CFCs such as CF2Cl2 are refrigerants whose use has been phased out because their destructive effect on Earth's ozone layer. The standard enthalpy of evaporation of CF2Cl2 is 17.4 kJ/mol, compared with change in H(vapor)=41kJ/mol for liquid water. How many grams of liquid CF2Cl2 are needed to cool 130.8g of water from 49.4 to 29.4 Celcius? The specific heat of water is 4.184 J/(g*C)
q needed to cool water is
q = mass water x specific heat water x (Tfinal-Tinitial)
Then 17,400 J/mol x ?mol = q to cool water by 20 C.
Solve for moles CF2Cl2 and convert to grams.
To solve this problem, we will use the principle of energy conservation and the formula:
q = mcΔT
Where:
q is the heat exchanged
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
First, we need to calculate the heat exchanged by the water using the formula above:
q_water = mcΔT
= 130.8g × 4.184 J/(g°C) × (29.4°C - 49.4°C)
= -8377.9632 J
Since the enthalpy of evaporation for CF2Cl2 is given in kJ/mol, we need to convert the heat exchanged by the water to kJ:
q_water = -8377.9632 J ÷ 1000
= -8.378 kJ
Next, we need to calculate the number of moles of CF2Cl2 required to release this much heat.
ΔH_vapor = 17.4 kJ/mol
Number of moles of CF2Cl2 = q_water ÷ ΔH_vapor
= -8.378 kJ ÷ 17.4 kJ/mol
= -0.481 mol
Since we are considering the mass of CF2Cl2, we can now calculate it using the molar mass of CF2Cl2.
Molar mass of CF2Cl2 = (1 * 12.01 g/mol) + (2 * 19.00 g/mol) + (1 * 35.45 g/mol)
= 120.91 g/mol
Mass of CF2Cl2 = Number of moles × Molar mass of CF2Cl2
= -0.481 mol × 120.91 g/mol
≈ -58.18 g
Note that we obtained a negative mass because the calculated value is the mass of CF2Cl2 that needs to be evaporated, which means removing that amount of CF2Cl2 from the system. Therefore, the value should be considered as 58.18 g.
So, approximately 58.18 grams of liquid CF2Cl2 are needed to cool 130.8 grams of water from 49.4 to 29.4 degrees Celsius.
To solve this problem, we need to calculate the amount of heat absorbed by the water during the cooling process, and then determine how many grams of CF2Cl2 are required to absorb the same amount of heat during its evaporation.
First, let's calculate the heat absorbed by the water using the equation:
q = m * c * ΔT
where:
q = heat absorbed or released
m = mass of the substance (water in this case)
c = specific heat capacity of the substance (water in this case)
ΔT = change in temperature
Given:
m_water = 130.8 g
c_water = 4.184 J/(g*C)
ΔT_water = (29.4 - 49.4) C = -20 C
Using the formula:
q_water = m_water * c_water * ΔT_water
q_water = 130.8 g * 4.184 J/(g*C) * (-20 C)
q_water = -109,663.04 J
Now, since we know the standard enthalpy of vaporization for CF2Cl2 (ΔH_vaporization = 17.4 kJ/mol), we can find out the number of moles of CF2Cl2 required to absorb the same amount of heat.
To do this, we use the equation:
q_vaporization = n * ΔH_vaporization
Given:
ΔH_vaporization = 17.4 kJ/mol
1 kJ = 1000 J
1 mol of CF2Cl2 weighs 120.91 g
Converting ΔH_vaporization to J/mol:
ΔH_vaporization = 17.4 kJ/mol * 1000 J/1 kJ
ΔH_vaporization = 17,400 J/mol
Now, let's calculate the number of moles of CF2Cl2:
n = q_water / ΔH_vaporization
n = -109,663.04 J / 17,400 J/mol
n = -6.31 mol
Since the enthalpy of vaporization corresponds to the phase change from liquid to gas, it means that 6.31 moles of CF2Cl2 are required to evaporate and absorb the same amount of heat. To calculate the mass, we multiply the number of moles by the molar mass:
mass_CF2Cl2 = n * molar mass_CF2Cl2
mass_CF2Cl2 = -6.31 mol * 120.91 g/mol
mass_CF2Cl2 = -761.5821 g
Since mass cannot be negative, we can disregard the negative sign. Therefore, approximately 761.6 grams of liquid CF2Cl2 are needed to cool 130.8 grams of water from 49.4 to 29.4 Celsius.