How much heat is associated with converting 100 g of steam at 112 degrees C to ice at -5.3 degrees C

To determine the amount of heat required to convert 100 g of steam at 112 degrees Celsius to ice at -5.3 degrees Celsius, we need to consider two separate phase changes: first, from steam to water, and then from water to ice.

The specific heat capacity (c) of water (in its liquid state) is 4.184 J/g°C. Therefore, to convert 100 g of steam at 112 degrees Celsius to water at 100 degrees Celsius, we need to calculate the heat using the formula:

Q1 = m × c × ΔT1

Where:
Q1 is the heat required
m is the mass of the steam (100 g)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT1 is the temperature change (100 - 112 = -12 degrees Celsius)

Plugging in the values, we get:

Q1 = 100 g × 4.184 J/g°C × (-12°C)
Q1 = -5014.08 J (since heat is lost in this case)

Next, we need to convert this water at 100 degrees Celsius to ice at -5.3 degrees Celsius. The latent heat of fusion (L) for water (the heat needed to convert a substance from a liquid to a solid state or vice versa without a change in temperature) is 334 J/g. The negative sign indicates energy being released during the phase transition.

Q2 = m × L

Where:
Q2 is the heat required
m is the mass of water (100 g)
L is the latent heat of fusion for water (334 J/g)

Plugging in the values, we get:

Q2 = 100 g × (-334 J/g)
Q2 = -33400 J (again, since heat is lost)

Now, to find the total heat required, we sum up the two amounts:

Total heat = Q1 + Q2
Total heat = -5014.08 J + (-33400 J)
Total heat = -38414.08 J

Therefore, it takes approximately -38414.08 Joules of heat to convert 100 g of steam at 112 degrees Celsius to ice at -5.3 degrees Celsius. The negative sign indicates that heat is being lost during the process.

To calculate the heat associated with converting steam to ice, we need to consider two steps: first, the heat required to cool down the steam from 112 degrees Celsius to 100 degrees Celsius, and second, the heat needed to convert the steam at 100 degrees Celsius to ice at -5.3 degrees Celsius.

Step 1: Cooling the steam from 112 degrees Celsius to 100 degrees Celsius
The specific heat capacity of steam is approximately 2.03 J/g°C.

The heat required to cool down 100 g of steam from 112 degrees Celsius to 100 degrees Celsius can be calculated using the formula:

q1 = mass × specific heat capacity × ΔT
q1 = 100 g × 2.03 J/g°C × (100°C - 112°C)

q1 = -1,616 J

Note: The negative value indicates heat loss during the cooling process.

Step 2: Converting steam at 100 degrees Celsius to ice at -5.3 degrees Celsius
The specific heat capacity of liquid water (steam at 100 degrees Celsius) is approximately 4.18 J/g°C.

The heat required to convert 100 g of steam at 100 degrees Celsius to ice at -5.3 degrees Celsius involves two phases: cooling the water to 0 degrees Celsius and then freezing it to ice.

Phase 1: Cooling water from 100 degrees Celsius to 0 degrees Celsius
q2 = mass × specific heat capacity × ΔT
q2 = 100 g × 4.18 J/g°C × (0°C - 100°C)

q2 = -41,800 J

Note: The negative value indicates heat loss during the cooling process.

Phase 2: Freezing water to ice
The heat of fusion (enthalpy of fusion) for water is approximately 334 J/g.

q3 = mass × heat of fusion
q3 = 100 g × 334 J/g

q3 = 33,400 J

In total, the heat associated with converting 100 g of steam at 112 degrees Celsius to ice at -5.3 degrees Celsius is,

Total heat = q1 + q2 + q3
Total heat = -1,616 J + -41,800 J + 33,400 J
Total heat = -9,016 J

Note: The negative value indicates overall heat loss during the process.