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158 J is required to vapourize 1 g of CF2Cl2
So for 235.kJ we need. 235.76 x 10^3 J/ 158 J = 1492.15g of CF2Cl2 = 1.492 kg of CF2Cl2
158 J is required to vapourize 1 g of CF2Cl2
So for 235.kJ we need. 235.76 x 10^3 J/ 158 J = 1492.15g of CF2Cl2 = 1.492 kg of CF2Cl2
Heat capacity can be illustrated as the amount of heat absorbed or released to modify the temperature of the compound. It can be represented as, q = mcΔT
Here, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
On the basis of the given values,
Mass of one ice cube = 30 g
Mass of 10 ice cubes = 30 × 18 = 540 g
Temperature of water = 22 °C
Temperature of ice = -5 °C
Specific heat capacity of water (s) = 2.08 J/g °C
Specific heat capacity of water (l) = 4.18 J/g °C
ΔHfusion (ice) = 6.02 kJ/g
ΔHvap (CF₂Cl₂) = 158 J/g
The heat change is calculated when temperature of water changes from 22°C to 0°C,
q1 = mwater.cwater.ΔT
= 540 g × 4.18 J/g °C * (0°C -22°C)
= -49.74 kJ
Heat change when 540 g of water freezes at O°C is termed as enthalpy of fusion that is given as,
q2 = ΔHfusion × mass of water
= -6.02 kJ/mol × 540 g / 18 g/mol
= -180.6 kJ
The heat change has to be calculated when temperature of 540 g ice changes from 0°C to -5°C (Tf) as,
q3 = mice.cice.ΔT
= 540 g × 2.08 J/g°C * (-5°C - 0°C)
= -5.62 kJ
Total heat lost by the system = q1 +q2 + q3
= (49.74 + 180.6 + 5.62)
= 235.96 kJ
Now, the heat gained by CF₂Cl₂ = Heat lost by the system
ΔHvap × mass of CF₂Cl₂ vaporized = 235.96 kJ
Mass of CF₂Cl₂ vaporized = 235.96 kJ / ΔHvap
= 235.96 kJ / 158 × 10⁻³ kJ/g
= 1.5 × 10³ g
Hence, the amount of CF₂Cl₂ vaporizes in the overall conversion is 1.5 × 10³ g
