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
Di unduh dari : Bukupaket.com
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
Kategori