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.
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)
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