Question

How many kilojoules are required to warm 25.0 g of water from 75.0 C to 100.0 C and convert it to steam at 100.0 C? The heat of vaporization of water is 2260 joules/g, and the the specific heat of water is 4.184 joules/g?

Answer 1

You have to add the energy required for the heating up and the energy for the vapourizing.

Explanation:

To heat 1 gram of water from 75 to 100 deg C means a temperture rise of 25 deg C.
Heat required: `25xx4.184=104.6J//g`
To vapourize that same gram: `2260J//g`

So total energy required to heat up and vapourize 1 gram:
`104.6+2260=2364.6J//g`

Multiply this by the mass of the water:
`Q=2364.6xx25=59115J~~59.1kJ`
(rounded to 3 SF)