Question

21. The heat of fusion for water is 6.009 kJ/mol. The heat of vaporization is 40.05 kJ/mol. The specific heat for water is 4.184 *b. What quantity of heat must be added to convert a 127g piece of ice at 0°C into water vapor at 100ºC?

Answer 1

`"377.5 kJ"`.

Explanation:

To answer this question, we can split it up into `3` different "steps":

1. `"127 g"` of ice melts into liquid water at `0 °C`.
2. Liquid water at `0 °C` becomes liquid water at `100 °C`.
3. Liquid water evaporates into water vapor at `100 °C`.

To find how much energy is required to melt `"127 g"` of water, we'll need to multiply the molar heat of fusion by number of moles:

`"heat of fusion" xx "number of moles" = "6.009 kJ/mol" xx "number of moles"`
`= "6.009 kJ/mol" xx "127 g"/"18.02 g/mol"`
`= "6.009 kJ/mol" xx "7.05 mol"`
`= 42.4 kJ`

Now, we can use `q=mCDeltaT` to find the heat required to convert `0°C` of water into `100 °C`.

`q = mCDeltaT = "127g" xx "4.184 J/g°C" xx (100-0) °C`
`= "127g" xx "0.004184 kJ/g°C" xx 100 °C`
`= 53.1 kJ`

Finally, to find the heat required to convert `127 g` of liquid water into water vapour, we just need to multiply the number of moles by the molar heat of vaporisation:

`"40.05 kJ/mol" xx "7.05 mol" = 282kJ`

Now, we can add all of these together to find the cumulative amount of heat that's needed:

`42.4 kJ + 53.1 kJ + 282kJ = 377.5 kJ`