Question

In: Chemistry

An ice cube tray contains enough water at 29.0°C to make 20 ice cubes that each...

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

Solutions

Expert Solution

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


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