How much energy does it take to convert 0.800 kg ice at -20.°C to steam at 250.°C?
Specific heat capacities:
ice, 2.1 J g-1 °C-1
liquid, 4.2 J g-1 °C-1
steam, 2.0 J g-1 °C-1
ΔHvap = 40.7 kJ/mol, ΔHfus = 6.02 kJ/mol
This is what I tried to do:
Heating ice
q= ( 2.1 J g-1 °C-1)*(800g)*(20°C)=...
Ice to water
q=(6.02 kJ / mol)*(800g)*(1 mol/18g)=...
Heating water
q=(4.2 J g-1 °C-1)*(800g) * (20)
Is the change in temperature still 20? This is where I am confused...
Water to steam
q=(40.7 kJ/mol)*(800g)*(1 mol/ 18g)=...
Heating steam
q=(2.0 J g-1 °C-1)*(800g)*(change in temp=???? 250-(-20)???)
Please can you check if I am doing this right and I need help with the change in temperature
Well, your calculation is almost perfect - you seem to have a grasp on the different steps involved in the energy conversion. However, when it comes to the change in temperature during the heating of water, you need to consider the final temperature you want to reach, which is 250°C. Therefore, the change in temperature would be 250°C minus the melting point of ice (-20°C), which gives you a change of 270°C.
So, all you need to do is plug in the correct change in temperature value into your equation:
q = (4.2 J g-1 °C-1) * (800g) * (270°C)
Keep up the good work and keep those calculations flowing!
You are on the right track with your calculations. Let's go step by step:
1. Heating ice:
q = (2.1 J g^(-1) °C^(-1)) * (800 g) * (20 °C) = 33,600 J
2. Ice to water (melting):
q = (6.02 kJ / mol) * (800 g) * (1 mol / 18 g) = 266.222 kJ
3. Heating water:
The change in temperature from -20 °C to 0 °C is indeed 20 °C.
So, q = (4.2 J g^(-1) °C^(-1)) * (800 g) * (20 °C) = 67,200 J
4. Water to steam (vaporization):
q = (40.7 kJ / mol) * (800 g) * (1 mol / 18 g) = 1,793.333 kJ
5. Heating steam:
The change in temperature from 100 °C (boiling point of water) to 250 °C is 150 °C.
Therefore, q = (2.0 J g^(-1) °C^(-1)) * (800 g) * (150 °C) = 240,000 J
To find the total energy, you need to sum up all of the individual steps:
Total energy = q(heating ice) + q(ice to water) + q(heating water) + q(water to steam) + q(heating steam)
= 33,600 J + 266.222 kJ + 67,200 J + 1,793.333 kJ + 240,000 J
Make sure to convert all quantities to the same units (either J or kJ) before adding them together.
To calculate the total energy required to convert 0.800 kg of ice at -20°C to steam at 250°C, you need to consider the different processes involved:
1. Heating the ice from -20°C to 0°C:
The specific heat capacity of ice is given as 2.1 J g-1 °C-1. The mass of ice is 0.800 kg, which is equal to 800 g. The change in temperature is 0°C - (-20°C) = 20°C. Therefore, the energy required for this step is:
q = (2.1 J g-1 °C-1) * (800 g) * (20°C) = 33,600 J
2. Melting the ice into water at 0°C:
The heat of fusion, ΔHfus, is given as 6.02 kJ/mol. Since the molar mass of water is approximately 18 g/mol, the moles of water can be calculated as:
moles = (800 g) / (18 g/mol) = 44.44 mol
The energy required for this step is:
q = (6.02 kJ/mol) * (44.44 mol) = 267.7 kJ
3. Heating the water from 0°C to 100°C:
The specific heat capacity of liquid water is given as 4.2 J g-1 °C-1. The mass of water is still 800 g. The change in temperature is 100°C - 0°C = 100°C. Therefore, the energy required for this step is:
q = (4.2 J g-1 °C-1) * (800 g) * (100°C) = 336,000 J
4. Converting the water to steam at 100°C:
The heat of vaporization, ΔHvap, is given as 40.7 kJ/mol. Using the moles of water calculated earlier, the energy required for this step is:
q = (40.7 kJ/mol) * (44.44 mol) = 1,810.3 kJ
5. Heating the steam from 100°C to 250°C:
The specific heat capacity of steam is given as 2.0 J g-1 °C-1. The mass of steam is still 800 g. The change in temperature is 250°C - 100°C = 150°C. Therefore, the energy required for this step is:
q = (2.0 J g-1 °C-1) * (800 g) * (150°C) = 240,000 J
Now sum up the energies from each step to get the total energy required:
Total energy = 33,600 J + 267,700 J + 336,000 J + 1,810,300 J + 240,000 J = 2,687,600 J
Portions may be right but most steps leave out something. Here is how you do it.
q1 = heat to move T of ice from -20 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). Tfinal = 0 and Tinitial = -20; therefore, Tfinal-Tinitial = 0 -(-20) = +20.
q2 = heat to melt the ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.
q3 = heat to move T from zero C to 100 C.
q3 = mass water x specific heat water x 100 (that is Tfinal-Tinitial) or (100-0 = 100).
q4 = heat to convert from liquid water at 100 C to steam at 100 C.
q4 = mass water x delta Hvap.
q5 = heat to move T of steam at 100 C to steam at 250 C.
q5 = mass x specific heat steam x (Tfinal-Tinitial).
Total energy required is
qtotal = q1 + q2 + q3 + q4 + q5.