Question

1. How much energy would be required to melt 450 grams of ice at 0°C?

Answer 1

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