Question

a. Determine the amount of heat needed to take a 34.5gram piece of ice (ΔHfus = 6.02 kJ/mol) to 32.0 oC liquid water. b.Determine the amount of heat release when 465.3g of 100.0oC steam condenses to 21oC liquid water. (100.0oC ΔHvap= 40.6 kJ/mol)

Answer 1

Given that the amount heat energy needed for melting one mole of ice is 6.02 kJ

34.5 g ice contains 34.5/18 = 1.92 moles of water

where 18 is the molar mass of water

Energy for melting is = 1.92 * 6.02 = 11.54 kJ

For raising the temperature by 1-degree celsius 4.184 J is needed for 1g water. Thus to increase temperature from 0 to 32-degree Celsius

= 34.5 x 4.184 * 32 = 4619.136 J or 4.619 kJ

Total energy required for the 34.5 g ice to reach to reach 32-degree celsius = 4.619 +11.54 = 16.15 kJ

We have supplied this much energy.

b) Similarly,

first 100-degree Celcius steam has to condense to 100 degree Celcius water

Given that for one mole system will release 40.6 kJ

465.3 g water contains 465.3/18 = 25.85 moles of water

Thus for this process total energy required = 25.85 x 40.6 = 1049.51 kJ

Then temperature has to decrease from 100 to 21

change in temperature = 79

as we discussed earlier, the energy released during this process is = 79 * 4.184 *465.3 = 153798 J or 153.798 kJ

the total energy released during this process is = 153.798 +1049.51 = 1203.308 kJ