Question

A thermos contains 0.200 liters of water that holds 80 ° C, but it should be cooled with zero-degree ice. What is the total volume of water when ice is no longer melting?
The melting point for ice at 0 ° C is 6.01 kJ mol-1 and the density of the water is
0.9999 g / cm 3 at 0 ° C and 0.9718 g / cm 3 at 80 ° C

Answer 1

given

decrease in temperature of water = from 80C to 0 C = 80 C

Heat of fusion = 6.01 kj/mol

specific heat capacity of water = 4.184 J/g

density of water at 0 C = 0.9999g/cm3

and at 80 C = 0.9718 g/cm3

now

D = M/V

M=DV

0.200 L =200cm3

m=200 x 0.9718 =194.36 g at 80 C

specific heat capacity of water = 4.184 J/g

total heat released in conversion of 0.200 l 9 194.36 g) of water from 80 to 0C

= 194.34 x Heat capacity x change in temperature

= 194.34 x 4.184 x 80

=65056.2 J

= 65.06 KJ

now that much head is to be absorbed by ice to gate converted into water

as heat of fusion of ice = 6.01 kJ /mol

number of moles of ice fused ( melt) = 65.06 / 6.01

                                         = 10.825 moles

                  mass of ice melted = mole x molar mass

                                                = 10.825 x 18

                                                 = 194.85 g

volume of ice melted = M x D

                             = 194.85 x 0.9999

                            = 194.83 cm3

now

Total volume of water = initial volume of water + volume of water of ice

                            = 200 + 194.83 cm3

                         = 394.83 cm3

                      V     = 0.3948 L

Total volume of water when no ice is melting is 0.3948 L