# 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?

Jul 6, 2017

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: $25 \times 4.184 = 104.6 J / g$
To vapourize that same gram: $2260 J / g$

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

Multiply this by the mass of the water:
$Q = 2364.6 \times 25 = 59115 J \approx 59.1 k J$
(rounded to 3 SF)