How much heat (in kJ) is evolved in converting 1.00 mol of steam at 145.0 degrees C to ice at -45.0 degrees C? The heat capacity of steam is 2.01 J/g* degrees C and of ice is 2.09 J/g*degrees C .

How much heat is evolved in converting 1.00 mol of steam at 145.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Express your answer in units of kilojoules Assume the system is at atmospheric pressure.

To determine the amount of heat evolved in converting steam to ice, we need to calculate the heat required to cool down the steam from 145.0 degrees C to 0 degrees C and the heat required to freeze the water at 0 degrees C and cool it down further to -45.0 degrees C. We will then add up these two values to get the total heat evolved.

First, let's determine the heat required to cool down 1.00 mol of steam from 145.0 degrees C to 0 degrees C:

1. Calculate the mass of the steam:
To do this, we need to know the molar mass of water. The molar mass of water is approximately 18.02 g/mol. Therefore, 1.00 mol of water is equivalent to 18.02 g.

2. Calculate the heat required to cool down the steam:
We can use the formula: q = m * C * ΔT, where q is the heat, m is the mass, C is the heat capacity, and ΔT is the change in temperature.

ΔT = (0 - 145.0) = -145.0 degrees C

We know the heat capacity of steam, C, is 2.01 J/g*degrees C. However, we need to convert it to kJ/mol*degrees C:

C' = C / molar mass of water
C' = 2.01 J/g*degrees C / 18.02 g/mol
C' ≈ 0.11148 J/mol*degrees C (rounded to 5 decimal places)

Now, apply the formula to calculate the heat required:

q1 = m * C' * ΔT
q1 = 18.02 g * 0.11148 J/mol*degrees C * (-145.0 degrees C)

The negative sign indicates that heat is being lost or evolved.

Next, let's determine the heat required to freeze the water at 0 degrees C and cool it down to -45.0 degrees C:

1. Calculate the heat capacity of ice:
The heat capacity of ice is given as 2.09 J/g*degrees C.

2. Calculate the heat required for the phase change:
The heat required to freeze water is given by the equation: q = m * ΔHf, where ΔHf is the heat of fusion (energy needed to convert a substance from a solid to a liquid or vice versa).

For water, ΔHf is 333.55 J/g.

Therefore, q2 = m * ΔHf
q2 = 18.02 g * 333.55 J/g

This gives you the energy required to convert 18.02 g of water to ice.

3. Calculate the heat required to cool down the ice:
Using the same formula as before, q = m * C * ΔT, we can calculate the heat required to cool down the ice from 0 degrees C to -45.0 degrees C.

ΔT = (-45.0 - 0) = -45.0 degrees C

q3 = m * C * ΔT
q3 = 18.02 g * 2.09 J/g*degrees C * (-45.0 degrees C)

Finally, we can find the total heat evolved by adding q1, q2, and q3:

Total heat evolved = q1 + q2 + q3

Make sure to perform the calculation to get the final answer in kJ.

q1 = heat removed to convert steam at 145 C to steam at 100 c.

q1 = mass steam x specific heat x delta T. Delta T is of course 45.

q2 = heat removed to condense steam at 100 to liquid water at 100 C.
q2 = mass steam x heat vaporization.

q3 = heat removed to cool water at 100 C to zero C.
q3 = mass water x specific heat x delta T.

q4 = heat to convert liquid water at zero C. to ice at zero C.
q4 = mass water x heat fusion.

q5 = heat removed to convert ice at zero C. to ice at -45 C.
q5 = mass ice x specific heat x delta T.

Total heat is q1 + q2 + q3 + q4 + q5.
Note: One mole of water is 18.015 grams. I see specific heats are in J/g so it's easier to convert mole water to grams first than to convert each of the specific heats to J/mol. Wben you finish with the total heat, convert J to kJ.

36.4 kJ