Answer:
The answer to your question is:
1.- Q = 53.95 kJ
2.- Q = 37450.2 J
Step-by-step explanation:
1.- Data
H2O(s) = 2.09 J/g · ◦C
H2O (ℓ) = 4.18 J/g · ◦C
H2O(g) = 2.03 J/g · ◦C
heat of fusion = 334 J/g
heat of vaporization = 2260 J/g
Heat = ? kJ
T1 = -15°C
T2 = 56°C
Process
Calculate heat from -15°C to 0°C
heat of fusion
heat from 0°C to 56°C
a) heat from -15°C to 0°C
Q = mCp(T2 - T1)
Q = (90)(2.09) (0 + 15)
Q = 2821.5 J
b) heat of fusion
Q = (334)(90)
Q = 30060 J
c) heat from 0° to 56°C
Q = (90)(4.18)(56 - 0)
Q = 21067.2 J
Total heat = Q = 2821.5 + 30060 + 21067.2
Q = 53948.7 J
Q = 53.95 kJ
2.-
Data
Energy = ?
mass = 12 g
T1 = 227°K
T2 = 382 °K
Specific heat of ice = 2.077 J/g·K
Specific heat of water (ℓ) = 4.180 J/g·K
Specific heat of steam = 2.042 J/g·K
H2O heat of fusion = 6.00 kJ/mol = 6000 J/ mol = 333.3 J/g
H2O heat of vaporization = 40.6 kJ/mol = 40600 J/ mol = 2255.6 J/g
Process
Calculate heat from 227 to 273°K
fusion heat
heat from 273 to 373°K
vaporization heat
heat from 373 to 382°K
heat from 227 to 273°K
Q = (12)(2.077)(273 - 227)
Q = 1146.5 J
fusion heat
Q = (12)(333.3)
Q = 4000 J
heat from 273 to 373 °K
Q = (12)(4.18)(373 - 273)
Q = 5016 J
vaporization heat
Q = (12)(2255.6)
Q = 27067.2 J
heat from 373 to 382
Q = (12)(2.042)(382 - 373)
Q = 220.5 J
Total heat
Q = 1146.5 + 4000 + 5016 + 27067.2 + 220.5
Q = 37450.2 J