Question

53.1 g of H2O exists as a liquid at 0 °C. How much energy is needed to turn the water into a solid at 0 °C ?

The heat of fusion for water = 333.55 J/g. The heat of vaporization for water= 2256 J/g.

Answer 1

To turn 53.1 grams of liquid water into ice at 0 °C, 17722.155 joules of energy are needed, as calculated using the heat of fusion for water (333.55 J/g).

Step-by-step explanation:

The student's question involves determining the amount of energy needed to turn 53.1 grams of liquid water at 0 °C into solid water (ice) at the same temperature using the heat of fusion of water. The heat of fusion represents the energy required to change the state of a substance from solid to liquid or vice versa without changing its temperature. In the case of water, the heat of fusion is 333.55 J/g.

Using the formula Q = m × Lf, where Q is the heat energy required, m is the mass of the substance (in grams), and Lf is the heat of fusion (in J/g), we can calculate the amount of energy needed for the phase change.

In this situation, Q = 53.1 g × 333.55 J/g = 17722.155 J. Therefore, 17722.155 joules of energy are needed to turn 53.1 grams of liquid water at 0 °C into ice at 0 °C.