How much heat (in kJ) is evolved in converting 1.00 mol of steam at 130.0 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.

You must go through all of the steps from steam at 130 to ice at -50.

q to move steam at 130 to steam at 100.
q1 = mass x specific heat x delta T.

q to convert steam at 100 to liquid water at 100.
q2 = mass x heat of vaporization.

q to move water at 100 to water at zero.
q3 = mass x specific heat x delta T.

q to convert water at zero to ice at zero.
q4 = mass x heat of fusion.

q to move ice at zero to ice at -50.
q5 = mass x specific heat x delta T.

Then total q =
q1+q2+q3+q4+q5.
I notice your specific heats are listed in joules so your answer will come out in Joules and you must convert to kJ since the problem asks for kJ.

55.078

57 kj

To determine the heat evolved in converting 1.00 mol of steam at 130.0°C to ice at -50.0°C, we need to follow a two-step process:

Step 1: Calculate the heat required to lower the temperature of steam from 130.0°C to 0.0°C.
Step 2: Calculate the heat released to freeze the water at 0.0°C into ice at -50.0°C.

Before we start, we need to gather the necessary information:

Given:
- The heat capacity of steam (Csteam) = 2.01 J/g°C
- The heat capacity of ice (Cice) = 2.09 J/g°C

Now let's calculate the heat for each step:

Step 1: Lowering the temperature of steam from 130.0°C to 0.0°C
To calculate the heat required, we need to know the mass of the steam. The molar mass of water is approximately 18.01 g/mol. Since we have 1.00 mol of steam, the mass of steam is 18.01 g.

The temperature change in this step is (0.0°C - 130.0°C) = -130.0°C.
Using q = m * C * ΔT, where q is the heat, m is the mass, C is the heat capacity, and ΔT is the temperature change, we can calculate the heat required:

q1 = m * Csteam * ΔT1
= 18.01 g * 2.01 J/g°C * (-130.0°C)
= -4805.26 J

To convert this to kJ, we divide by 1000:

q1 = -4805.26 J / 1000
= -4.805 kJ

Step 2: Freezing steam at 0.0°C into ice at -50.0°C
Since we are converting all of the steam to ice, the mass remains the same (18.01 g). The temperature change in this step is (-50.0°C - 0.0°C) = -50.0°C.

Using the same formula as above, we can calculate the heat released:

q2 = m * Cice * ΔT2
= 18.01 g * 2.09 J/g°C * (-50.0°C)
= -1894.79 J

Converting to kJ:

q2 = -1894.79 J / 1000
= -1.895 kJ

Finally, to find the total heat evolved, we add the two calculated heats:

Total heat evolved = q1 + q2
= -4.805 kJ + (-1.895 kJ)
= -6.70 kJ

Therefore, the heat evolved in converting 1.00 mol of steam at 130.0°C to ice at -50.0°C is -6.70 kJ.