( 0,1173 kg . 2,2 . 103 J kg-1 K-1 . -31°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 . 87°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 ) = smeltwarmte smeltwarmte=J kg-1 = rs
the answer must be in this form, but i do not know how do to it.
3 . 3 = 9
. = x
the calculations:
Formules:
Qijs = Mijs . Cijs . ΔT
Qijswater = Mijs . Ckokendwater . ΔT
Qsmeltijs = Mijs . Ckokendwater
Qkokendwater = Mkokendwater . Ckokendwater . ΔT
Gegevens:
Mijs = 117,3 gram
Mkokendwater = 134,8 gram
Maluminiumbakje = 78 gram
Mtotaal = 225,21 gram
Tijs = -18°C
Tkokendwater = 100°C
Tomgeving = 21°C
Teind = 13°C
Uitwerking:
1)
Qijs = Mijs . Cijs . ΔT
-Mijs = 117,3 gram = 0,1173 kg
-Cijs = 2,2 . 103 J kg-1 K-1
-ΔT = ?
ΔT = Tijs – Teind ( ΔT = Tbegin – Teind )
-Tijs = -18°C
-Teind = 13°C
ΔT = -18°C - 13°C
-ΔT = -31°C
Qijs = 0,1173 kg . 2,2 . 103 J kg-1 K-1 . -31°C
-Qijs = -7999,86 J
2)
Qijswater = Mijs . Ckokendwater . ΔT
-Mijs = 117,3 gram = 0,1173 kg
-Ckokendwater = 4,18 . 103 J kg-1 K-1
-ΔT = ?
ΔT = Tkokendwater – Teind
-Tkokendwater = 100°C
-Teind = 13°C
ΔT = 100°C - 13°C
-ΔT = 87°C
Qijswater = 0,1173 kg . 4,18 . 103 J kg-1 K-1 . 87°C
-Qijswater = 42657,318 J
3)
Qsmeltijs = Mijs . Cwater
-Mijs = 117,3 gram = 0,1173 kg
-Ckokendwater = 4,18 . 103 J kg-1 K-1
Qsmeltijs = 0,1173 kg . 4,18 . 103 J kg-1 K-1
- Qsmeltijs = 490,314 J K-1
4)
Qkokendwater = Mkokendwater . Ckokendwater . ΔT
- Mkokendwater = 134,8 gram = 0,1348 kg
-Ckokendwater = 4,18 . 103 J kg-1 K-1
-ΔT = ?
ΔT = Tkokendwater – Teind
-Tkokendwater = 100°C
-Teind = 13°C
ΔT = 100°C - 13°C
-ΔT = 87°C
Qkokendwater = 0,1348 kg . 4,18 . 103 J kg-1 K-1 . 87°C
-Qkokendwater = 49021,368 J
De energie bij 1, 2 en 3 samen moet net zo groot zijn als de energie bij 4:
-7999,86 J + 42657,318 J + 490,314 J K-1 = Qkokendwater
Qkokendwater = 35147,772 J K-1 ( Ik had 49021,368 J )
Bepaal de smeltwarmte per kg:
( 0,1173 kg . 2,2 . 103 J kg-1 K-1 . -31°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 . 87°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 ) = smeltwarmte
Smeltwarmte =
Smeltwarmte water = 334 .103 J kg-1
( 0,1173 kg . 2,2 . 103 J kg-1 K-1 . -31°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 . 87°C ) + ( 0,1173 kg . 4,18 . 103 J kg-1 K-1 ) = smeltwarmte smeltwarmte=J kg-1 = rs
What are the units of the 2.2 4.18 ? Also, it appears you need to convert the C degrees to Kelvin.