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

1 Answer
Dec 14, 2014

Answer:

#"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#