Calculation of final temperature after mixing ice and steam

Step 1: Calculate the heat absorbed by ice to reach its melting point

- Heat absorbed by ice, q1 = m1 × Lf, where m1 = mass of ice and Lf = latent heat of fusion of ice
- m1 = 100 g, as given in the problem
- Lf = 334 J/g, as per the table of latent heat values
- Therefore, q1 = 100 × 334 = 33,400 J

Step 2: Calculate the heat absorbed by ice to melt completely

- Heat absorbed by ice, q2 = m1 × Cp × ΔT, where Cp = specific heat of ice and ΔT = change in temperature
- Cp = 2.09 J/g°C, as per the table of specific heat values
- ΔT = 0 - (-1) = 1°C, as the ice is at 0°C and needs to be cooled to -1°C before melting
- Therefore, q2 = 100 × 2.09 × 1 = 209 J

Step 3: Calculate the heat absorbed by water to reach 100°C

- Heat absorbed by water, q3 = m3 × Cp × ΔT, where m3 = mass of water and Cp = specific heat of water
- m3 = m1 = 100 g, as the ice melts completely to form water
- Cp = 4.18 J/g°C, as per the table of specific heat values
- ΔT = 100 - 0 = 100°C, as the water needs to be heated to 100°C
- Therefore, q3 = 100 × 4.18 × 100 = 41,800 J

Step 4: Calculate the heat absorbed by water to vaporize completely

- Heat absorbed by water, q4 = m4 × Lv, where m4 = mass of water and Lv = latent heat of vaporization of water
- m4 = m3 = 100 g, as the water vaporizes completely to form steam
- Lv = 2260 J/g, as per the table of latent heat values
- Therefore, q4 = 100 × 2260 = 226,000 J

Step 5: Calculate the heat released by steam to reach 100°C

- Heat released by steam, q5 = m5 × Cp × ΔT, where m5 = mass of steam and Cp = specific heat of steam
- m5 = 10 g, as given in the problem
- Cp = 1.84 J/g°C, as per the table of specific heat values
- ΔT = 100 - 100 = 0°C, as the steam needs to be cooled to 100°C
- Therefore, q5 = 10 × 1.84 × 0 = 0 J

Step 6: Calculate the final temperature of the mixture

- Heat gained by water and steam, q6 = q3 + q4 + q5 = 41,800 + 226,000 + 0 = 267,800 J
- Heat lost by ice, q7 = q1 + q2 = 33,