Question

Answer the Q.'s I the spaces below. 1. 'Antifreeze' (ethylene glycol, (HOCH,CH,OH)) is much more viscous that water. Why? 2. Tell why small droplets of water on an oily surface are almost spherical 3. Tell why you may be able to survive a bitter cold day in 'snow-cave. 4. Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a heating curve for the process. The specific heat of ice is 2.11 J/( g. °C); 4.18 J/(g. °C) for water, 2.00 J/g. °C. AH fus for water is 6.01 kJ/mol; 4H, for water = 40.67 kJ/mol.

Answer 1

1. Ethylene glycol (HO-CH₂-CH₂-OH) has two hydroxyl groups at either end whereas water (H-OH) has a single hydroxyl group. For ethylene glycol, longer chains and branches of molecules can be formed by hydrogen bonding using these two hydroxyl groups. Hence, a more entangled network will form, which prevents the free flow of the liquid. Whereas for water, due to a single hydroxyl group, the entanglement will be reduced by many folds. Hence, ethylene glycol is more viscous than water.
2. Water adheres weakly to wax in comparison to itself. If we look closer at the surface of the liquid, the molecule at the surface is laterally pulled upward and downward due to attracting forces of other adjacent molecules. The molecule is pulled downward more, as there is no molecule above to counterbalance it. Due to this inward pull, the molecules on the surface tends to achieve the minimum surface area, which is that of a sphere. Hence, small droplets of water on an oil surface are almost spherical.
3. Ice and the air inside the snow-cave are not good conductors. As we are inside the snow-cave, so the air inside the snow-cave cannot conduct heat to the outside either can the air outside the snow-cave conduct heat to the inside of the cave. Hence, the air around our body can be a bit warmer than the walls. So, people can survive a chilly day in a snow-cave.
4. Mass of ice = 25.0 g.

Within a phase, the amount of heat required = Mass × Specific heat × Temperature difference.

When a phase change occurs, the amount of heat required = Mass × Heat of fusion/vaporization at the transition point.

Now, for raising the temperature of ice from -25℃ to 0℃,

The amount of heat required = 25 g × 2.11 J/(g℃) × (0℃ - (-25℃))

                                               = 1318.75 J = 1.32 kJ.

For the melting of ice at 0℃ to water at 0℃,

The amount of heat required = 25 g × 6.01 kJ/mol / 18 g

                                               = 8.35 kJ.

For raising the temperature of water from 0℃ to 100℃,

The amount of heat required = 25 g × 4.18 J/(g℃) × (100℃ - 0℃)

                                               = 10450 J = 10.45 kJ.

For the boiling of water i.e. phase change of water at 100℃ to water vapour at 100℃,

The amount of heat required = 25 g × 40.67 kJ/mol / 18 g

                                               = 56.49 kJ.

For raising the temperature of water vapour from 100℃ to 105℃,

The amount of heat required = 25 g × 2.00 J/(g℃) × (105℃ - 100℃)

                                               = 250 J = 0.25 kJ.

So, the total amount of heat required to heat ice at -25℃ to water vapour at 105℃ will be

        = (1.32 + 8.35 + 10.45 + 56.49 + 0.25) kJ = 76.86 kJ.

The heating curve for the process is given below:

0.25 kJ 56.49 KJ gas (water vapour) L boiling Temperature (°C) .14.01 8.35 kJ I liguid (water) melting =- 1.32 solid (ice) E-_- -22 Heat absorbed (kJ)