Calculate the amount of energy in kilojoules needed to change 225 g of water ice at -10 C to steam at 125 C. The following constants may be useful:
Cm(ice)=36.57 J/(mol*C)
Cm(water)=75.40 J/(mol*C)
Cm(steam)=36.04 J/(mol*C)
Delta H(fus)=+6.01 kJ/mol
Delta H(vap)=+40.67 kJ/mol
These are the numbers I calculated:
Cm(ice)=4.57 kJ
Delta H(fus)=75.06 kJ
Cm(water)=94.17 kJ
Delta H(vap)=507.97 kJ
Cm(steam)=11.25 kJ
Total= 693 kJ
The only one I disagree with slightly is vaporization. I obtained 507.95 kJ. I don't round anything until the last calculation; i.e., (225/18.015) x 40.67 = 507.95171 which rounds to 507.95 kJ.
Well, well, well, looks like we've got a chemistry question here. Let's break it down!
First, we need to calculate the energy required to heat up the ice from -10 °C to 0 °C. Since we know that the specific heat capacity of ice (Cm(ice)) is 36.57 J/(mol*C), we can use the formula:
Q = m * Cm * ΔT
where Q is the energy (in J), m is the mass (in g), Cm is the specific heat capacity (in J/(mol*C)), and ΔT is the change in temperature (in °C).
So, for the ice phase, the energy required will be:
Q(ice) = 225 g * (36.57 J/(mol*C)) * (0 - (-10) °C)
Now, to get the energy required to melt the ice, we can use the enthalpy of fusion (ΔH(fus)):
Q(fusion) = moles * ΔH(fus)
We can find the number of moles of ice by using the molar mass of water (18.015 g/mol):
moles = 225 g / 18.015 g/mol
Finally, to get the energy required to heat up the resulting water from 0 °C to 100 °C, we can use the specific heat capacity of water (Cm(water)):
Q(water) = moles * Cm(water) * ΔT
Now comes the fun part. We need to calculate the energy required to vaporize the water. We can use the enthalpy of vaporization (ΔH(vap)):
Q(vaporization) = moles * ΔH(vap)
And last but not least, we need to calculate the energy required to heat up the resulting steam from 100 °C to 125 °C, using the specific heat capacity of steam (Cm(steam)):
Q(steam) = moles * Cm(steam) * ΔT
Finally, to find the total energy required, we just sum up all these energies:
Total energy = Q(ice) + Q(fusion) + Q(water) + Q(vaporization) + Q(steam)
Are you still with me, or did your brain melt like ice on a hot summer day? Take your time with the calculations, and if you have any specific questions, just let me know!
To calculate the amount of energy needed to change 225 g of water ice at -10°C to steam at 125°C, we need to consider the following steps and use the given constants:
1. Calculate the energy required to raise the temperature of ice from -10°C to 0°C.
- This can be calculated using the formula Q = m * Cm * ΔT, where Q is the energy, m is the mass, Cm is the specific heat capacity, and ΔT is the temperature change.
- Cm(ice) = 36.57 J/(mol*C), but we need to convert it to J/(g*C) since the mass is in grams.
- So, Cm(ice) = 36.57 J/(mol*C) * (1 mol/18 g) = 2.03 J/(g*C)
- ΔT = 0°C - (-10°C) = 10°C
- Q1 = 225 g * 2.03 J/(g*C) * 10°C
2. Calculate the energy required to melt the ice into water at 0°C.
- This can be calculated using the formula Q = m * ΔH(fus), where Q is the energy, m is the mass, and ΔH(fus) is the heat of fusion.
- ΔH(fus) is given as +6.01 kJ/mol, so we need to convert it to J/g.
- ΔH(fus) = +6.01 kJ/mol * (1000 J/1 kJ) * (1 mol/18 g)
3. Calculate the energy required to raise the temperature of water from 0°C to 100°C.
- This can be calculated using the same formula as step 1, but now using the specific heat capacity of water, Cm(water).
- Cm(water) = 75.40 J/(mol*C) * (1 mol/18 g) = 4.19 J/(g*C)
- ΔT = 100°C - 0°C = 100°C
- Q3 = 225 g * 4.19 J/(g*C) * 100°C
4. Calculate the energy required to vaporize the water into steam at 100°C.
- This can be calculated using the formula Q = m * ΔH(vap), where Q is the energy, m is the mass, and ΔH(vap) is the heat of vaporization.
- ΔH(vap) is given as +40.67 kJ/mol, so we need to convert it to J/g.
- ΔH(vap) = +40.67 kJ/mol * (1000 J/1 kJ) * (1 mol/18 g)
5. Calculate the energy required to raise the temperature of steam from 100°C to 125°C.
- This can be calculated using the same formula as step 1, but now using the specific heat capacity of steam, Cm(steam).
- Cm(steam) = 36.04 J/(mol*C) * (1 mol/18 g) = 2.00 J/(g*C)
- ΔT = 125°C - 100°C = 25°C
- Q5 = 225 g * 2.00 J/(g*C) * 25°C
To find the total energy required, we sum up the energies from all these steps:
Total Energy = Q1 + Q2 + Q3 + Q4 + Q5
Remember to convert the units of energies (kJ to J or J to kJ) if necessary.