Question

An ice cube tray contains enough water at 28.0°C to make 18 ice cubes that each has a mass of 30.0 g. The tray is placed in a freezer that uses CF2Cl2 as a refrigerant. The heat of vaporiztion of CF2Cl2 is 158 J/g. What mass of CF2Cl2 must be vaporized in the refrigeration cycle to convert all the water at 28.0°C to ice at –5.0°C? The heat capacities forH2O(s) and H2O(l) are 2.03 J/g·°C and 4.18 J/g·°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

Mass= _______g

Answer 1

Total mass of water = 18 x 30.0 g = 540 g

Total mass of water in moles will be = (540 g * 1 mole / 18 g) = 30 mol water

Now, q₁ = m_water * C * T = 540 g * 4.18 J/g⁰C * ( 28 -0)⁰C = 63201 J = 63.2 KJ

q₂ = m_water* _fusion = 30 mol * 6.02 KJ/mol = 180.6 KJ

and q₃ = m_water*C*T = 540 g * 2.03 J/g⁰C* [ 0-(-5)]⁰C = 5481 J = 5.481 KJ

Total q = q₁ + q₂ + q₃ = 63.2 KJ + 180.6 KJ + 5.481 KJ = 249.281 KJ

Given that heat of vaporization of CF₂Cl₂ is 158 J/g

This means that 158 J is needed to vaporize 1 g of CF₂Cl₂

Therefore, for 249.281 KJ = 249281 J we need =(249281 / 158 ) g = 1577.73 g CF₂Cl₂ =1.578 Kg CF₂Cl₂