An ice cube tray contains enough water at 22.0 C to make 18 ice cubes that each have a mass of 30.0g. The tray is placed in a freezer that uses CF2CL2 as a refridgerant. The heat of vap of CF2CL2 is 158 J/g. What mass of CF@CL2 must be vaporized in the refridgeration cycle to convert all the water at 22.0 C to ice at -5.o C.
this is all i have so far. please help? thanks
H capacity H20 (s) = 2.08 J/g C
H capacity H20 (l) = 4.18 J/g C
ethanply of fusion of ice =6.02 KJ/mol.
mass of water = 30g x 18 cubes = 540 g.
heat required to lower T of water from 22 to zero is mass x 4.18 x 22 = xx
heat required to freeze 540 g water is
mass x heat of fusion water = yy
heat required to lower T of 540 g ice from zero to -5 is mass x 2.08 x 5 = zz
Total heat required to lower T of 540 g water from 22 to -5 C is xx + yy + zz.
What do you have to do this job? You have a refrigerant which soaks up 158 J for each gram of the stuff used. So how many grams does it take to equal the total heat required from above?
Check my thinking. Check my work.
To find the mass of CF2CL2 that needs to be vaporized in the refrigeration cycle, we can use the equation:
Q = mcΔT
Where:
Q is the heat transferred (in this case, heat required to freeze the water)
m is the mass of the substance (CF2CL2) we want to find
c is the specific heat capacity of the substance (CF2CL2)
ΔT is the change in temperature (temperature difference)
First, let's calculate the heat required to freeze the water:
Q = mcΔT
Q = (18 ice cubes) * (30.0g ice cube) * (2.08 J/g C) * (22.0 C - (-5.0 C))
Q = 18 * 30.0 * 2.08 * 27.0 J
Now, let's convert the heat to match the units of the heat of vaporization (J/g) of CF2CL2:
1 kJ = 1000 J
1 mol CF2CL2 = 122.9 g CF2CL2 (molar mass of CF2CL2)
Heat required in kJ = (18 * 30.0 * 2.08 * 27.0 J) / 1000
Next, we need to calculate the moles of CF2CL2 needed to provide this heat:
moles = (heat required in kJ) / (enthalpy of fusion of ice in kJ/mol)
moles = (18 * 30.0 * 2.08 * 27.0 J / 1000) / (6.02 kJ/mol)
Finally, we can calculate the mass of CF2CL2 needed:
mass = moles * molar mass
mass = (moles) * (molar mass of CF2CL2)
mass = (moles) * (122.9 g CF2CL2)
These calculations will give you the mass of CF2CL2 that needs to be vaporized in the refrigeration cycle to convert all the water at 22.0 C to ice at -5.0 C.
To solve this problem, we need to calculate the amount of heat that needs to be transferred from the water to freeze it and then determine how much CF2CL2 needs to be vaporized to absorb this heat.
First, let's calculate the heat required to lower the temperature of the water from 22.0°C to 0°C (the freezing point of water) using the specific heat capacity of water (4.18 J/g°C):
Q1 = m × c × ΔT
Q1 = 18 × 30.0 g × 4.18 J/g°C × (0 - 22.0°C)
Q1 = -56,412 J
Next, let's calculate the heat required to solidify the water at 0°C using the heat of fusion of ice:
Q2 = m × ΔHfus
Q2 = 18 × 30.0 g × (-6.02 kJ/mol) × (1 mol/18 g)
Q2 = -1003 J
Now, let's calculate the heat required to lower the temperature of the ice from 0°C to -5.0°C using the specific heat capacity of ice (2.08 J/g°C):
Q3 = m × c × ΔT
Q3 = 18 × 30.0 g × 2.08 J/g°C × (-5.0 - 0°C)
Q3 = -18,720 J
The total heat required to convert all the water at 22.0°C to ice at -5.0°C is the sum of Q1, Q2, and Q3:
Qtotal = Q1 + Q2 + Q3
Qtotal = -56,412 J + (-1003 J) + (-18,720 J)
Qtotal = -76,135 J
Now, let's determine the amount of CF2CL2 that needs to be vaporized to absorb this amount of heat. We can use the heat of vaporization of CF2CL2 (158 J/g):
mCF2CL2 = Qtotal / ΔHvap
mCF2CL2 = -76,135 J / (-158 J/g)
mCF2CL2 ≈ 482 g
Therefore, approximately 482 grams of CF2CL2 need to be vaporized in the refrigeration cycle to convert all the water at 22.0°C to ice at -5.0°C.
You have the mass of the water to be frozen, figure the heat lost to convert it to ice at -5C.
You will have to watch units, on the heat of fusion, convert that to J/g.
Then, the heat lost has to equal the heat gained, or
Massfreon*Heatvap= heat lost