how much heat is required to completely vapporize 2.3 grams of ice starting at -50 degrees Celsius? (the specific heat of ice is 2.0 J/g.C; the specific heat of water is 4.184 J/g.C; the heat of fusion is 333 J/g; and the heat of vaporization is 22601 J/g)

Question

how much heat is required to completely vapporize 2.3 grams of ice starting at -50 degrees Celsius? (the specific heat of ice is 2.0 J/g.C; the specific heat of water is 4.184 J/g.C; the heat of fusion is 333 J/g; and the heat of vaporization is 22601 J/g)

Answer 1

q1 = heat to move solid ice from -50 C to zero C.

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

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

q3 = heat to move liquid water from zero C to 100 C
q3 = mass water x specific heat water x (Tfinal-Tinitial)

q4 = heat to convert liquid water at 100 C to vapor at 100 C.
q4 = mass water x heat vap.

qTotal = q1 + q2 + q3 + q4

Answer 2

To determine the amount of heat required to completely vaporize 2.3 grams of ice starting at -50 degrees Celsius, we need to consider the various steps involved in the process:

Step 1: Heating the ice from -50°C to 0°C
Step 2: Melting the ice at 0°C
Step 3: Heating the water from 0°C to its boiling point
Step 4: Vaporizing the water at its boiling point

Let's calculate the heat required for each step and then sum them up to find the total heat required.

Step 1: Heating the ice from -50°C to 0°C
To calculate the heat required for this step, we will use the specific heat of ice, which is 2.0 J/g°C.
ΔQ = m * C * ΔT
ΔQ = 2.3 g * 2.0 J/g°C * (0°C - (-50°C))

Step 2: Melting the ice at 0°C
To calculate the heat required for this step, we will use the heat of fusion, which is 333 J/g.
ΔQ = m * Hfus
ΔQ = 2.3 g * 333 J/g

Step 3: Heating the water from 0°C to its boiling point
To calculate the heat required for this step, we will use the specific heat of water, which is 4.184 J/g°C.
ΔQ = m * C * ΔT
ΔQ = 2.3 g * 4.184 J/g°C * (100°C - 0°C)

Step 4: Vaporizing the water at its boiling point
To calculate the heat required for this step, we will use the heat of vaporization, which is 22601 J/g.
ΔQ = m * Hvap
ΔQ = 2.3 g * 22601 J/g

Now, let's sum up all the heat values:

Total heat required = ΔQ_1 + ΔQ_2 + ΔQ_3 + ΔQ_4

Finally, calculate the total heat required to vaporize 2.3 grams of ice starting at -50°C by summing up the heat values from each step.