you wish to combine two samples, one sample contains is solid water at 0 degree celsius.The other sample contains liquid water at 0 degree celsius,you want to combine both raise the temperature to 100 degree celcius but boil off only 50 ml to gas phase keep the rest as liquid.How much energy is neededfor this?Both is 100ml(iceand liquid water)
here liquid water initial temperatureis 0 degree celcius.final is 100
I posted the answer then re-read the problem and saw I had not included energy to remove that 50 mL H2O at the boiling point. Therefore, I deleted that post. Here is the complete answer.
energy to melt ice + energy to raise melted ice to 100 from zero C + energy to raise 100 g H2O initially at zero to 100 + energy to evaporate 50 mL (which I assume is 50 g but you can look up the density and change that 50 mL to mass).
Ice mass = volume x density
If you need to convert 50 mL water @ 100, then mass water = volume x density.
q = [mass ice x heat fusion] + [mass melted ice x specific heat liquid water x (Tfinal-Tinitial)] + [mass liquid water initially x (Tfinal-Tinitial)] + [mass H2O at 100 x heat vaporization]
Here Tfinal is 100 and Tinitial is 0.
To calculate the amount of energy needed to raise the temperature of the samples and boil off a portion of it, we need to consider the heat required for each step separately.
1. Raising the temperature of the solid water:
The specific heat capacity of solid water (ice) is 2.09 J/g°C. We have 100 mL of ice, which has a density of 0.92 g/mL (convert mL to grams by multiplying by density).
Mass of ice = 100 mL * 0.92 g/mL = 92 g
The temperature needs to be raised from 0°C to 100°C, so the change in temperature is 100°C - 0°C = 100°C.
The energy required is given by the formula:
Energy = mass * specific heat capacity * change in temperature
Energy = 92 g * 2.09 J/g°C * 100°C = 19,088 J
2. Raising the temperature of the liquid water:
The specific heat capacity of liquid water is 4.18 J/g°C. We have 100 mL of liquid water, which also has a density of 0.92 g/mL (convert mL to grams).
Mass of liquid water = 100 mL * 0.92 g/mL = 92 g
The temperature needs to be raised from 0°C to 100°C, so the change in temperature is 100°C - 0°C = 100°C.
The energy required is given by the same formula as before:
Energy = mass * specific heat capacity * change in temperature
Energy = 92 g * 4.18 J/g°C * 100°C = 38,536 J
3. Boiling off 50 mL of the liquid water:
To calculate the energy required to boil off a portion of the liquid water, we need to consider the heat of vaporization of water. The heat of vaporization of water is 40.7 kJ/mol.
First, we need to find the number of moles of water in 50 mL of liquid. The molar mass of water is 18.015 g/mol.
Number of moles = mass / molar mass = 50 mL * 0.92 g/mL / 18.015 g/mol ≈ 2.31 mol
The energy required to boil off the liquid water is given by the formula:
Energy = number of moles * heat of vaporization
Energy = 2.31 mol * 40.7 kJ/mol = 94 kJ = 94,000 J
To find the total energy required, we add up the energies from each step:
Total Energy = Energy for raising the temperature of ice + Energy for raising the temperature of liquid water + Energy for boiling off liquid water
Total Energy = 19,088 J + 38,536 J + 94,000 J = 151,624 J