A quantity of steam (750 g) at 122°C is condensed, and the resulting water is frozen into ice at 0°C. How much heat was removed?
0.678
To calculate the amount of heat removed, we need to determine the change in heat energy during the phase changes from steam to water and from water to ice.
First, let's calculate the heat required to condense the steam into water:
1. Calculate the heat required to heat the steam from 122°C to 100°C (the boiling point of water):
Q1 = m1 * c1 * ΔT1
where:
m1 = mass of steam = 750 g
c1 = specific heat capacity of steam = 2.03 J/g°C (approximate value for steam)
ΔT1 = change in temperature = 100°C -122°C
Q1 = heat required for this step
2. Calculate the heat required to condense the steam into water at 100°C:
Q2 = m2 * ΔHv
where:
m2 = mass of steam = 750 g
ΔHv = heat of vaporization = 40.7 kJ/g (approximate value for water vapor)
Q2 = heat required for this step
Next, let's calculate the heat required to freeze the water into ice:
3. Calculate the heat required to cool the water from 0°C to 100°C (the freezing point of water):
Q3 = m3 * c3 * ΔT3
where:
m3 = mass of water = 750 g
c3 = specific heat capacity of water = 4.18 J/g°C
ΔT3 = change in temperature = 0°C - 100°C
Q3 = heat required for this step
4. Calculate the heat required to freeze the water into ice at 0°C:
Q4 = m4 * ΔHf
where:
m4 = mass of water = 750 g
ΔHf = heat of fusion = 333.5 kJ/kg (approximate value for water)
Q4 = heat required for this step
Finally, calculate the total heat removed by summing up all the individual heat values:
Total heat removed = Q1 + Q2 + Q3 + Q4
Note: It's important to be consistent with units throughout the calculations. In this case, the specific heat capacity is in J/g°C, and the heat of vaporization and heat of fusion are in kJ/g and kJ/kg, respectively.