Question

If 2.0 g of water at 0.00°C is to be vaporized, how much heat must be added to it? The specific heat of water is 1.0 cal/g∙K, its heat of fusion is 80 cal/g, and its heat of vaporization is 539 cal/g.

Answer 1

Answer:

$Q_{vap} = 1438\,cal$

Explanation:

The heat needed to vaporize the water is:

$Q_{vap} = (2.0\,g)\cdot[80\,\frac{cal}{g} +(1\,\frac{cal}{g\cdot K} )\cdot (100\,K)+539\,\frac{cal}{g} ]$

$Q_{vap} = 1438\,cal$