Question

10 kg of liquid water is in a container maintained at atmospheric pressure, 101325 Pa. The water is initially at 373.15 K, the boiling point at that pressure.The latent heat of water -> water vapor is 2230 J/g. The molecular weight of water is 18 g.103 J of heat is added to the water.1)How much of the water turns to vapor?mass(vapor)=

Answer 1

Answer:

$m=0.0462\ g$ of water is converted into vapour.

Explanation:

Given:

• mass of water, $m_w=10\ kg$

• pressure conditions, $P=101325\ Pa$

• temperature conditions, $T=373.15\ K$

• latent heat of vapourization of water, $L=2230\ J.g^{-1}$

• amount of heat supplied to the water, $103\ J$

Now using the equation of heat considering latent heat only:

(since water already at boiling point at atmospheric temperature)

$Q=m.L$

$103=m\times 2230$

$m=0.0462\ g$ of water is converted into vapour.