How much heat is required to warm 229 g of ice from -45.3°C to 0.0°C, melt the ice, warm the water from 0.0°C to 100.0°C, boil the water, and heat the steam to 173.0°C?

To find the total amount of heat required, we need to calculate the heat required for each step and then sum them up.

Step 1: Warming the ice from -45.3°C to 0.0°C
The heat required to warm a substance is given by the formula:

Q = m * C * ΔT

Where:
Q = heat energy
m = mass of the substance
C = specific heat capacity of the substance
ΔT = change in temperature

The specific heat capacity of ice is 2.09 J/g°C.

The change in temperature (ΔT) is given by:
ΔT = final temperature - initial temperature
= 0°C - (-45.3°C)
= 45.3°C

Let's calculate the heat required for this step:

Q1 = 229 g * 2.09 J/g°C * 45.3°C
= 229 g * 96.177 J/g
≈ 22,023 J

Step 2: Melting the ice
The heat required to melt a substance can be calculated using the formula:

Q = m * ΔHf

Where:
Q = heat energy
m = mass of the substance
ΔHf = enthalpy of fusion

The enthalpy of fusion for water is 334 J/g.

Let's calculate the heat required for this step:

Q2 = 229 g * 334 J/g
= 76,786 J

Step 3: Warming the water from 0.0°C to 100.0°C
The specific heat capacity for water is 4.18 J/g°C.

Let's calculate the heat required for this step:

Q3 = 229 g * 4.18 J/g°C * 100.0°C
= 96,062 J

Step 4: Boiling the water
The heat required to boil a substance can be calculated using the formula:

Q = m * ΔHv

Where:
Q = heat energy
m = mass of the substance
ΔHv = enthalpy of vaporization

The enthalpy of vaporization for water is 2260 J/g.

Let's calculate the heat required for this step:

Q4 = 229 g * 2260 J/g
= 517,540 J

Step 5: Heating the steam from 100.0°C to 173.0°C
The specific heat capacity for steam is 2.03 J/g°C.

Let's calculate the heat required for this step:

Q5 = 229 g * 2.03 J/g°C * (173.0°C - 100.0°C)
= 28,551 J

Now, let's sum up the heat required for each step to find the total amount of heat:

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 22,023 J + 76,786 J + 96,062 J + 517,540 J + 28,551 J
≈ 741,962 J

Therefore, the total amount of heat required to warm 229 g of ice from -45.3°C to 0.0°C, melt the ice, warm the water from 0.0°C to 100.0°C, boil the water, and heat the steam to 173.0°C is approximately 741,962 J.

To calculate the total amount of heat required, we need to consider each step individually:

Step 1: Heating the ice from -45.3°C to 0.0°C

To calculate the heat required to warm the ice, we use the formula: Q = m * ΔT * c, where Q is the heat, m is the mass, ΔT is the change in temperature, and c is the specific heat capacity.

Given:
- Mass (m) = 229 g
- Initial Temperature (T1) = -45.3°C
- Final Temperature (T2) = 0.0°C
- Specific Heat Capacity (c) = 2.09 J/g°C (specific heat of ice)

Using the formula, the heat required to warm the ice is:
Q1 = m * ΔT * c
Q1 = 229 g * (0.0°C - (-45.3°C)) * 2.09 J/g°C
Q1 = 229 g * 45.3°C * 2.09 J/g°C

Step 2: Melting the ice at 0.0°C

To calculate the heat required to melt the ice, we use the formula: Q = m * ΔHf, where Q is the heat, m is the mass, and ΔHf is the heat of fusion (also known as latent heat).

Given:
- Mass (m) = 229 g
- Heat of Fusion (ΔHf) = 334 J/g (heat required to melt ice)

Using the formula, the heat required to melt the ice is:
Q2 = m * ΔHf
Q2 = 229 g * 334 J/g

Step 3: Heating the water from 0.0°C to 100.0°C

To calculate the heat required to warm the water, we use the formula: Q = m * ΔT * c, where Q is the heat, m is the mass, ΔT is the change in temperature, and c is the specific heat capacity.

Given:
- Mass (m) = 229 g
- Initial Temperature (T1) = 0.0°C
- Final Temperature (T2) = 100.0°C
- Specific Heat Capacity (c) = 4.18 J/g°C (specific heat of water)

Using the formula, the heat required to warm the water is:
Q3 = m * ΔT * c
Q3 = 229 g * (100.0°C - 0.0°C) * 4.18 J/g°C

Step 4: Boiling the water and heating the steam from 100.0°C to 173.0°C

To calculate the heat required to boil the water and heat the steam, we use the formula: Q = m * ΔT * c, where Q is the heat, m is the mass, ΔT is the change in temperature, and c is the specific heat capacity.

Given:
- Mass (m) = 229 g
- Initial Temperature (T1) = 100.0°C
- Final Temperature (T2) = 173.0°C
- Specific Heat Capacity (c) = 2.03 J/g°C (specific heat of water vapor)

Using the formula, the heat required to boil the water and heat the steam is:
Q4 = m * ΔT * c
Q4 = 229 g * (173.0°C - 100.0°C) * 2.03 J/g°C

Finally, to calculate the total heat required, we add up the heat from each step:

Total heat required = Q1 + Q2 + Q3 + Q4

You can substitute the values you have for each quantity (mass, temperatures, specific heat capacities) and calculate the total heat required.

There are two formulas you use.

If you are within the same phase, use
q = mass x specific heat in that phase x (Tfinal-Tinitial).
Example. q for 25g H2O from 5 C to 95 C is
q = mass x specific heat x (95-5) = ?

The other is at a phase change; i.e., ice melting or water boiling to steam.
q = mass x heat fusion for melting.
q = mass x heat vaporization for liquid water turning to steam.

The easiest way to do these problems is in pieces.
ice from -45.3 to zero C.
melt ice
raise T from zero C to 100 C
vaporize water.
raise T of steam from 100 to 173.0C.
Then add all of the q values together to find the total Q.