the heat of fusion water is335 J/g. The heat of evaporization of water is 2.26 kJ/g, the specific heat of ice is 2.05 J/Deg/g, the specific heat of steam is 2.08 J/deg/g and the specific heat of liquid water is 4.184 J/deg/g. How much heat would be needed to convert 12.09 g of ice at -17 degree C to steam at 135 C

Work this problem in pieces.

q with a single phase (liquid to liquid, vapor to vapor, solid to solid) = mass x specific heat in that phase x (Tfinal-Tinitial).

q at a phase change is
mass x heat fusion at zero C.
mass x heat vaporization at 100 C.

Then add each segment of q to find total Q.

To determine the total heat needed to convert 12.09 g of ice at -17 degrees Celsius to steam at 135 degrees Celsius, we need to consider the different phases and temperature ranges involved in the process.

Here's a step-by-step breakdown to calculate the total heat needed:

1. Calculate the heat needed to heat the ice from -17 degrees Celsius to its melting point of 0 degrees Celsius:
- The specific heat of ice is given as 2.05 J/Deg/g.
- The temperature change is 0 - (-17) = 17 degrees Celsius.
- The mass of ice is 12.09 g.
- The heat needed can be calculated using the formula: heat = mass x specific heat x temperature change.
- Therefore, heat = 12.09 g x 2.05 J/deg/g x 17 deg = 442.4635 J.

2. Calculate the heat needed to melt the ice into liquid water at 0 degrees Celsius:
- The heat of fusion for water is given as 335 J/g.
- The mass of ice is still 12.09 g.
- The heat needed can be calculated using the formula: heat = mass x heat of fusion.
- Therefore, heat = 12.09 g x 335 J/g = 4048.15 J.

3. Calculate the heat needed to heat the liquid water from 0 degrees Celsius to its boiling point of 100 degrees Celsius:
- The specific heat of liquid water is given as 4.184 J/deg/g.
- The temperature change is 100 - 0 = 100 degrees Celsius.
- The mass of liquid water is still 12.09 g.
- The heat needed can be calculated using the formula: heat = mass x specific heat x temperature change.
- Therefore, heat = 12.09 g x 4.184 J/deg/g x 100 deg = 5051.196 J.

4. Calculate the heat needed to convert the liquid water into steam at 100 degrees Celsius:
- The heat of vaporization for water is given as 2.26 kJ/g, which can be converted to 2260 J/g.
- The mass of liquid water is still 12.09 g.
- The heat needed can be calculated using the formula: heat = mass x heat of vaporization.
- Therefore, heat = 12.09 g x 2260 J/g = 27363.4 J.

5. Calculate the heat needed to heat the steam from 100 degrees Celsius to 135 degrees Celsius:
- The specific heat of steam is given as 2.08 J/deg/g.
- The temperature change is 135 - 100 = 35 degrees Celsius.
- The mass of steam is still 12.09 g.
- The heat needed can be calculated using the formula: heat = mass x specific heat x temperature change.
- Therefore, heat = 12.09 g x 2.08 J/deg/g x 35 deg = 880.464 J.

6. Add up all the calculated heats to obtain the total heat needed:
- Total heat = heat for heating ice + heat for melting ice + heat for heating liquid water + heat for vaporizing water + heat for heating steam
- Total heat = 442.4635 J + 4048.15 J + 5051.196 J + 27363.4 J + 880.464 J
- Total heat = 36685.6735 J

Therefore, to convert 12.09 g of ice at -17 degrees Celsius to steam at 135 degrees Celsius, approximately 36685.67 J of heat is required.