To calculate the total amount of heat absorbed during the entire process, you need to consider the energy required for several steps:
1. Heating the ice from -30.0°C to 0°C.
2. Melting the ice at 0°C to water at 0°C.
3. Heating the water from 0°C to 100.0°C.
4. Vaporizing the water at 100.0°C to steam at 100.0°C.
5. Heating the steam from 100.0°C to 140.0°C.
Let's calculate the energy for each step:
1. Heating the ice from -30.0°C to 0°C:
q1 = (mass of ice) x (specific heat of ice) x (temperature change)
q1 = (2.00 mol x 18.015 g/mol) x (2.06 J/g°C) x (0 - (-30.0°C))
q1 = 2.00 x 18.015 g x 2.06 J/g°C x 30.0°C
q1 = 11,439 J
2. Melting the ice at 0°C to water at 0°C:
q2 = (mass of ice) x (heat of fusion of water)
q2 = (2.00 mol x 18.015 g/mol) x (334 J/g)
q2 = 2.00 x 18.015 g x 334 J/g
q2 = 12,010 J
3. Heating the water from 0°C to 100.0°C:
q3 = (mass of water) x (specific heat of water) x (temperature change)
q3 = (2.00 mol x 18.015 g/mol) x (4.18 J/g°C) x (100.0°C - 0°C)
q3 = 2.00 x 18.015 g x 4.18 J/g°C x 100.0°C
q3 = 15,078 J
4. Vaporizing the water at 100.0°C to steam at 100.0°C:
q4 = (mass of water) x (heat of vaporization of water)
q4 = (2.00 mol x 18.015 g/mol) x (40.7 kJ/g) x 1000 J/kJ
q4 = 2.00 x 18.015 g x 40.7 x 1000 J/g
q4 = 2.92 x 10^6 J
5. Heating the steam from 100.0°C to 140.0°C:
q5 = (mass of steam) x (specific heat of steam) x (temperature change)
q5 = (2.00 mol x 18.015 g/mol) x (2.01 J/g°C) x (140.0°C - 100.0°C)
q5 = 2.00 x 18.015 g x 2.01 J/g°C x 40.0°C
q5 = 2.89 x 10^3 J
Now, add up all the energy changes to find the total heat absorbed:
Total q = q1 + q2 + q3 + q4 + q5
Total q = 11,439 J + 12,010 J + 15,078 J + 2.92 x 10^6 J + 2.89 x 10^3 J
Total q ≈ 2.93 x 10^6 J
To convert this to kilojoules (kJ), divide by 1000:
Total q ≈ 2.93 x 10^6 J / 1000 = 2930 kJ
So, the total amount of heat absorbed when 2.00 moles of ice at -30.0°C is converted to steam at 140.0°C is approximately 2930 kJ.