how many kcal of heat must be withdrawn from 4.56 kg of steam at 125 degrees C to change it to ice at -44.5 degrees C???
To calculate the amount of heat needed to change steam to ice, we need to consider the latent heat of vaporization and the latent heat of fusion.
First, we need to calculate the heat required to cool the steam from 125 degrees Celsius to 100 degrees Celsius:
1. Calculate the heat needed to cool the steam from 125 degrees C to 100 degrees C:
Q1 = m × cp × (T2 - T1)
Q1 = 4.56 kg × 4.186 J/g°C × (100 - 125) °C
Q1 = -244.16 kJ (Negative sign represents heat removed from the system)
Next, we need to calculate the heat required to convert the steam at 100 degrees Celsius to water at 100 degrees Celsius:
2. Determine the heat of vaporization:
The heat of vaporization for water at 100 degrees Celsius is 2260 kJ/kg.
Q2 = m × Hvap
Q2 = 4.56 kg × 2260 kJ/kg
Q2 = 10305.6 kJ
Now, we need to calculate the heat required to cool the water from 100 degrees Celsius to 0 degrees Celsius:
3. Calculate the heat needed to cool the water from 100 degrees C to 0 degrees C:
Q3 = m × cp × (T2 - T1)
Q3 = 4.56 kg × 4.186 J/g °C × (0 - 100) °C
Q3 = -1903.2 kJ (Negative sign represents heat removed from the system)
Lastly, we need to calculate the heat required to freeze the water at 0 degrees Celsius to ice at -44.5 degrees Celsius:
4. Determine the heat of fusion:
The heat of fusion for water is 334 kJ/kg.
Q4 = m × Hfusion
Q4 = 4.56 kg × 334 kJ/kg
Q4 = 1525.04 kJ
Finally, we sum up all the calculated quantities to get the total heat required:
Total Heat (Q) = Q1 + Q2 + Q3 + Q4
Total Heat (Q) = -244.16 kJ + 10305.6 kJ - 1903.2 kJ + 1525.04 kJ
Total Heat (Q) = 9103.28 kJ
Therefore, approximately 9103.28 kcal of heat must be withdrawn in order to change 4.56 kg of steam at 125 degrees Celsius to ice at -44.5 degrees Celsius.