BAB IV

4.1 Data Pengamatan

NO Massa (m)

Arus (I) Tegangan ᵼ₁ ᵼ₂ ᴧᵼ T₁ T₂ ᴧT

1 0,053 1,7 2,4 0 26 26 31 33 2˚C2 0,053 1,8 2,6 26 215 189 33 35 2˚C3 0,053 1,8 2,6 215 423 208 35 37 2˚C4 0,053 1,8 2,7 423 676 253 37 37 2˚C5 0,053 1,9 2,6 676 886 210 39 39 2˚C

Massa calorimeter =167,27 gr = 0,167 Kg

Massa calorimeter + air = 221,18 gr = 0,221 Kg

Massa air = 53,91 gr = 0,053 gr

4.2 Analisis Data

4.2.1 Perhitungan Tanpa KTP

4.2.1.1 Energi Listrik

W = V.Ī.t

W₁ = V.Ī.t

= 2,4 x 1,7 x 26

= 106, 08 ĵ

W₂ = V.Ī.t

= 2,6 x 1,8 x189

= 884,52 ĵ

W₃ = V.Ī.t

= 2,6 x 1,8 x 208

= 973,44 ĵ

W⁴= V.Ī.t

= 2,7 x 1,9 x 253

= 1229,58 ĵ

W⁵ = V.Ī.t

= 2,6 x 1,9 x 210

= 1037,4ĵ

4.2.1.2 Daya Listrik

P = V.Ī

P₁ = V.Ī.

= 2,4 . 1,7

= 4,08 Watt

P₂= V.Ī.

= 2,6 . 1,8

= 4,68 Watt

P₃= V.Ī.

= 2,6 . 1,8

=4,68 Watt

P₄= V.Ī.

= 2,7 . 1,8

= 4,86 Watt

P₅= V.Ī

= 2,6 . 1,9

= 4,94 Watt

4.2.1.3 Hambatan

R =VĪ

R₁ = V1,

= 2,41,7

= 1,41 Ω

R₂ = VĪ

= 2,61,8

= 1,44 Ω

R₃ = VĪ

= 2,61,8

= 1,44 Ω

R₄ = VĪ

= 2,71,8

= 1,5 Ω

R₅ = VĪ

= 2,61,9

= 1,37 Ω

4.2.1.4 Energi Kalor

Q= (M .C .∆T+C .∆T )

Q ₁=(M ₁.C . ∆T ₁+C .∆T ₁ )

=(0,053 . 1 . 2 + 75,2 . 2)

= (0,106 + 150,4)

= 150,506 kal

Q ₂=(M ₂.C . ∆T ₂+C .∆T ₂ )

= (0,053 . 1 . 2 + 75,2 .2)

=0,106 + 150,4

=150,506 kal

Q ₃=(M ₃.C . ∆T ₃+C .∆T ₃ )

= (0,053 . 1 . 2 + 75,2 .2)

= 0,106 + 150,4

= 150,506 kal

Q ₄=(M ₄.C . ∆T ₄+C .∆T ₄ )

=(0,053 . 1 . 2 + 75,2 .2)

= 0,106 + 150,4

= 150,506 kal

Q5=(M 5 .C .∆T 5+C .∆T5 )

=(0,053 . 1 . 2 + 75,2 .2)

= 0,106 + 150,4

= 150,506 kal

4.2.2 Perhitungan dengan KTP

∆V=13xnst voltmeter

= 13x 0,2

= 0,067

∆ I=13x nst Amperemeter

= 13x 0,1

= 0,033

∆ t=13x nst stopwatch

= 13x 0,1

=0,033

4.2.2.1 Energi Listrik

∆W=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

∆W ₁=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

= √ (1,7 .26 .0,067 )2+ (2,4 .26 .0.033 )2+(¿2,4 .1,7 .0,033) ² ¿

= √ (2,96 )2+(2,05 )2+(0,13)²

= √8,76+4,20+0,02

= √12,98

= 3,60 Joule

∆W ₂=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

= √ (1,8 .189 .0,067 )2+(2,6 .189 .0.033 )2+(¿2,6 .1,8 .0,033)² ¿

= √ (22,79 )2+(16,22 )2+(0,15) ²

= √519+263,09+0,02

= √782,49

= 27,97 Joule

∆W ₃=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

= √ (1,8 .208 .0,067 )2+(2,6 .208 .0.033 )2+(¿2,6 .1,8 .0,033) ² ¿

= √ (25,08 )2+(17,85 )2+(0,15) ²

= √629,01+17,85+0,02

= √947,65

= 30,78 Joule

∆W ₄=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

= √ (1 ,8. 253 .0,067 )2+(2,7 .253 .0.033 )2+(¿2,7 .1,8 .0,033)² ¿

= √ (30,51 )2+ (22,54 )2+(0,16) ²

= √930,86+508,05+0,02

= √143,93

= 37,93 Joule

∆W ₅=¿

= ¿

= √ ( I . t .∆V )2+(V . t .∆ I )2+(V . I .∆ t )2

= √ (1,9 .210 .0,067 )2+(2,6 .210 .0.033 )2+(¿2,9 .1,9 .0,033)² ¿

= √ (26,73 )2+(18,02 )2+(0,16) ²

= √714,49+324,72+0,02

= √1039,23

= 32,24 Joule

4.2.2.2 Daya Listrik

∆ p={( ∆P∆V )

2

(∆V )2+( ∆P∆ I )

2

(∆ I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

∆ p₁={( ∆ P∆V )

2

(∆V )2+(∆ P∆ I )

2

(∆I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

=√ (1,7 )2 (0,067 )2+(2,4 )2 (0,033 )2

=√(2,89 .4,49 x 10¯ ³¿)+(5,76 .1,09 x10¯ 3)¿

=√0,013+6,28 x10¯ ³

=√0,02

= 0,14 Watt

∆ p₂={( ∆ P∆V )

2

(∆V )2+(∆ P∆ I )

2

(∆I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

=√ (1,8 )2 (0,067 )2+(2,6 )2 (0,033 )2

=√(3,24 .4,489x 10¯ ³¿)+(6,76 .1,09 x10¯ 3)¿

=√0,014+7,36 x 10¯ ³

=√0,02

= 0,14 Watt

∆ p₃={( ∆ P∆V )

2

(∆V )2+(∆ P∆ I )

2

(∆I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

=√ (1,8 )2 (0,067 )2+(2,6 )2 (0,033 )2

=√(3,24 .4,489x 10¯ ³¿)+(6,76 .1,09 x10¯ 3)¿

=√0,014+7,36 x 10¯ ³

=√0,02

= 0,14 Watt

∆ p₄={( ∆ P∆V )

2

(∆V )2+(∆ P∆ I )

2

(∆I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

=√ (1,8 )2 (0,067 )2+(2,6 )2 (0,033 )2

=√(3,24 .4,489x 10¯ ³¿)+(6,76 .1,09 x10¯ 3)¿

=√0,014+7,36 x 10¯ ³

=√0,02

= 0,14 Watt

∆ p₅={( ∆ P∆V )

2

(∆V )2+(∆ P∆ I )

2

(∆I ) ² } 12

= √ ( I )2 (∆V )2+(V )2(∆I ) ²

=√ (1,8 )2 (0,067 )2+(2,6 )2 (0,033 )2

=√(3,24 .4,489x 10¯ ³¿)+(6,76 .1,09 x10¯ 3)¿

=√0,014+7,36 x 10¯ ³

=√0,02

= 0,14 Watt

4.2.2.3 Hambatan

∆ R=¿

= ¿

= ¿

∆ R₁=¿

= ¿

= ¿

={( 0,0671,7 )

2

+( 2,4.0,0331,7 ²

) ² } 12

={(0,04)²+(0,02)²}12

={16 x 10¯ ⁴+4 x 10¯ ⁴ }12

=√{16 x10¯ ⁴+4 x10¯ ⁴ }

=√2x 10¯ ³

=0,04Ω

∆ R₂=¿

= ¿

= ¿

={( 0,0671,8 )

2

+( 2,6.0,0331 ,8²

) ² } 12

={(0,04)²+(0,02)²}12

={16 x 10¯ ⁴+9 x10¯ ⁴ }12

=√{16 x10¯ ⁴+9x 10¯ ⁴ }

=√25 x10¯ ⁴

=0,05Ω

∆ R₃=¿

= ¿

= ¿

={( 0,0671,8 )

2

+( 2,6.0,0331 ,8²

) ² } 12

={(0,04)²+(0,02)²}12

= {16 x 10¯ ⁴+9 x10¯ ⁴ }12

=√{16 x10¯ ⁴+9x 10¯ ⁴ }

=√25 x10¯ ⁴

=0,05Ω

∆ R₄=¿

= ¿

= ¿

={( 0,0671,8 )

2

+( 2,7.0,0331 ,8²

) ² } 12

={(0,04)²+(0,03)²}12

=√{16 x10¯ ⁴+9x 10¯ ⁴ }

=√25 x10¯ ⁴

=0,05Ω

∆ R₅=¿

= ¿

= ¿

={( 0,0671,9 )

2

+( 2,6.0,0331 ,9²

) ² } 12

={(0,03)²+(0,02)²}12

={9x 10¯ ⁴+4 x 10¯ ⁴ }12

=√{9 x 10¯ ⁴+4 x 10¯ ⁴ }

=√13 x10¯ ⁴

=0,04Ω

4.2.2.4 Energi Kalor

∆ m=13xnst ohauss

= 13x 0,01

=3,33x10¯³ gr

=3,33x10¯⁶Kg

∆T=13x nst termometer

=13x10

=3,3℃

∆Q={( ∂Q∂m x∆ m)2

+( ∂Q∆Tx ∆T )

2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

∆Q ₁={( ∂Q∂m x ∆m)2

+( ∂Q∆T x∆T )2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

={(1.3,3)x3,33x10¯⁶)²+(0,053.1+75,2)x3,3)²}12

={(1,098x10¯⁵)²+(248,33)²}12

={(1,21x10¯¹°)+(6617,79)}12

=√61667,79

=248,33 kal

∆Q ₂={( ∂Q∂m x ∆m)2

+( ∂Q∆T x∆T )2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

={(1.3,3)x3,33x10¯⁶)²+(0,053.1+75,2)x3,3)²}12

={(1,098x10¯⁵)²+(248,33)²}12

={(1,21x10¯¹°)+(6617,79)}12

=√61667,79

=248,33 kal

∆Q ₃={( ∂Q∂m x ∆m)2

+( ∂Q∆T x∆T )2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

={(1.3,3)x3,33x10¯⁶)²+(0,053.1+75,2)x3,3)²}12

={(1,098x10¯⁵)²+(248,33)²}12

={(1,21x10¯¹°)+(6617,79)}12

=√61667,79

=248,33 kal

∆Q ₄={( ∂Q∂m x ∆m)2

+( ∂Q∆T x∆T )2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

={(1.3,3)x3,33x10¯⁶)²+(0,053.1+75,2)x3,3)²}12

={(1,098x10¯⁵)²+(248,33)²}12

={(1,21x10¯¹°)+(6617,79)}12

=√61667,79

=248,33 kal

∆Q ₅={( ∂Q∂m x ∆m)2

+( ∂Q∆T x∆T )2

}12

={((C.∆T ¿ x∆m ¿ ²+((m.c+C ) x ∆T ) ² }12

={(1.3,3)x3,33x10¯⁶)²+(0,053.1+75,2)x3,3)²}12

={(1,098x10¯⁵)²+(248,33)²}12

={(1,21x10¯¹°)+(6617,79)}12

=√61667,79

=248,33 kal

4.2.3 KTP Mutlak

4.2.3.1 Energi Listrik

W ±∆W

(W₁ ±∆W ₁) = (106,08 ±3,60¿ ĵ

(W₂ ±∆W ₂) = (884,52 ±27,97¿ ĵ

(W₃ ±∆W ₃) = (973,44 ±30,78¿ ĵ

(W₄ ±∆W ₄) = (1229,58 ±37,93¿ ĵ

(W₅ ±∆W ₅) = (1037,4 ±32,24¿ ĵ

4.2.3.2 Daya Listrik

P ±∆ P

(P₁ ±∆ P₁¿ = (4,08 ±0,14¿Watt

(P₂ ±∆ P₂¿ = (4,68 ±0,14¿Watt

(P₃ ±∆ P₃¿ = (4,86 ±0,14¿Watt

(P₄ ±∆ P₄¿ = (4,86 ±0,14¿Watt

(P₅ ±∆ P₅¿ = (4,94 ±0,14¿Watt

4.2.3.3 Hambatan

R ±∆ R

(R₁ ±∆ R₁¿ = (1,41 ±0,04¿Ω

(R₂ ±∆ R₂¿ = (1,44 ±0,05¿Ω

(R₃ ±∆ R₃¿ = (1,44 ±0,05¿Ω

(R₄ ±∆ R₄¿ = (1,5 ±0,05¿Ω

(R₅ ±∆ R₅¿ = (1,37 ±0,04¿Ω

4.2.3.4 Energi Kalor

Q₁ ±∆Q ₁=(150,506±248,33 ) kal

Q₂ ±∆Q ₂=(150,506±248,33 ) kal

Q₃ ±∆Q ₃=(150,506±248,33 ) kal

Q₄ ±∆Q ₄=(150,506±248,33 ) kal

Q₅ ±∆Q ₅=(150,506±248,33 ) kal

4.2.4 KTP Relatif

4.2.4.1 Energi Listrik

∆Ww

x 100%

∆W ₁w₁

x100% =3,60

106,08x100%

= 3,39

∆W ₂w₂

x100% =27,97

884,52x100%

= 3,16

∆W ₃w₃

x100% =30,78

973,44x100%

= 3,16

∆W ₄w₄

x100% =37,93

1229,58x100%

= 3,08

∆W ₅w₅

x100% =32,24

1037,4x 100%

= 3,11

4.2.4.2 Daya Listrik

∆ PP

x 100%

∆ P₁P₁

x100% = 0,144,08

x100%

= 3,43

∆ P₂P₂

x100% = 0,144,6 8

x100%

= 2,99 = 3

∆ P₃P₃

x100% = 0,144,6 8

x100%

= 2,99 = 3

∆ P₄P₄

x100% = 0,144 ,8 6

x100%

= 2,88

∆ P₅P₅

x100% = 0,144,94

x100%

= 2,83

4.2.4.3 Hambatan

∆ RR

x 100%

∆ R₁R₁

x 100% = 0,041,41

x 100%

= 2,83

∆ R₂R₂

x 100% = 0,051,44

x 100%

= 3,47

∆ R₃R₃

x 100% = 0,051,44

x 100%

= 3,47

∆ R₄R₄

x 100% = 0,051,5

x 100%

= 3,33

∆ R₅R₅

x 100% = 0,041,37

x 100%

= 2,92

4.2.4.4 Energi Kalor

∆QQ

x 100%

∆Q ₁Q₁

x 100% = 248,33

150,506 x 100%

= 165

∆Q ₂Q₂

x 100% = 248,33

150,506 x 100%

= 165

∆Q ₃Q₃

x 100% = 248,33

150,506 x 100%

= 165

∆Q ₄Q₄

x 100% = 248,33

150,506 x 100%

= 165

∆Q ₅Q₅

x 100% = 248,33

150,506 x 100%

= 165