How much heat (in kJ) is required to warm 10.0 g of ice, initially at -13.0 ∘C, to steam at 109.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C. The heat capacity of liquid water is 4.18 J/g⋅∘C. This substance has an enthalpy of fusion of 6.01 kJ/mol and an enthalpy of vaporization of 40.7 kJ/mol.

To calculate the total heat required to warm the ice to steam, we need to consider the following steps:

1. Heating the ice from -13.0 °C to 0 °C.
2. Melting the ice at 0 °C to liquid water at 0 °C.
3. Heating the liquid water from 0 °C to 100 °C.
4. Boiling the liquid water at 100 °C to steam at 100 °C.
5. Heating the steam from 100 °C to 109 °C.

Step 1: Heating the ice from -13.0 °C to 0 °C.
The heat required = mass × specific heat capacity × change in temperature
= 10.0 g × 2.09 J/g⋅∘C × (0 - (-13.0))
= 10.0 g × 2.09 J/g⋅∘C × 13.0 °C
= 270.70 J

Step 2: Melting the ice at 0 °C to liquid water at 0 °C.
The heat required = mass × enthalpy of fusion
= 10.0 g × (6.01 kJ/mol ÷ molar mass of water)
= 10.0 g × (6.01 kJ/mol ÷ 18.015 g/mol)
= 3.34 kJ

Step 3: Heating the liquid water from 0 °C to 100 °C.
The heat required = mass × specific heat capacity × change in temperature
= 10.0 g × 4.18 J/g⋅∘C × (100 - 0)
= 4,180 J

Step 4: Boiling the liquid water at 100 °C to steam at 100 °C.
The heat required = mass × enthalpy of vaporization
= 10.0 g × (40.7 kJ/mol ÷ molar mass of water)
= 10.0 g × (40.7 kJ/mol ÷ 18.015 g/mol)
= 90.13 kJ

Step 5: Heating the steam from 100 °C to 109 °C.
The heat required = mass × specific heat capacity × change in temperature
= 10.0 g × 2.01 J/g⋅∘C × (109 - 100)
= 2,010 J

Now, let's add up the individual heats to find the total heat required:
270.70 J + 3.34 kJ + 4,180 J + 90.13 kJ + 2,010 J

Converting all the units to kJ:
= 0.27070 kJ + 3.34 kJ + 4.180 kJ + 90.13 kJ + 2.010 kJ
= 100.93 kJ

Therefore, the total heat required to warm 10.0 g of ice, initially at -13.0 °C, to steam at 109.0 °C is 100.93 kJ.

To calculate the amount of heat required to warm the ice to steam, we will need to consider a few steps.

Step 1: Find the heat to raise the temperature of ice from -13.0 °C to 0 °C.
To determine the heat required to raise the temperature, we use the formula:

q = m * C * ΔT

Where:
q = heat (in J)
m = mass of the substance (in g)
C = specific heat capacity (in J/g⋅°C)
ΔT = change in temperature (in °C)

Here, the mass (m) is 10.0 g, the specific heat capacity (C) is 2.09 J/g⋅°C, and the change in temperature (ΔT) is (0 - (-13)) = 13 °C.

q1 = 10.0 g * 2.09 J/g⋅°C * 13 °C = 270.7 J

Step 2: Find the heat required to melt the ice at 0 °C.
To determine the heat required for the phase change (melting), we use the formula:

q = m * ΔHf

Where:
q = heat (in J)
m = mass of the substance (in g)
ΔHf = enthalpy of fusion (in J/g)

Here, the mass (m) is 10.0 g, and the enthalpy of fusion (ΔHf) is 6.01 kJ/mol. We need to convert ΔHf to J/g by dividing it by the molar mass of water (18.015 g/mol).

ΔHf = 6.01 kJ/mol * (1 mol / 18.015 g) = 333.8 J/g

q2 = 10.0 g * 333.8 J/g = 3338 J

Step 3: Find the heat to raise the temperature of liquid water from 0 °C to 100 °C.
Using the same formula as in step 1:

q = m * C * ΔT

Here, the mass is still 10.0 g, but the specific heat capacity is now 4.18 J/g⋅°C, and the change in temperature is (100 - 0) = 100 °C.

q3 = 10.0 g * 4.18 J/g⋅°C * 100 °C = 4180 J

Step 4: Find the heat required to vaporize the water at 100 °C.
To determine the heat required for the phase change (vaporization), we use the formula:

q = m * ΔHv

Where:
q = heat (in J)
m = mass of the substance (in g)
ΔHv = enthalpy of vaporization (in J/g)

Here, the mass (m) is 10.0 g, and the enthalpy of vaporization (ΔHv) is 40.7 kJ/mol. We need to convert ΔHv to J/g by dividing it by the molar mass of water (18.015 g/mol).

ΔHv = 40.7 kJ/mol * (1 mol / 18.015 g) = 2257 J/g

q4 = 10.0 g * 2257 J/g = 22570 J

Step 5: Find the heat to raise the temperature of steam from 100 °C to 109 °C.
Using the same formula as in step 1:

q = m * C * ΔT

Here, the mass is still 10.0 g, but the specific heat capacity is now 2.01 J/g⋅°C, and the change in temperature is (109 - 100) = 9 °C.

q5 = 10.0 g * 2.01 J/g⋅°C * 9 °C = 181.89 J

Step 6: Sum up all the heats from each step.

Total heat = q1 + q2 + q3 + q4 + q5
= 270.7 J + 3338 J + 4180 J + 22570 J + 181.89 J
= 29741.59 J

To convert this to kJ, divide by 1000:

Total heat in kJ = 29741.59 J / 1000 = 29.74 kJ

Therefore, it requires approximately 29.74 kJ of heat to warm 10.0 g of ice, initially at -13.0 °C, to steam at 109.0 °C.

add the heats:

heat to warm up ice to 0C
heat to melt ice at 0C
heat to warm water to 100C
Heat to convert water to steam at 100C
Heat to warm steam from 100 to 109C