AC7C?0300;:7;;418714A0308?03008
48:80=70;=G098:0?030 0830= 0838148:0=:0; G0=6A0010=G0:=G0:4=08:0=AC7C?030 08;418714A0308
?030 0848AB8E0B4A41CB4=C=9C::0=70;70;A410608148:CB 0 0;G0=638148:0=?030H0BA410=38=634=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=AC7CH0B1460=BC=694=8AH0B
42000B40B8A38BC;8AA410608148:CB 4B40=60=
10=G0:=G0:0;G0=638148:0=:0;80B0C9C;4 0AA0H0B60B0C:6
:0;94=8A:0;6I0B0C :6I ?4C1070=AC7CI
34. 30?0B38BC;8AA410608148:CB J
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
Joseph Black
1728–1799
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149
Kalor
3. Kapasitas Kalor
030;07... 0
42000B40B8A?4=G0B00=B4A41CB
38BC;8AA410608148:CB 038:0;94=8A30?0B383458=8A8:0=A41060810=G0:=G0:0;G0=638?4
;C:0=AC0BCH0BC=BC:4=08::0=AC7C
:6H0BB4A41CBA414A0
I- 4=C=9C::0=:0;94=8A14140?0H0B?030AC7C I30=B4:0=0=
B4B0?
0BA54
4B8:0=301490;0=380B0A?0A838?0=B083810E07B48:0B0708=3030?0B 40A0:0=?0A8:48=6G0=6B48=90:B40A0?0=0A4C380==30140;87
4=68=90:08;0CB8;0CB4;0B85;418738=68=1C:0=03070;?0A830=08;0CB A00A00B4:4=0A8=00B07088A0;=G0=30B8=90C
:608;0CB30=
:6
?0A8?0=B08380=B00=G04=CCB=300=0:07G0=648;8:8:0;94=8A;4187 B8=668080B0C?0A8
Mari Mencari Tahu
J 03034.
030;07:0?0A8B0A:0;:0;I0B0C
430?0B?414300=?4=64B80=0=B00:0?0A8B0A:0;
30=:0;94=8A B4B0?8A42000B40B8A:43C0=G048;8:87C1C=60=A410608148:CB
0834.
30=34.38?4;47
J 4B40=60=
0AA0H0B:6 :0;94=8A :6I
4. Hukum Kekekalan Energi untuk Kalor
=30B4;074?4;0908107E04=468B830:30?0B3828?B0:0=30= B830:?4=07CA=07B4B0?84=46830?0B14C10714=BC:308A0BC14=BC:
4=468 :4 14=BC: 4=468 ;08= 410608 2=B7 4=468 4:0=8: 30?0B 14C1074=90384=468;8AB8:4C380=4=468;8AB8:30?0B14C107
;0684=90384=4682070G030=A4B4CA=G048:80=9C604=46830;0 14=BC::0;140A0;30814=BC:4=468;08=
Sumber:
Physics, 1980
- 0; 4=8A4140?0-0B?030C7C I30=4:0=0=
0B
-03+4 ,;
55 -03+4
,; 53
;:7;4B8; 0:A0
8 4AJI
208
I C0?
I
030=0=CA80 B48=
C=8=60= ;C8=8C
41060 4A80B0C1090
807 04
40: 0GC
4=6
Kesetaraan kalori dan joule adalah 1 kalori
= 4,2 joule 1 joule
= 0,24 kalori 1 kkal
= 1.000 kalori
Ingatlah
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?
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150
Mudah dan Aktif Belajar Fisika untuk Kelas X
410=G0: 600838?0=0A:0=308 I4=9038 I 8:00AA094=8A08030;07 :0;6I0B0C
:6B4=BC:0=
0 10=G0:=G0:0;G0=638B48008B4A41CB30;0:0;8 1 10=G0:=G0:0;G0=638B48008B4A41CB30;09C;4
8
8:4B07C8 6
:0;6I IJ I I
6
:0;6I I :0;8 03810=G0:=G0:0;G0=638B48008B4A41CB030;07 :0;8
1 08:4A4B000=:0;830=9C;438:4B07C8107E0 :0;8
9C;4A478=660
K
9C;4 9C;4
Contoh
7.2
;4?0A B480
;4?0A B480
4;0=9CB=G034.
38:4=0;A410608 J
termometer pengaduk
kalorimeter
isolator air
4?B=60;C8=8C140AA0 630=14AC7C I380AC::0=:430;0
608 G0=614AC7C I4=60=4=60108:0=?4BC:00=:0;34=60=;8=6:C=60=78BC=6
AC7C0:7820?C0=98:0:0;94=8A0;C8=8C :630=:0;94=8A08
:6 8
8:4B07C8
08
6
08
I
;
6
;
I
08
:6
;
:6
;4?0A B480
08 08
6
:6 IJ 6 :6J I JJ I
I I
038AC7C0:78A4B4;07B49038:4A4B810=60=B40;030;07I
Contoh
7.1
C:C4:4:0;0==468C=BC::0;30?0B3800B834=60=4=6 6C=0:0=A4?4B8B0?0:?030.3 49C;070AA0
H0B20838?0=0A:0=4C380=H0B208B4A41CB380AC::0=:430;0 :0;84B4C7CH0B208B4A41CB38C:C34=60=B44B430=3820B0B
4;0=9CB=G0A49C;070AA0H0B20834=60=AC7CG0=6;41874=307 380AC::0=:430;0:0;84B430=3803C:34=60=?4=603C:78=660
:43C0H0B2081420?C30=40B04C380=AC7CH0B208G0=6B4 20?C 38C:C 30= 3820B0B 4;00 ?4=603C:0= 14;0=6AC=6 B49038
?4?8=3070=:0;-0B208G0=614AC7C;4187B8=6684;4?0A:0=:0;=G0 A430=6:0=H0B208G0=614AC7C;41874=3074=G40?:0;308H0B
208 G0=6 14AC7C B8=668 A478=660 ?030 0:78=G0 4=20?08 AC7C :4 A4B810=60=G0=638A41CB34=60=
4=CCB 041 -,
J
9C;07:0;G0=638;4?0A
;4?0A
A00 34=60= 9C;07 :0; G0=6 38B480
B480
4200 0B40B8A?4=G0B00=B4A41CB38BC;8AA410608148:CB
Gambar 7.1
Kalorimeter
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151
Kalor
5. Perubahan Wujud Zat
4;07=30?4;0908107E0H0B30?0B1414=BC:B860EC9C3G08BC ?030B20830=60A:810B?4=60C7AC7CEC9C3H0B30?0B14C107
308EC9C3?030B4=903820830=3082084=903860A0B0CA410;8:=G0 -0BA4;0;C4=4800B0C4;4?0A:0=:0;A4;00?4C1070=EC9C3
14;0=6AC=6B4B0?8B830:38A4B0834=60=:4=08:0=0B0C?4=CC=0=AC7C 4106082=B7:4B8:04AA430=64=208B49038?4C1070=EC9C3308
?030B4=9038208030?48AB8E0B4A41CB381CBC7:0=A49C;07:0; B4B0?8B830:38A4B0834=60=:4=08:0=AC7C
4;090838060?030.3
4C1070=EC9C3308208:460A 38A41CBA4AA410;8:=G030860A:420838A41CB
A4A ?4C1070= EC9C3 308 ?030B :4 60A 38A41CB . A4A A410;8:=G030860A:4?030B38A41CB.
a. Kalor Uap dan Kalor Embun
8:0=304=4B4A:0=A?88BCA?030:C;8BB0=60=10680=:C;8BG0=6 38B4B4A8A?88BCA0:0=B40A038=68=0;B4A41CB38A4101:0=:0;?030
10680= :C;8B B0=60= G0=6 38B4B4A8 A?88BCA 38A40? ;47 A?88BCA C=BC: 4=6C0?0=G0:=G0:0;G0=638?4;C:0=C=BC:4=6C0?:0=A?88BCA
A410=38=634=60=0AA0A?88BCA30=1460=BC=6?030:0;;0B4=C0? :0;C0?A?88BCAB4A41CB
030;07., .30?C=030;07:.
, . 0;C0?AC0BCH0BA0014A034=60=:0;41C=H0BB4A41CB8A0;=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=G00:0=44;C:0=:0;A410=G0:
Gambar 7.2
Diagram perubahan wujud zat Gas
Cair Padat
mencair membeku
menyublim menyublim
menguap mengembun
34=60= 030;07:0;C0? :6
-
4?4;870B:0=B8B8:3838730=:0;C0?1410608H0B ?030B4:0=0=
0BA544;0908B014;B4A41CBH0B0=0:07G0=64
8;8:8:0;C0?B414A0 J
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?410608-0B0304:0=0=
0B
5 -03
,
J J
J J
5, ;
K K
K K
K K
K K
K K
K K
K 4;8C
8364= 8B64=
:A864= ;:7;
0:A0 8
C;5C 80778B0
=B8= 40:
0A 41060
Sumber:
Physics, 1980
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152
Mudah dan Aktif Belajar Fisika untuk Kelas X
c. Perubahan Wujud Es Menjadi Uap
8:0 A49C;07 0AA0 4A G0=6 AC7C=G0 38 10E07 I 38?0=0A:0= 38148:0=:0;78=660AC7C=G0380B0A
IA49C;070AA04AB4A41CB
14C107EC9C3A4;CC7=G04=9038C0?4C1070=EC9C34A4=9038 C0?30?0B3800B8?030.3
4B40=60= J 49C;07:0;38148:0=:4?0304AA478=660AC7C4A=08:308J
I 4=9038 I4;CC70AA00A871414=BC:H0B?030BG08BC4A
J 49C;07:0;38148:0=C=BC:4=6C107EC9C3H0BA478=6604A 4=90380834=60=AC7CB4B0?G08BC I
J 49C;07:0;38148:0=C=BC:4=08::0=AC7C308 I78=66008 4=38387?030AC7C
I4;CC70AA00A871414=BC:EC9C3
H0B208G08BC08
Gambar 7.3
Pemanasan es pada tekanan 1 atmosfer.
Energi A
B C
D E
Suhu °C
100
t
1
b. Kalor Lebur dan Kalor Beku
41=6:074A38?0=0A:0=?030AC7C I30=B4:0=0=C300
0BA54 4AC;084=208030A00B4A4=208381CBC7:0=:0;C7C:4B8:04A4=
20838A41CBB8B8:;41C410;8:=G098:008B4A41CB=3038=68=:0=08 8BC0:0=414:C48AB8E0?414:C0=8=838A4B0834=60=?4;4?0A0=
:0;C7C:4B8:008414:C38A41CBB8B8:14:C0=G0:=G0:0;G0=6 381CBC7:0= 0B0C 38;4?0A:0= ;47 AC0BC H0B C=BC: 4=208 ?030 B8B8:
;41C=G00B0C414:C?030B8B8:;41C=G0A410=38=634=60=0AA0H0B 8BC30=1460=BC=6?030:0;;41C14:CH0BB4A41CB
030;07., .30?C=030;07:
., 0; ;41CAC0BCH0BA0014A034=60=:0;14:CH0BB4A41CB
8:0 :0; ;41C :0; 14:C AC0BC H0B 030;07 C=BC: 4;41C 414:CH0BG0=60AA0=G0?030B8B8:;41CB8B8:14:C0:0=4
4;C:0=:0;A410=G0: J
-4=C=9C::0=B8B8:;41C30=:0;;41C1410608H0B?030 B4:0=0=
0BA544;0908B014;B4A41CBH0B0=0:07G0=648;8:8
B8B8:;41CB44=307
5 5,63
;
4;8C 8364=
8B64= :A864=
;:7; 0:A0
8 C;5C
80778B0 =B8=
40: 0A
41060 J
J J
J J
- 8B8:41C30=0;41C410608-0B?0304:0=0=
0B
-0363,
K K
K K
K K
K K
K K
K K
K
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 kalg dan kalor penguapan air 540 kalg, maka
banyaknya air mula-mula adalah ... gram.
a.
4 d. 65
b. 10 e. 79
c. 35
SPMB , 2002
Pembahasan Diketahui:
m
es + air
= 320 gram, T = 0°C m
uap
= 79 gram, T = 100°C Kalor lebur es, L = 79,0 kalgram
Kalor penguapan air , L = 540 kalgram T
akhir
= 79°C Kalor yang di lepas uap air 100°C
Q
lepas
= m
uap
L + m
uap
c T
= 79 540 + 79 1 100 – 79 = 79 561 kalori
Q
terima
= m
es
L + m
es
c T + m
air
c T
= 320 – m
air
79 + 320 – m
air
1 79 – 0 + m
air
1 79 – 0 = 640 – m
air
79 Asas black : Q
lepas
= Q
terima
79 561 = 640 – m
air
79 m
air
= 79 gram
Jawaban : E
Pembahasan
Soal
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153
Kalor
830;0A41C071490=0B430?0B :60814AC7C I430;01490=0B4A41CB 380AC::0=A4?B=64A14AC7CJ IA410=G0:
:68:4B07C8:0;;41C4A
K :6 8:0?4BC:00=:0;70=G0B490380=B000830=4AB4=BC:0=AC7C
0:78:4A4B810=60=B40; 8
8:4B07C8
08
:6 K
:6
08
I
08
:6I
4A
J I
4A
:6I 8A0;:0=
030;07:0;G0=638;4?0A:0=08308 IA0?08 I
08 08
:6
:6I I 9C;4
030;07:0;G0=6381CBC7:0=C=BC:4=6C1074A14AC7CJ I4=90384A 14AC7C I
4A 4A
:6
:6I JJ 030;07:0;G0=638A40?4AC=BC:4;41C
4A 4A
:6K :6
0;G0=638A40?4A 4=G0B0:0;G0=638A40?4A;418714A0308?030:0;G0=638;4?0A:0=0840B8
:0;G0=6381CBC7:0=4AC=BC:4=208;418714A0308?030:0;G0=638;4?0A:0=08 A0?08 I 03870=G0A410680=4AG0=64=20830=AC7C:4A4B810=60=B40;
030;07 I
Contoh
7.3
4. Pemuaian Zat
C0BC14=30108:?030B2080C?C=60AB43880B0A?0B8:4;?0B8:4; A0=60B:428;G0=6A4;0;C1464B038A41CB;4:C; 00:0=B0;4:C;
H0B?030BA0=60B1434:0B0=030H0B208900:0=B0;4:C;=G0060: 4=660=6A430=6:0=?03060A900:0=B0;4:C;=G0A0=60B4=660=6
470B8:0=.3
8:0AC0BC14=3038?0=0A:0=;4:C;;4:C;8BC1464B0A40:8= 24?0B4B00=0=B0;4:C;B4A41CB4=G4101:0=;4:C;;4:C;
A0;8=63=6 a.
Pemuaian Zat Padat
4C080=?030H0B?030B30?0B3800B84;0;C8?4C1070=?0=90=6 ;C0A30=D;C4
.61+ 4CB0A:0E0B;60G0=6?0=90=6=G0
30=14AC7C 38?0=0A:0=
A0?08AC7C0:0:0E0B;608BC0:0=4C08A478=660?0=90=6=G0 4=9038
J 49C;07:0;38148:0=C=BC:4=6C107EC9C3H0BA478=660H0B 208G0=64=3838714C1074=9038C0??030AC7CB4B0?G08BC
I
Gambar 7.4
Partikel-partikel pada a zat padat
b cair c gas.
a b
c
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154
Mudah dan Aktif Belajar Fisika untuk Kelas X
430A0:0=,5754 4, 38 0B0A 140?0 :458A84= C08 ?0=90=6:0E0BB410600=38=6:0=70A8;=G034=60=-148:CB
8=8G0=64C0B:458A84=C08?0=90=614140?01070=?030AC7C I 470B8:0=.3
J
4B01070=?0=90=6:0E0B30?0B38B4=BC:0=34=60=?4A000= A410608148:CB
J
03034.B4A41CB 030;07:458A84=C08?0=90=6 0834.30=34.38?4;47
J
J
0B0C .
J
60=30;418740708?4C080=?0=90=6;0:C:0=;07:4680B0= 148: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 T
kabel dan posisi beban x
dan y .
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 berbedasama?
10. Apa kesimpulan yang Anda peroleh dari percobaan tersebut?
20 40
60 80
100
amperemeter sumber tegangan
x y
+ – + –
kabel mistar
statif beban
statif
termometer digital
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.
untuk Anda
Information for You
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Kalor
41CBC70= ?4=64B07C0= 4=64=08 :458A84= C08 ?0=90=6 AC0BC 1070= 030;07 C=BC: 4?478BC=6:0= ?4=66C=00= 1070= B4A41CB
8A0;=G0?48;870=1070=30=C:C0=G0=6386C=0:0=C=BC::=ABC:A8 9410B0=
030A0;07A0BCC9C=6:=ABC:A89410B0=34=38148:0=301090 G0=630?0B14?CB01410A4B8:09410B0=4C080:810B?0=0A30A0
9410B0=30?0B4=6640::0=301090B4A41CB030C9C=6G0=6;08= 9C6038148:0=24;07G0=64C=6:8=:0=30A09410B0=30?0B14640:
Gambar 7.6
Pada salah satu ujung jembatan ini, dipasang roda dan diberi celah
untuk memberi ruangan ketika jembatan memuai.
sungai baja
roda
- 458A84=C080=90=6410608-0B?030C7C I
5 046+ ;
;C8=8C C=8=60=30=?4C=66C
4B=30=10BC 020180A0
020?G4F 807
4A8 E0A0
090 04
41060 K
J
K
J
K
J
K
J
K
J
K
J
K
J
K
J
K
J
K
J
JK
J
K
J
Sumber:
Physics, 1980
030AC7C I?0=90=6:0E0B14A8030;07 40?0:07?0=90=6:0E0B14A8 B4A41CB?030AC7C
I98:0:458A84=C08?0=90=614A8
K
J
I
8
8:4B07C8 I
I
K
J
I
J
.
J .
K
J
I
IJ I 038?0=90=6:0E0B14A8B4A41CB?030AC7C
I030;07
Contoh
7.4
.664
4:4?8=6;60G0=6?0=90=6=G0-30=;410=G0.0:0=4=60;08 C08;C0A98:038?0=0A:0=4C080=;C0AAC0BCH0B1460=BC=6:4?030
:458A84=C08;C0AG0=638148;010=6 C08;C0AB414=BC:308 3C0 ?4C080= G08BC ?4B01070= ?0=90=6 30= ?4B01070= ;410
:810B=G014A0:458A84=C08;C0A A0034=60=3C0:0;8:458A84= C08?0=90=6G08BC
8:0A4:4?8=6;60G0=6;C0A=G0 30=AC7C=G0
38?0=0A:0= A0?08AC7C;60B4A41CB0:0=4C08A478=660;C0A=G04=9038
4A0=G0?4B01070=;C0A:4?8=6;60 B4A41CB30?0B38BC;8A:0=
30;0?4A000=148:CB 8A0;=G0;C0A?4A468
J
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156
Mudah dan Aktif Belajar Fisika untuk Kelas X
4:4?8=60;C8=8C34=60=?0=90=6
230=;410 238?0=0A:0=308
I A0?08
I 8:0:458A84=C08?0=90=60;C8=8CB4A41CB 030;07K
I B4=BC:0=;C0A:4?8=60;C8=8CA4B4;0738?0=0A:0=
8
8:4B07C8 2K 2
2
2
K
J
IK
J
I IJ
I
I0:0
2 .
K
J
IK
I
2 038;C0A:4?8=60;C8=8CA4B4;0738?0=0A:0=030;07
2
Contoh
7.5
.6 0-6. 4C080=D;C414=301460=BC=6:4?030:458A84=C08D;C4
G0=638148;010=6 4C080=D;C4B414=BC:308B860?4C080= G08BC?4B01070=?0=90=6?4B01070=;41030=?4B01070=B8=668
:810B=G014A0:458A84=C08D;C4 A0034=60=B860:0;8:4 58A84=C08?0=90=6G08BC
J 8:0A41C0714=301414=BC:10;:G0=6D;C4=G0
30=AC7C=G0 38?0=0A:0=A0?08AC7C14=30B4A41CB0:0=4C08A478=660D;C4
=G04=9038 4A0=G0?4B01070=D;C414=301414=BC:C0=630?0B38BC;8A
30;014=BC:?4A000=148:CB 8A0;=G0D;C4:C1CA
J
4=60=40AC::0=7060
V V
V
0:034.
4=9038
4=60=40AC::0=7060 A J
0:034. 4=9038
J
0B0C J
J
J
Gambar 7.7
Pemuaian luas pada keping logam.
2
Gambar 7.8
Pemuaian volume pada benda berbentuk balok.
2 2
2
3 2
2
2
J
0B0C J
J
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157
Kalor
41C0714A814D;C4 38?0=0A:0=308 IA0?08
I 8:00AA094=8A14A8
?030AC7C I030;07 :6 30=:458A84=C08?0=90=6=G0
K
J
I 78BC=6;070AA094=8A14A8?030AC7C
I
8
8:4B07C8 :6
3
K
J
K
J
I I
+;C414A8A4B4;0738?0=0A:0=030;07 .
K
J
I
I 4B4;0738?0=0A:0=D;C414=3014C107B4B0?80AA0=G0B4B0?
3
7.200 kg 1,033 m
m V
:6 0380AA094=8A14A84=9038:6
Contoh
7.6
b. Pemuaian Zat Cair
030CC=G0?4C080=H0B20870=G030?0B3800B84;0;C8?4C107 0=D;C4=G0 8:0A41C071490=064;0AG0=6148A80870?8?4=C7
38?0=0A:0=A4B4;07:4=08:0=AC7C080:0=BC?0748AB8E0B4A41CB 30?0B38B40=6:0=A410608148:CB
C0=6640:?0B8:4;?030H0B208;418714A0308?030C0=6640: ?0B8:4;?030H0B?030B 8:0:43C0H0B8BC4=60;08?40=0A0=A4200
14A000=?0B8:4;?030H0B208;4187;4;C0A014640:3810=38=6:0= 34=60=?0B8:4;H0B?030B;47:04=08BCD;C408;418724?0B14
B0107308?030D;C41490=064;0AA478=660080:0=BC?0748AB8E0 B4A41CB4=C=9C::0=107E0:458A84=C08D;C4H0B208;418714A0
308?030:458A84=C08D;C4H0B?030B c.
Anomali Air
4;07=30:4B07C8107E098:0H0B?030BH0B20830=60A38?0=0A:0= H0BH0B B4A41CB 0:0= 4C08 A488=6 34=60= :4=08:0= AC7C 030
?40=0A0=0803070;4=08:G0=630?0B3800B88A0;:0==30 40=0A:0=0814AC7C IA410=G0:
2
80=B00AC7C I 30=
ID;C4080:0=4=GCACBB4B0?80AA0=G0B4B0?A478=6600AA0
94=8A=G0=08: 8:0?40=0A0=38B4CA:0=A0?08380B0A
ID;C408 0:0=4C08A4?4B8H0BH0BG0=6;08=850B?4C080=08G0=6B830:
B40BC 8=8 38A41CB A4?4B8 38?4;870B:0= 6058: ?030
.3
086058:B4A41CBB0?0:107E0308 IA0?08
ID;C408 B4CA4=GCACBA0?08:C0=6308
2
4C380=380B0AAC7C
I D;C4084C08
=BC: 0AA0 G0=6 A00 ?030 AC7C I D;C4 4A ;4187 14A0 308?030D;C40840B80AA094=8A4A;4187:428;308?0300AA0
94=8A 08 :04=0 0AA0 94=8A 1410=38=6 B410;8: 34=60= D;C4 BC;07A4101=G01=6:070=4A30?0B4=60?C=6380B0A08
Gambar 7.9
Grafik peristiwa anomali air
1.0000 1.0002
1.0004 1.0006
1.0008 1.0010
1.0012 1.0014
1.0016
18 16
12 8
4 Suhu °C
V olume cm
3
Sumber:
Physics, 1980
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
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158
Mudah dan Aktif Belajar Fisika untuk Kelas X
d. Pemuaian Gas 030
2643-44;0:C:0=?42100=G0=64=C=9C::0=
107E0A4C060A4C0834=60=:458A84=C08G0=6A00G08BCA4 14A0
4=CCB=G0 98:0 A4C0 60A 38?0=0A:0= D;C4 30= B4:0=0==G014C107
.6415,551 470B8:0= .3 41C07B01C=6:0;4=6BCBC?=G038148
?8?0:428;G0=6C9C=6=G038?0A0=6810;=:4?4A 8:0B01C=6B4A41CB 38?0=0A:0=AC7C60AG0=6140303830;0B01C=60:0==08:030:4
0300=8BC?0B8:4;60A14640:A0;8=61434A0:0=:4A460;000730= 0AC::430;010;=4;0;C8?8?0A478=66010;=414A048AB8E0
B4A41CB4=C=9C::0=B4;07B49038?4C080=60A3830;0B01C=6?030 B4:0=0=B4B0?A0034=60=B4:0=0=C30038;C0B01C=6
058:AC7CB47030?D;C4?030?A4A?40=0A0=60A34=60= B4:0=0=B4B0?B0?0:?030.3
086010B4A41CBB0?0:107E098:0B4:0=0=60AB4B0?D;C4 60A1410=38=6;CCA34=60=AC7C=G04;0=9CB=G0?4=G0B00=8=838A4
1CBC:C70;4A 42000B40B8A7C:C8=838BC;8A
0B0C J
4B40=60= D;C40E0;
D;C460AA4B4;0738?0=0A:0= AC7C0E0;
AC7C60AA4B4;0738?0=0A:0= 4A0=G0?4B01070=D;C460AA00B38?0=0A:0=44=C78?4
A000=G0=6A0034=60=14A0=G0?4B01070=D;C4H0B?030BG08BC J
;47:04=0 30=
1 273
0:0
.44170-6.551 470B8:0= .3
030 10680= C9C=6 ?8?0 38BCBC? 0?0B
34=60= AC10B ?00;= 8:0 B01C=6 ?030 .3
38?0=0A:0= D;C460AB4B0?:810B=G0B4:0=0=60A14B010714A0 038?030
?40=0A0=8=8D;C460AB4B0?A430=6:0=B4:0=0==G014C107 =3030?0B4=64B07C87C1C=60=0=B00:4=08:0=AC7C30=B4
:0=0= ?030 ?40=0A0= 60A ?030 D;C4 B4B0? 34=60= 4;0:C:0= ?42100=148:CB
J
Gambar 7.10
Gas dipanaskan pada tekanan tetap pipa
balon
tabung kaleng
balon sebelum
memuai
bunsen
Gambar 7.11
Volume gas sebagai fungsi dari suhu pada pemanasan gas dengan tekanan
tetap. 100
200 300
400 500
V
TK
Gambar 7.12
Gas dipanaskan pada volume tetap pipa
sumbat
tabung
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Kalor
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.
Gambar 7.13
Perangkat percobaan untuk mengetahui hubungan antara
kenaikan suhu dan tekanan. Bourdon
termometer raksa
bunsen kaki
tiga
42100=A4C?034=60=:4680B0=B4A41CB?4=0738;0:C:0=?00 07;88A8:00870A8;?42100=G0=644:0;0:C:0=38?4;476058:
B4:0=0=60AB47030?AC7C?030D;C4B4B0?1414=BC:608A;CCA;870B :410;8.3
42000B40B8:0608A8=844=C78?4A000=
0B0C 4B40=60=
B4:0=0=60A0B AC7C60A
:=AB0=B0 08?4A000=
38:4B07C8107E0B4:0=0=60A1410=38=6
;CCA 34=60= AC7C=G0 ?030 D;C4 B4B0? 4;0=9CB=G0 ?4=G0B00= B4A41CB38A41CBC:C0GCAA02
J
Gambar 7.14
Grafik tekanan terhadap suhu
TK pPa
Tes Kompetensi Subbab
A
3+,--.6,6-5 8BC=610=G0::0;G0=638?4;C:0=C=BC:4=6C107
0
:64A308 I4=903808 I 1 64A308 I4=9038C0?
I
41C0764;0A148A8 608?030 I40?0604A G0=6AC7C=G0J I70CA3820?C:430;008
A478=660AC7C0:7820?C0=4A30=084=9038 I 38:4B07C8:0;94=8A08
:6108:0=
?4BC:00=:0;34=60=64;0A 0301490=0G0=6148A8 60814AC7C I380AC:
:0=4A10;:G0=60AA0=G0 614AC7CJ
I 8BC=6;07AC7C0:7820?C0=30=0AA04AG0=64;41C
4B8=6680=08B49C=38:4B07C8
8:0A4;CC74=468 4:0=8:38C1074=90384=468:0;:0;94=8A08
:630=
A 78BC=6;07:4=08:0=AC7C
08B49C=A4B4;0790BC7 4BC=9C:6C=0:0=?4A000=4=468?B4=A80;
4=468:0;C=BC:4=08::0=AC7C 40?0 10=G0:=G0 :0; G0=6 38?4;C:0= C=BC:
4=6C1074AJ G0=60AA0=G0
604=9038
08 98:0:0;94=8A4A :0;6I:0;94=8A08
:0;6I30=:0;;41C4A :0;6470B8:0= 6010148:CB
–5°C 0°C
0°C 20°C
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160
Mudah dan Aktif Belajar Fisika untuk Kelas X
B. Perpindahan Kalor
8:08=68=40A0:0838?4;C:0=:0;G0=6140A0;3080?8:? 030?40=0A0=A4?0=2808C;0C;0:0;3080?8:?14?8=307
:4?0=28B4?0B08:4C380=:0;308?0=28B4A41CB14?8=307:408 4?8=3070=:0;B4903830814=3014AC7CB8=668?0=0A:414=30
14AC7C;41874=30738=68=4=30G0=6?0=0A4148:0=:0;:4 ?03014=30G0=638=68=40=308=G0:0;B830:30?0B14?8=3070:0=
AC;8BC=BC:40A0:08A4101:0;3080?8:?B830:30?0B14?8=307 :408
30 B860 14=BC: ?4?8=3070= :0; G08BC 70=B00= +0;80=30=?0=200=
1. Perpindahan Kalor Secara Konduksi
8A0;:0==3040=0A:0=A0;07A0BCC9C=610B0=6;60A4?4B8 .3
4=BC=G0?0B8:4;?0B8:4;?030C9C=6;60G0=638?0=0A8
1464B0;418724?0B40:8=14A09C;07:0;G0=638148:0=:4?030 C9C=6 ;60 8BC A40:8= 24?0B 64B00= ?0B8:4;=G0 410680= 4=468
:8=4B8:G0=6388;8:8?0B8:4;G0=61464B0B4A41CB38148:0=:4?030 ?0B8:4;?0B8:4; 38 34:0B=G0 4;0;C8 BC1C:0= :810B=G0 ?0B8:4;
?0B8:4;G0=638BC1C:8BC8:CB1464B0348:80=A4B4CA=G078=660 64B00=?0B8:4;A0?08:4C9C=6;60G0=6B830:38?0=0A8
4010B0=64B00=?0B8:4;B4A41CB38A4B0834=60=?4010B0= :0;308C9C=6;60G0=638?0=0A8A0?08:4C9C=6;60G0=6B830:
38?0=0A8:810B=G0C9C=6;60G0=6B830:38?0=0A84=9038?0=0A 010B0=:0;8=8B830:38A4B0834=60=?4?8=3070=?0B8:4;?0B8:4;
;60 38A41CB
0B0C .3 4?4;870B:0=A410B0=6;6034=60=;C0A?4=0?0=6
30=?0=90=610B0=6 38?0=0A:0=A0;07A0BCC9C=6=G0030?48AB8E0 B4A41CB:0;0:0=4010B:4C9C=6A414;07:0=0=G0=6AC7C=G0
;41874=307430A0:0=?4=4;8B80=G0=6B4;0738;0:C:0=38?4;47 107E0?4?8=3070=:0;A4200:=3C:A81460=BC=6?03094=8A;60
;C0A ?4=0?0=6 ?4=670=B0 :0; ?414300= AC7C 308 C9C=6C9C=6 ;60B4?0B:0;4010BA4B0?0=90=690;0=G0=638;0;C8;47:0;
4?8=3070=:0;A4B80?A0BC0=E0:BC38CCA:0=A410608148:CB
Gambar 7.15
Contoh perpindahan kalor secara konduksi.
Gambar 7.16
Hantaran kalor T
1
T
2
A
J J
J 4B40=60=
:0;G0=64010B?4A0BC0=E0:BC A0B0CE0BB :458A84=:=3C:A8B40;H0B A0B0C,
;C0A?4=0?0=610B0=6 ?0=90=610B0=6
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161
Kalor
-
30=-4?4;870B:0=:458A84=:=3C:A8B40; H0B1C:0=;6030=;60
. .4; A
0B0407 4B=
020 0GC:48=6
01CA BG50
08=B410; 300
- 458A84==3C:A840;-0B
C:0=600B00B0 . .4;
40: 41060
;C8=8C C=8=60=
810; 090
0:A0 -
458A84==3C:A840;60 0B00B0
2. Perpindahan Kalor Secara Konveksi
030 A00B =30 40=0A:0= 08 38 30;0 ?0=28 :0; 0:0= 14?8=307 308 30A0 ?0=28 :4 ?4C:00= 08 A4200 :=D4:A8
4?8=3070=:0;A4200:=D4:A838A4B08640:0=0AA00B0C640:0= ?0B8:4;?0B8:4;H0B?40=B00=G0=D4:A870=G0B49038?030H0BG0=6
30?0B4=60;85;C830 303C0200?4?8=3070=:0;4;0;C870=B00=:=D4:A8G08BC
:=D4:A8A42000;080730=:=D4:A8?0:A0 a.
Konveksi Alamiah 8A0;:0==3040=0A:0=A4?0=2808A4?4B8?030.3
4B4;07 08 38 10680= 10E07 ?0=28 4=480 :0; 08 B4A41CB 0:0= 4C08A478=6600AA094=8A=G0;4187:428;308?0300AA094=8A0838
10680= 0B0A 414300= 0AA0 94=8A B4A41CB 4=60:810B:0= ?0B8:4; ?0B8:4;08G0=6140AA094=8A;4187:428;0:0=14640::40B0A
4?0BG0=638B8=660;:0=?0B8:4;08G0=6140AA094=8A;4187:428; 0:0=B48A8;47?0B8:4;08G0=6140AA094=8A;418714A048AB8E0
B4A41CB14;0=6AC=6B4CA4=4CAA478=660?0B8:4;?0B8:4;0830;0 ?0=28 14?CB0 =08: 30= BCC= ;80=0;80= ?0B8:4; G0=6 14640:
B4A41CB38A4B0834=60=?4?8=3070=:0;4?8=3070=:0;34=60= 4=60;8:0=?0B8:4;?0B8:4;08A4?4B88=838A41CB+
=D4:A8 0;0807 10=G0: 389C?08 38 ?018:?018: G0=6 4=6 6C=0:0=241=60A0?0A70A8;?410:00=48;8:80AA094=8A;4187
:428; 308?030 0AA0 94=8A C300 38 A4:8B0=G0 :810B=G0 60A 70A8; ?410:00=0:0=4=60;8:40B0A4?0BG0=638B8=660;:0=;4760A
70A8;?410:00=0:0=388A8;47C300A4:8B0G0=648;8:80AA094=8A ;418714A0308?0300AA094=8A60A70A8;?410:00=
=68=;0CB30=0=68=300BB490381430A0:0=:=D4:A80;0807 C300=68=;0CB30=0=68=300BG0=6380=500B:0==4;0G0=C=BC:
14;0G0B490384;0;C8:=D4:A80;0807C300380=0?0=0A38?8=307 :0=308AC0BCB4?0B:4B4?0B;08=34=60=?4640:0=?0B8:4;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.
Tantangan
untuk Anda
Gambar 7.17
Peristiwa konveksi alamiah
Sumber:
Fisika Universitas, 2002
Sumber:
Fisika Universitas, 2002
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162
Mudah dan Aktif Belajar Fisika untuk Kelas X
38?8=307:0=030A80=6708300B0=;418724?0B?0=0A308?030;0CB A478=660C300?0=0A380B0A300B0==08:30=B4?0B=G03860=B8:0=
C30038=68=3080B0A;0CB8=8G0=638A41CB0=68=;0CB0300;0708 B49038A410;8:=G000B0=;418724?0B38=68=308?030;0CBA478=660
C300380B0A;0CB=08:30=B4?0B=G03860=B8:0=;47C3003080B0A 300B0=G0=638A41CB0=68=300B
b. Konveksi Paksa
=D4:A8 ?0:A0 10=G0: 386C=0:0= ?030 A8AB4 ?4=38=68= 4A8= 8A0;=G0?0304A8=18;4A8=:0?0;;0CB4A8=384A4;AB0A8=430=
:8?0A0=68= =D4:A8 ?0:A0 A4?4B8 ?030 .3 38?0:08 30;0 A8AB4
?4=38=68=4A8=18;84=60;838A4:8B0C0=64A8=4;0;C8?8?0 ?8?03810=BC;47A41C07??008,0;G0=638B480
4A8=18;30870A8;?A4A?410:00=4=20?08AC7C
I030 AC7C8=84C=6:8=:0=4A8=18;4C084;4187810B0A:400=0=
30=4=60:810B:0=10680=10680=4A8=18;4=9038;4074CA0:0= ?4B00G0=6A48=6389C?08030;07?030:?A8;8=344A8=4=9038
4;4=6:C=64=60C7148:CB=G0D8A:A8B0A8=G0:?4;C0A4=9038 4=3074=24
0=0A?0304A8=18;14?8=307;47A8:C;0A8084=C9C:40 380B 300 38=68= 308 ;C0 4A8= 38B08: ;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= 14A4=BC70= 34=60= 1;:1;: 4A8= C=BC: 4=6C;0=6
A8:C;0A8148:CB=G0 0385C=6A80380B030;074=9060AC7C4A8= 060B830:4;0?0C810B0A?0=0AG0=6388H8=:0=
C;07 4=468 :0; ?4 A0BC0= E0:BC G0=6 38B480 ;47 5;C830 A4:8B0=G0 A4200 :=D4:A8 030;07 A410=38=6 34=60= ;C0A ?4C:00=
14=30G0=614A4=BC70=34=60=5;C83034=60=1430AC7C
t
4200 0B40B8A38BC;8A
Q t
0B0C J
4B40=60= 9C;07:0;?4A0BC0=E0:BC A
:458A84=:=D4:A8B40; A I
;C0A?4=0?0=6?4?8=3070=:0;
t
?414300=AC7CG0=638?0=0A834=60=AC7C5;C830I 8;08:458A84=:=D4:A8B40;1460=BC=6?03014=BC:30=:43C3C:
0=?4C:00=0894=8A5;C830G0=614A4=BC70=34=60=?4C:00= J
C7CC30030;0A41C07C0=60=A414A0 IA430=6:0=AC7C?4C:00=94=34;0 ?030C0=60=B4A41CB I40?0;09C:0;G0=638B480;4794=34;0:020A4;C0A
98:0:458A84=:=D4:A8C300A00B8BCK
J
:0;A I
Contoh
7.7
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?
Tantangan
untuk Anda
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163
Kalor
3. Perpindahan Kalor secara Radiasi
0=0A0B07084C?0:0=AC144=468B414A01068:4;0=6AC=60= 783C?0:7;C:381C80=0A3080B0708A0?08:4C8B830:4;0;C8
70=B00= :=3C:A8 0B0C?C= 0;80= :=D4:A8 A4101 :=3C:A8 30= :=D4:A8 44;C:0= H0B ?40=B00 A430=6:0= 0=B000B0708 30=
C8B430?0BC0=670?00B0C34=60=:0B0;08=B830:030H0B?40=B00 0=0A :0; 308 0B0708 A0?08 :4 C8 B49038 A4200 0380A8
?0=200=0;3080B0708A0?08:4C830;014=BC:64;10=6 4;4:B06=4B8: 038 030;07
4C:00=14=30G0=614E0=078B030?0B4=G40?30=4 0=20:0=4=468:0;0380A834=60=108:A430=6:0=?4C:00=14=30
14E0=0?CB874=G40?30=40=20:0=:0;0380A834=60=1CC: 0=4;ACG0386C=0:0=C=BC:4=G40?30=40=20:0=
0380A8A8=00B0708;47:04=08BC1830=6;6014=660=G038148 E0=078B0=468:0;0380A8380=500B:0=C=BC:40=0A:0=08
18;18;B0=6:8?4=60=6:CB8=G0:?03010680=0B0AB0=6:83820B 34=60=E0=0?CB870;B4A41CB380:AC3:0=6C=04=678=308?4=G4
0?0=4=468?0=0AA4200:=D4:A8;478=G0: 4C:00=14=3078B0;418710=G0:4=G40?30=40=20:0=
4=468:0;A410;8:=G0?4C:00=14=3014E0=0?CB87;4187A438:8B 4=G40?30=40=20:0=:0;0380A8
a. Api Unggun
4=07:07=304;0:C:0=3830407?46C=C=60=C307 4=903870;G0=6180A03830;0?4:4070=41C0B0?8C=66C=?8
C=66C=381C0B3080=B8=60=B8=6?7=:48=6G0=63810:04B4;0738 1C0B0?8C=66C=B4=BC=304=903840A070=60BE0;0C?C=BC1C7
=30B830:14A4=BC70=34=60=0?80=0A0?84=60;84;0;C8C300 ;418710=G0:A42000380A83080?8C=66C=4=64=08BC1C7=30308
?030A4200:=D4:A8 b.
Rumah Kaca
830:A4C064;10=60380A8A8=00B070830?0B4010B0AC: :430;0C07:0200=G02070G0B0?0:G0=630?0B4010B38=38=6
:0200B0C?;0AB8:A430=6:0=A8=0C;B0D8;4B30=A8=08=5040738 ?0=BC;:0=:410;8;4738=38=6:020=468:0;0380A83082070G0B0?0:
38A40?;47B0=0730=B0=00=3830;0C0=6C07:020410;8:=G0 B0=0730=B0=00=0:0=40=20:0=:410;864;10=60380A814C?0
A8=0 8=50407 0=90=6 64;10=6 G0=6 ;4187 14A0 4=G4101:0= 64;10=68=50407B4?40=6:0?;4738=38=6:020A478=660AC7CC0=60=
4=9038;418770=60B30=B0=00=30?0B783C?34=60=A460
8
J IJ I
I
K
J
:0;A I
K
J
:0;A I
I :0;
038;09C:0;G0=638B480;4794=34;0:020
:0;
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
untuk Anda
Information for You
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164
Mudah dan Aktif Belajar Fisika untuk Kelas X
0?08G0=638:4;C0:0=;47B0=0730=B0=00=4=G4101:0= ?4=6C0?0=B4?40=6:0?;4738=38=6:020;80=C0?08:40B0AA4200
:=D4:A8B830:1490;0=A478=660C0?08:410;890BC7:430;0B0=07 30=B0=00=30;014=BC:H0B2080300:78=G0:4;41010=C300
30;0C07:02030?0BB49060A4?0=90=60380A8:0;0B070814 ;0=6AC=6
c. Radiasi Benda Hitam
0415
J
30=68 0-5:.
J 4=G0B0:0= 107E0 14A0=G0 4=468 G0=6 38?0=20:0= ;47 AC0BC
?4C:00=?4A0BC0=E0:BC?4A0BC0=;C0AA410=38=634=60=?0=6:0B 4?0BAC7C?4C:00=8BC30?0B38BC;8A34=60=?4A000=
4B40=60= 4=468G0=638?0=20:0=0B0C38A40??4A0BC0=E0:BC?4A0BC0=
;C0A A 0B0CE0BB
:=AB0=B0CCB450=;BH0==K
J
E0BB AC7CCB;0:
48A8D8B0A?4C:00=B830:14A0BC0= =BC:14=3078B0A4?C=07060
14=3014=30;08=7060
:458A84=48A8D8B0A=G0;4187:428;308?030A0BCA430=6:0=C=BC:14=30 14E0=0?CB87A4?C=0 0601460=BC=6?030:40300=?4
C:00=14=30G08BC:4:0A00==G0A4B0E0=030814=30=468G0=6 38?0=20:0=;47A41C0714=3030;0A0BC0=9C;438B4=BC:0=34=60=
?4A000= J
J 4B40=60=
4=468G0=638?0=20:0=;47?4C:00=14=30 ;C0A?4C:000=14=30
;00E0:BC48A84=468A4:= 8:03810=38=6:0=B47030?;8=6:C=60=G0=614AC7C
?4A000= =G00:0=4=9038
J
41C0714=3048;8:8?4C:00=78B0A4?C=014AC7C
IC0A?4C:00= 2
40=20:0=4=468:4;8=6:C=60=G0=614AC7CI4=BC:0=4=468?4 A0BC0=E0:BCG0=638?0=20:0=14=30B4A41CB
8 14=3078B0
2 K
J
Contoh
7.8
J 4=40?0=:=A4?0380A830;0:4783C?0=A470870810=G0:38
9C?082=B7=G0?410:00=?0300;0B?40=660=6+?4=648=60= :?830=B410:0C30=A8AB4?4=38=68=0B0C?40=0A0=C07
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 m
2
?
Tantangan
untuk Anda
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165
Kalor
K
J
E0BB
K
J
. J
K
J
, 03814A04=468?4A0BC0=E0:BCG0=638?0=20:0=;4714=30B4A41CB030;07
,
030AC7C A41C0714=3040=20:0=4=468A414A0
A40?04=468 G0=638?0=20:0=14=30B4A41CB?030AC7C
8 8:4B07C8
A
A 0384=468G0=638?0=20:0=14=30?030AC7C 030;07
A
Contoh
7.9
Tes Kompetensi Subbab
B
3+,--.6,6-5 47?0=0A14AC7CI38BC0=6:0=:430;03C0
1C07B4:G0=6A0BCB4::408:E0=078B0 30=A0BC=G0;068B4:4=68;0B
8:0AC7CC0=6I140?0:424?0B0=78;0=6=G0 :0; 308 A4B80? B4: :4 ;8=6:C=60= 98:0 ;C0A
?4C:00=B4:030;07K
J
4=603C: B41C0B 308 0;C8=8C 34=60= ;C0A ?4=0?0=6
30=?0=90=6
2 8:0AC7C08 30;01490=0 IAC7C38C9C=6?4=603C:I
30=:=3C:B8D8B0AB40; ,IB4=BC:0= 0 :424?0B0=0;80=:0;?0300;80=:0;?030
?4=603C: 1 :0;G0=64=60;84;0;C8?4=603C:A4;00
90 ?8?0=0A14AC7CI38BC0=6:0=:430;0
20=6:8:408:E0=02:4;0B 8:0AC7C C0=6IB4=BC:0=:424?0B0=78;0=6=G0:0;308
20=6:8:4;8=6:C=60=;C0A20=6:8 2
A0A;02:4=G0B0:0=107E09C;07:0; G0=638;4?0A?030AC0BCA8AB40:0=A00
34=60=9C;07:0;G0=638B480 4C080=?030H0B?030B30?0B3800B84
;0;C8?4C1070=?0=90=6;C0A30=D;C4 030 CC=G0 ?4C080= H0B 208 70=G0
30?0B 3800B8 4;0;C8 ?4C1070= D;C4 A090
Rangkuman
0;030;074=468G0=638?8=307:0=308 AC0BC14=30:414=30G0=6;08=
0; 94=8A 30?0B 383458=8A8:0= A410608 10=G0:=G0:0;G0=638?4;C:0=AC0BCH0B
C=BC:4=08::0=AC7C
:6H0BB4A41CB A414A0
I
0?0A8B0A :0; 030;07 10=G0:=G0 G0=6 38?4;C:0=C=BC:4=08::0=AC7C14=30
A414A0
I
Kata Kunci
• konduksi
• konveksi
• radiasi
• rumah kaca
• radiasi benda hitam
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166
Mudah dan Aktif Belajar Fisika untuk Kelas X
0;030;074=468G0=638?8=307:0=308 AC0BC14=30:414=30G0=6;08=
30B86014=BC:?4?8=3070=:0;G08BC :=3C:A8:=D4:A830=0380A8
4?8=3070= :0; 4;0;C8 H0B ?40=B00 34=60=B830:38A4B08?4?8=3070=?0B8:4;
?0B8:4;H0BB4A41CBA0200?40=4=38A41CB 70=B00=0B0C:=3C:A8
4?8=3070=:0;A4200:=D4:A80;80= 38A4B08640:0=0AA00B0C640:0=?0B8:4;
?40=B00=G05;C830 0380A8 030;07 ?4?8=3070= :0; 30;0
14=BC:64;10=64;4:B06=4B8:
30?0B 14?8=307
A4200
=3C:A80=B00= =D4:A8;80=
0380A80=200=
30?0B 4=G4101:0=
-03
4C1070= ,C9C3
2=B7 ?48AB8E0=G0
4C080= 4=30
A478=660 B49038
4C1070=C7C 4;41C
414:C 4=6C0?
4=641C= 4=GC1;8
94=8A=G0
4C080=+;C4 4C080=0=90=6
4C080=C0A
Peta Konsep
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?
Refleksi
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.
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167
Kalor
0;0A8AB4A0BC0=8=B4=0A8=0;14A00=AC7C 48;8:8A0BC0=
0 =4EB= 3 :4;D8=
1 E0BB 4 :0;8
2 9C;4 0;01B;B4AB430?0B 60:?8?030AC7C
I8B0107:0=ACACA410=G0: 6014AC7C I 8:0B830:030:0;20?C0=0C?C=:0;G0=6
B4A40?1B;B4A
08 :?8
ACAC
:0;6I C7C20?C0==G0030;07
0 I 3 I
1 I 4 I
2
I C7CA410B0=61090G0=6?0=90=6=G0
38B8=6:0B:0=
308 I4=9038
IA478=660?0=90=6=G014B0107 8;84B44B01070=?0=90=6AC0BC10B0=61090
G0=6?0=90=6=G0 298:038?0=0A:0=308 IA0?08 I030;07
3 1
4 2
41C071;014=660B41C0B308?4C=66C34=60= :458A84=C08;8=40
K
J
I030 I 908908=G0
8:01;0B4A41CB38?0=0A:0=A0?08
I?4B01070=;C0A?4C:00=1;0030;07A414A0 30;0
0 K
J
3
K
J
1
K
J
4
K
J
2
K
J
30=14BCCBBCCB030;07:458A84=C08?0=90=6 :458A84=C08;C0A30=:458A84=C08D;C448:CB
8=8G0=64=C=9C::0=7C1C=60=G0=614=0308 :4B860=G0030;07
3 1
4 2
410B0=614A8C;0C;0?0=90=6=G0 38?0=0A:0=
308AC7C 4=9038
8:0:458A84=C08?0=90=6 10B0=6B4A41CB ?4B01070=?0=90=610B0=6
030;07 1
J
2
J
3
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Tes Kompetensi Bab 7
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168
Mudah dan Aktif Belajar Fisika untuk Kelas X
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A. Arus Listrik dan Muatan
B. Hukum Ohm dan Hambatan
C. Rangkaian Seri dan Paralel
D. Hukum II Kirchhoff
E. Sumber Arus Searah dari
Proses Kimiawi
F. Tegangan Listrik Searah dan Bolak-
Balik
Listrik Dinamis
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3870?8A4C0AC3CB:B0AC307B4?0A0=60?C;0?C8=84C?0:0= 2=B7?40;0B0=30;0:4783C?0=A4708708F0=640=500B:0=4=468
;8AB8: A4B4;07 38C107 4=9038 4=468 2070F0 0;C 106080=0 4=468 ;8AB8:8=830?0B4=F0;0:0=;0?C=300:0=4?4;0908=F0?030
1018=8 303C094=8A0CA;8AB8:F08BC0CA;8AB8:1;0:10;8:20,2,
300,230=0CA;8AB8:A400702300,20300CA ;8AB8:1;0:10;8:C0B0=;8AB8:4=60;830;03C00071;0:10;8:
30?C=?0300CA;8AB8:A4007C0B0=;8AB8:70=F04=60;830;0 A0BC007=B7?40;0B0=;8AB8:F0=64=66C=0:0=0CAA4007F08BC
:0;:C;0B0+-2 -,20- 90 30= ;0?C A4=B4 030 101 8=8 0:0= 38?4;0908 0CA ;8AB8: A4007 430=6:0= 0CA ;8AB8: 1;0:10;8: 0:0=
38?4;0908;418710=F0:38:4;0A.
169
• memformulasikan besaran-besaran listrik rangkaian tertutup sederhana satu loop;
• mengidentifikasi penerapan listrik AC dan DC dalam kehidupan sehari-hari;
• menggunakan alat ukur listrik.
Setelah mempelajari bab ini, Anda harus mampu:
menerapkan konsep kelistrikan dalam berbagai penyelesaian masalah dan berbagai produk teknologi.
Hasil yang harus Anda capai:
Nyala lampu pada malam hari, selain berfungsi sebagai penerangan juga menjadi bagian dari keindahan kota.
Bab
8
Sumber:
Young Scientist,1994
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170
Mudah dan Aktif Belajar Fisika untuk Kelas X
2
A. Arus Listrik dan Muatan
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