Question

What amount of energy is used when 33.3 grams of ice at 0.00 °C is converted to steam at 150.0 °C?

Answer 1

`"103.4 kJ"` is the total amount of energy needed to convert that much ice to steam.

Explanation:

The answer is `103.4kJ`.

We need to determine the total energy required to go from ice to water, and then from water to vapor - the phase changes underwent by the water molecules.

In order to do this, you'll need to know:

Heat of fusion of water: `DeltaH_f` = `334` `J`/`g`;
Heat of fusion vaporization of water: `DeltaH_v` = `2257` `J`/`g`;
Specific heat of water: `c` = `4.18` `J`/`g^@C`;
Specific heat of steam: `c` = `2.09` `J`/`g^@C`;

So, the following steps describe the overall process:

1. Determine the heat required to convert `0^@C` ice to `0^@C` water:

`q_1 = m * DeltaH_(f) = 33.3 g * 334 J/(g) = 11122.2J`

2. Determine the heat required to go from water at `0^@C` to water at `100^@C`:

`q_2 = m * c_(water) * DeltaT = 33.3g * 4.18 J/(g*^@C) * (100^@C - 0^@C) = 13919.4J`

3. Determine the heat required to convert `100^@C` water to `100^@C` vapor:

`q_3 = m * DeltaH_(v) = 33.3g * 2257 J/(g) = 75158.1 J`

4. Determine the heat required to go from `100^@C` vapor to `150^@C` vapor:

`q_4 = m * c_(vap o r) * DeltaT = 33.3g * 2.09 J/(g*^@C) * (150^@C - 100^@C) = 3479.9J`

Therefore, the total heat required is

`q_(TOTAL) = q_1+q_2+q_3+q_4 = 103679.6J = 103.4kJ`