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

2. How many joules of energy would be release as 325 grams of steam condenses to liquid water at 100°C?

3. Calculate the number of joules of energy necessary to vaporize 85 grams of water at 100°C.

4. How much energy would be released by 225 grams of water solidifying at 0°C?

5. How much energy must be absorbed by 20.0 g of water to increase its temperature from 283.0 °C to 303.0 °C?

6. When 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, how much heat energy is released?

7. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

8. What is the specific heat capacity of silver metal if 55.00 g of the metal absorbs 198 joules of heat and the temperature rises 15.0°C?

9. A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0 °C to 28.5 °C. Find the
mass of the water, in grams.

10. The specific heat of ethanol is 2.46 J/g oC. Find the heat required to raise the temperature of 193 g of ethanol
from 19oC to 35oC.

11. How much energy would be necessary to convert ice at -23.0°C to steam at 118°C? It may help to sketch the heating curve to see the energy changes required.

WILL BE MARKED BRAINLIEST

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

Yep!

Explanation: