1. Given that 50 grams of ice is heated at -20.0 °C to steam at 135.0 °C.

i. Show the graph of the changes from ice to steam
ii. Calculate the energy needed to change the ice to steam
Please use these values:
Heat of fusion = 334.16 J g¯1
Heat of vaporization = 2259 J g¯1
specific heat capacity for solid water (ice) = 2.06 J g¯1 K¯1
specific heat capacity for liquid water = 4.184 J g¯1 K¯1
specific heat capacity for gaseous water (steam) = 2.02 J g¯1K¯1

To show the graph of the changes from ice to steam, we need to plot the temperature on the y-axis and the amount of heat added on the x-axis.

i. Here is the graph:

Temperature (°C)
|
135.0 | Steam
|
|
           |
|
           | Water (liquid)
  0.0     |        
   -20.0   |          Ice
|
+----------------
Amount of Heat Added

ii. Calculating the energy needed to change the ice to steam involves three steps:

Step 1: Heating the ice from -20.0 °C to 0.0 °C
- We need to calculate the energy needed to raise the temperature of the ice, using the specific heat capacity for solid water (ice).
- The formula is Q = m * c * ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- Let's calculate this energy:
Q1 = 50 g * 2.06 J g¯1 K¯1 * (0.0 °C - (-20.0 °C))
= 50 g * 2.06 J g¯1 K¯1 * 20.0 °C
= 2060 J

Step 2: Melting the ice at 0.0 °C to water at 0.0 °C
- We need to calculate the energy needed to change the ice to water, using the heat of fusion.
- The formula is Q = m * Hf, where Q is the energy, m is the mass, and Hf is the heat of fusion.
- Let's calculate this energy:
Q2 = 50 g * 334.16 J g¯1
= 16708 J

Step 3: Heating the water from 0.0 °C to 135.0 °C
- We need to calculate the energy needed to raise the temperature of the water, using the specific heat capacity for liquid water.
- The formula is Q = m * c * ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- Let's calculate this energy:
Q3 = 50 g * 4.184 J g¯1 K¯1 * (135.0 °C - 0.0 °C)
= 35272 J

Total Energy:
- The total energy needed is the sum of the energies calculated in each step.
- Let's calculate the total energy:
Total energy = Q1 + Q2 + Q3
= 2060 J + 16708 J + 35272 J
= 54040 J

Therefore, the energy needed to change the ice to steam is 54040 J.

To show the graph of the changes from ice to steam, we need to plot the temperature on the x-axis and the amount of heat added on the y-axis.

Here's how the graph will look like:

i. Graph of the Changes from Ice to Steam:

|
T=135°C| Steam
|_____________________________
| |
| |
T=0°C | Liquid Water
|_____________________________
| |
T=-20°C| Ice
|_____________________________
|
|
-------------------------
Heat Added

ii. To calculate the energy needed to change the ice to steam, we need to consider the energy required for each step of the process:

A. Heating the ice from -20.0°C to 0°C:
Energy = mass x specific heat capacity x temperature change

Energy_1 = 50 g x 2.06 J g¯1 K¯1 x (0 - (-20.0))
= 50 g x 2.06 J g¯1 K¯1 x 20.0 °C
= 2060 J

B. Melting the ice at 0°C:
Energy_2 = mass x heat of fusion

Energy_2 = 50 g x 334.16 J g¯1
= 16708 J

C. Heating the liquid water from 0°C to 100°C:
Energy_3 = mass x specific heat capacity x temperature change

Energy_3 = 50 g x 4.184 J g¯1 K¯1 x (100 - 0)
= 50 g x 4.184 J g¯1 K¯1 x 100 °C
= 20920 J

D. Boiling the water at 100°C:
Energy_4 = mass x heat of vaporization

Energy_4 = 50 g x 2259 J g¯1
= 112950 J

E. Heating the steam from 100°C to 135.0°C:
Energy_5 = mass x specific heat capacity x temperature change

Energy_5 = 50 g x 2.02 J g¯1 K¯1 x (135.0 - 100)
= 50 g x 2.02 J g¯1 K¯1 x 35.0 °C
= 3557 J

Total Energy needed to change ice to steam = Energy_1 + Energy_2 + Energy_3 + Energy_4 + Energy_5
= 2060 J + 16708 J + 20920 J + 112950 J + 3557 J
= 155195 J

q1 = heat needed to raise T ice froom -20 to zero C.

q1 = mass ice x specific heat ice x (Tfinal-Tinitial)

q2 = heat needed to melt ice @ zero C to liquid water @ zero C.
q2 = mass ice x heat fusion ice.

q3 = heat needed to raise T water at zero to 100 C.
q3 = mass water x specific heat H2O x (Tfinal-Tinitial)

q4 = heat needed to convert water @ 100 C to steam at 100 C.
q4 = mass water x heat vaporization

q5 = heat needed to raise T of steam at 100 to steam at 135.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)

Total q = q1 + q2 + q3 + q4 + q5

33kj