Question

How much heat is necessary to vaporize 200 grams of water at 100°C to form steam at 100°C?

Answer 1

Given:

The mass of water is: m = 200 g

To find:

The heat necessary to vaporize 200 g of water at 100°C to form steam at 100°C.

Explanation:

The heat Q, mass m, and the latent heat L are related as:

$Q=mL$

The latent heat of the water to convert it from 100°C to form steam at 100°C is 2260 J/g. Thus,

L = 2260 J/g

Substituting the values in the above equation, we get:

$\begin{gathered} Q=200\text{ g}\times2260\text{ J/g} \\ \\ Q=452000\text{ J} \\ \\ Q=452\text{ kJ} \end{gathered}$

Final answer:

Thus, 452 kJ of heat will be required to vaporize 200 g of water.