How much energy is needed to convert 10kg of ice at -15 degrees centigrade to 125 degrees centigrade of steam?

Express your answer in joules.

To calculate the energy needed to convert ice at -15 degrees Celsius to steam at 125 degrees Celsius, we need to consider the energy required for three different processes:

1. Heating the ice from -15°C to 0°C (Phase change from solid to liquid).
2. Melting the ice at 0°C (Phase change from solid to liquid).
3. Heating the liquid water from 0°C to 100°C (Phase change from liquid to gas).
4. Heating the water vapor from 100°C to 125°C.

Here are the steps to calculate the total energy:

1. Calculate the energy required to heat the ice from -15°C to 0°C.
- The specific heat capacity of ice is 2.09 J/g°C.
- The temperature change is 0°C - (-15°C) = 15°C.
- The mass of ice is 10 kg.
- Energy = mass * specific heat capacity * temperature change.

Energy = 10 kg * 2.09 J/g°C * 15°C
= 313.5 kJ (Convert to Joules by multiplying by 1000)
= 313,500 J

2. Calculate the energy required to melt the ice at 0°C (latent heat of fusion).
- The latent heat of fusion of ice is 334 J/g.
- The mass of ice is 10 kg.
- Energy = mass * latent heat of fusion.

Energy = 10 kg * 334 J/g
= 3340 kJ (Convert to Joules by multiplying by 1000)
= 3,340,000 J

3. Calculate the energy required to heat the liquid water from 0°C to 100°C.
- The specific heat capacity of liquid water is 4.18 J/g°C.
- The temperature change is 100°C - 0°C = 100°C.
- The mass of liquid water is the same as the mass of ice because they are at the same temperature.
- Energy = mass * specific heat capacity * temperature change.

Energy = 10 kg * 4.18 J/g°C * 100°C
= 4180 kJ (Convert to Joules by multiplying by 1000)
= 4,180,000 J

4. Calculate the energy required to heat the water vapor from 100°C to 125°C.
- The specific heat capacity of water vapor is 2.03 J/g°C.
- The temperature change is 125°C - 100°C = 25°C.
- The mass of water vapor can be calculated using the equation:
mass = (10 kg + 333.55 g) * (volume ratio of steam at 125°C to liquid water at 100°C)

The volume ratio of steam at 125°C to liquid water at 100°C is approximately 1687:1.
Therefore, mass ≈ 10 kg * 1687 ≈ 16,870 g

- Energy = mass * specific heat capacity * temperature change.

Energy = 16,870 g * 2.03 J/g°C * 25°C
= 8607.75 kJ (Convert to Joules by multiplying by 1000)
= 8,607,750 J

Finally, to calculate the total energy required, sum up the energies calculated in each step:

Total Energy = Energy Step 1 + Energy Step 2 + Energy Step 3 + Energy Step 4
= 313,500 J + 3,340,000 J + 4,180,000 J + 8,607,750 J
= 16,441,250 J

Therefore, the energy needed to convert 10 kg of ice at -15 degrees Celsius to 125 degrees Celsius of steam is approximately 16,441,250 Joules.