What is the final temperature and physical state if water when 47.50 kJ of energy is added to 15.0 g of ice at -10.0 C?

Helpful information:
Cs (ice) = 2.1 J/g C
Heat of fusion (water) = 6.02 kJ/mol
Cs (water) = 4.184 J/g C
Heat of vaporization (water) = 40.7 kJ/mol
Cs (steam) = 2.0 J/g C

Please check my answers below.
My answer:
Tf = 167.9 C
Physical state of water = gas

I get 167 but that difference may be due to rounding errors.

To find the final temperature and physical state of water after adding 47.50 kJ of energy to 15.0 g of ice at -10.0°C, follow these steps:

1. Calculate the amount of energy required to raise the temperature of the ice from -10.0°C to 0°C. We can use the specific heat capacity formula: q = m * Cs * ΔT.
- q = energy (in joules)
- m = mass of the substance (in grams)
- Cs = specific heat capacity of the substance (in joules per gram Celsius)
- ΔT = change in temperature (in Celsius)

q1 = 15.0 g * 2.1 J/g°C * (0°C - (-10.0°C))
q1 = 15.0 g * 2.1 J/g°C * 10.0°C
q1 = 315 J

2. Convert the energy consumed (315 J) into kilojoules.
q1 = 315 J * (1 kJ / 1000 J)
q1 = 0.315 kJ

3. Calculate the energy required to melt the ice at 0°C into water. The formula for heat of fusion is energy = moles * heat of fusion.
- The molar mass of water (H2O) = 18.015 g/mol
- The heat of fusion of water = 6.02 kJ/mol

moles of ice = mass of ice (in grams) / molar mass of water
moles of ice = 15.0 g / 18.015 g/mol ≈ 0.832 mol

q2 = 0.832 mol * 6.02 kJ/mol
q2 = 4.99824 kJ

4. Calculate the energy required to raise the temperature of the water from 0°C to the final temperature (Tf).

q3 = 15.0 g * 4.184 J/g°C * Tf

5. Calculate the energy required to vaporize the water at the final temperature (Tf). Use the formula for heat of vaporization: energy = moles * heat of vaporization.
- The heat of vaporization of water = 40.7 kJ/mol

moles of water = mass of water (in grams) / molar mass of water
moles of water = 15.0 g / 18.015 g/mol ≈ 0.832 mol

q4 = 0.832 mol * 40.7 kJ/mol
q4 = 33.85024 kJ

6. Calculate the total energy required:

Total energy = q1 + q2 + q3 + q4
Total energy = 0.315 kJ + 4.99824 kJ + q3 + 33.85024 kJ

7. Equate the total energy required to the energy added (47.50 kJ) to find the final temperature (Tf).

Total energy = 47.50 kJ

0.315 kJ + 4.99824 kJ + q3 + 33.85024 kJ = 47.50 kJ

Solve for q3:

q3 ≈ 47.50 kJ - (0.315 kJ + 4.99824 kJ + 33.85024 kJ)
q3 ≈ 8.33652 kJ

Substitute q3 into the heat capacity equation:

8.33652 kJ = 15.0 g * 4.184 J/g°C * Tf

Solve for Tf:

Tf ≈ 8.33652 kJ / (15.0 g * 4.184 J/g°C)
Tf ≈ 0.3336°C

8. Determine the physical state of the water based on the final temperature (Tf).

Since the final temperature (0.3336°C) is below the boiling point, the physical state of the water will be a liquid.

Therefore, the final temperature of water is approximately 0.3336°C, and the physical state of the water is liquid.