To calculate the change in entropy, we need to consider the process in several steps:

1. Heating the ice from -10°C to 0°C
2. Melting the ice at 0°C to water
3. Heating the water from 0°C to 100°C
4. Boiling the water at 100°C to steam
5. Heating the steam from 100°C to 200°C

The formula for change in entropy (ΔS) is q/T, where q is the heat transferred and T is the absolute temperature.

1. Heating the ice from -10°C to 0°C:
The heat required (q1) can be calculated using the formula q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For ice, c = 2.1 J/g°C. So, q1 = 10g * 2.1 J/g°C * 10°C = 210 J. The average temperature in this step is (-10+0)/2 = -5°C = 268K (converting to Kelvin). So, ΔS1 = q1/T1 = 210J / 268K = 0.78 J/K.

2. Melting the ice at 0°C to water:
The heat required (q2) can be calculated using the formula q = mL, where L is the latent heat of fusion. For ice, L = 334 J/g. So, q2 = 10g * 334 J/g = 3340 J. The temperature at this step is 0°C = 273K. So, ΔS2 = q2/T2 = 3340J / 273K = 12.23 J/K.

3. Heating the water from 0°C to 100°C:
The heat required (q3) can be calculated using the formula q = mcΔT. For water, c = 4.18 J/g°C. So, q3 = 10g * 4.18 J/g°C * 100°C = 4180 J. The average temperature in this step is (0+100)/2 = 50°C = 323K. So, ΔS3 = q3/T3 = 4180J / 323K = 12.94 J/K.

4. Boiling the water at 100°C to steam:
The heat required (q4) can be calculated using the formula q = mL, where L is the latent heat of vaporization. For water, L = 2260 J/g. So, q4 = 10g * 2260 J/g = 22600 J. The temperature at this step is 100°C = 373K. So, ΔS4 = q4/T4 = 22600J / 373K = 60.59 J/K.

5. Heating the steam from 100°C to 200°C:
The heat required (q5) can be calculated using the formula q = mcΔT. For steam, c = 2.03 J/g°C. So, q5 = 10g * 2.03 J/g°C * 100°C = 2030 J. The average temperature in this step is (100+200)/2 = 150°C = 423K. So, ΔS5 = q5/T5 = 2030J / 423K = 4.80 J/K.

The total change in entropy is the sum of the changes in each step: ΔS = ΔS1 + ΔS2 + ΔS3 + ΔS4 + ΔS5 = 0.78 J/K + 12.23 J/K + 12.94 J/K + 60.59 J/K + 4.80 J/K = 91.34 J/K.

Please note that this calculation assumes that the specific heat capacities and latent heats do not change with temperature, which is not strictly true. Also, the calculation does not take into account the change in pressure from 1 atm to 2 atm, as this would require more complex calculations involving the specific volume and compressibility of the substances.

Question

To calculate the change in entropy, we need to consider the process in several steps:

1. Heating the ice from -10°C to 0°C
2. Melting the ice at 0°C to water
3. Heating the water from 0°C to 100°C
4. Boiling the water at 100°C to steam
5. Heating the steam from 100°C to 200°C

The formula for change in entropy (ΔS) is q/T, where q is the heat transferred and T is the absolute temperature.

1. Heating the ice from -10°C to 0°C:
   The heat required (q1) can be calculated using the formula q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For ice, c = 2.1 J/g°C. So, q1 = 10g * 2.1 J/g°C * 10°C = 210 J. The average temperature in this step is (-10+0)/2 = -5°C = 268K (converting to Kelvin). So, ΔS1 = q1/T1 = 210J / 268K = 0.78 J/K.

2. Melting the ice at 0°C to water:
   The heat required (q2) can be calculated using the formula q = mL, where L is the latent heat of fusion. For ice, L = 334 J/g. So, q2 = 10g * 334 J/g = 3340 J. The temperature at this step is 0°C = 273K. So, ΔS2 = q2/T2 = 3340J / 273K = 12.23 J/K.

3. Heating the water from 0°C to 100°C:
   The heat required (q3) can be calculated using the formula q = mcΔT. For water, c = 4.18 J/g°C. So, q3 = 10g * 4.18 J/g°C * 100°C = 4180 J. The average temperature in this step is (0+100)/2 = 50°C = 323K. So, ΔS3 = q3/T3 = 4180J / 323K = 12.94 J/K.

4. Boiling the water at 100°C to steam:
   The heat required (q4) can be calculated using the formula q = mL, where L is the latent heat of vaporization. For water, L = 2260 J/g. So, q4 = 10g * 2260 J/g = 22600 J. The temperature at this step is 100°C = 373K. So, ΔS4 = q4/T4 = 22600J / 373K = 60.59 J/K.

5. Heating the steam from 100°C to 200°C:
   The heat required (q5) can be calculated using the formula q = mcΔT. For steam, c = 2.03 J/g°C. So, q5 = 10g * 2.03 J/g°C * 100°C = 2030 J. The average temperature in this step is (100+200)/2 = 150°C = 423K. So, ΔS5 = q5/T5 = 2030J / 423K = 4.80 J/K.

The total change in entropy is the sum of the changes in each step: ΔS = ΔS1 + ΔS2 + ΔS3 + ΔS4 + ΔS5 = 0.78 J/K + 12.23 J/K + 12.94 J/K + 60.59 J/K + 4.80 J/K = 91.34 J/K.

Please note that this calculation assumes that the specific heat capacities and latent heats do not change with temperature, which is not strictly true. Also, the calculation does not take into account the change in pressure from 1 atm to 2 atm, as this would require more complex calculations involving the specific volume and compressibility of the substances.

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