Answer:
Explanation:
1. The amount of heat needed to melt ice at 0°C is equal to the mass of the ice times the latent heat of fusion.
q = mL
q = (450 g) (334 J/g)
q = 150,300 J
q = 150 kJ
2. The amount of heat released by the condensation of steam at 100°C is equal to the mass of the steam times the latent heat of vaporization.
q = mL
q = (325 g) (2260 J/g)
q = 734,500 J
q = 735 kJ
3. q = mL
q = (85 g) (2260 J/g)
q = 192,100 J
q = 190 kJ
4. q = mL
q = (225 g) (334 J/g)
q = 75,150 J
q = 75.2 kJ
5. Above 100°C, water is steam. The amount of heat needed to increase the temperature of steam is equal to its mass times its specific heat times the change in temperature.
q = mCΔT
q = (20.0 g) (2.03 J/g/°C) (303.0°C − 283.0°C)
q = 812 J
6. q = mCΔT
q = (15.0 g) (2.03 J/g/°C) (250.0°C − 275.0°C)
q = -761 J
7. q = mCΔT
q = (10.0 g) (0.90 J/g/°C) (55°C − 22°C)
q = 297 J
8. q = mCΔT
198 J = (55.0 g) C (15°C)
C = 0.24 J/g/°C
9. q = mCΔT
41,840 J = m (4.184 J/g/°C) (28.5°C − 22.0°C)
m = 1540 g
10. q = mCΔT
q = (193 g) (2.46 J/g/°C) (35°C − 19°C)
q = 7600 J
11. First, the temperature of the ice must be raised to 0°C.
q = mCΔT
q = m (2.09 J/g/°C) (0°C − (-23.0°C))
q/m = 48.1 J/g
Next, the ice must be melted.
q = mL
q/m = 334 J/g
Then, the water must be heated to 100°C.
q = mCΔT
q = m (4.184 J/g/°C) (100°C − 0°C)
q/m = 418.4 J/g
The water is then vaporized.
q = mL
q/m = 2260 J/g
Finally, the steam is heated to its final temperature.
q = mCΔT
q = m (2.03 J/g/°C) (118°C − 100°C)
q/m = 36.5 J/g
So the total amount of energy needed is:
q/m = 48.1 J/g + 334 J/g + 418.4 J/g + 2260 J/g + 36.5 J/g
q/m = 3100 J/g
