In: Chemistry
An ice cube tray contains enough water at 29.0°C to make 20 ice cubes that each has a mass of 30.0 g. The tray is placed in a freezer that uses as a refrigerant. The heat of vaporiztion of is 158 J/g. What mass of must be vaporized in the refrigeration cycle to convert all the water at 29.0°C to ice at –5.0°C? The heat capacities for and 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.
What is the mass in g
mass of ice = number of cube * mass of each cube
= 20*30.0 g
= 600 g
1st calculate the heat required by ice
Ti = 29.0 oC
Tf = -5.0 oC
Cl = 4.18 J/g.oC
Heat released to convert liquid from 29.0 oC to 0.0 oC
Q1 = m*Cl*(Ti-Tf)
= 600 g * 4.18 J/g.oC *(29-0) oC
= 72732 J
Lf = 6.02KJ/mol =
6020J/mol
Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 600.0/18.016
= 33.3037 mol
Heat released to convert liquid to solid at 0.0 oC
Q2 = n*Lf
= 33.3037 mol *6020 J/mol
= 200488.4547 J
Cs = 2.03 J/g.oC
Heat released to convert solid from 0.0 oC to -5.0 oC
Q3 = m*Cs*(Ti-Tf)
= 600 g * 2.03 J/g.oC *(0--5) oC
= 6090 J
Total heat released = Q1 + Q2 + Q3
= 72732 J + 200488.4547 J + 6090 J
= 279310 J
This heat must be supplied by refrigerant
use:
Q = mass of refrigerant * heat of vaporisation
279310 J = m*158 J/g
m = 1768 g
Answer: 1768 g