A quantity of steam (390 g) at 113 degrees Celsius is condensed, and the resulting water is frozen into ice at 0 degree Celsius. How much heat, in kcal, was removed a) from the steam to reach 100 degrees Celsius?b) to condense the steam from 100 degrees Celsius to water at the same temperature? c) from the water at 100 degrees Celsius to reach 0 degree Celsius?d) to freeze the water into ice at 0 degrees Celsius?
The Correct Answer and Explanation is :
To determine the total heat removed during the transformation of 390 grams of steam at 113°C to ice at 0°C, we need to consider each phase change and temperature change separately. The specific heat capacities and latent heats involved are as follows:
- Specific heat of steam: 0.48 kcal/kg·°C
- Latent heat of vaporization: 540 kcal/kg
- Specific heat of water: 1 kcal/kg·°C
- Latent heat of fusion: 80 kcal/kg
a) Heat removed from the steam to reach 100°C:
The steam must be cooled from 113°C to 100°C. The heat removed is calculated using the formula:
[ Q = m \times c_{\text{steam}} \times \Delta T ]
Where:
- ( m = 390 \, \text{g} = 0.39 \, \text{kg} )
- ( c_{\text{steam}} = 0.48 \, \text{kcal/kg·°C} )
- ( \Delta T = 113°C – 100°C = 13°C )
Calculating:
calculatorturn0calculator0
b) Heat removed to condense the steam at 100°C:
The steam condenses at 100°C, releasing heat equal to its latent heat of vaporization:
[ Q = m \times L_{\text{vaporization}} ]
Where:
- ( L_{\text{vaporization}} = 540 \, \text{kcal/kg} )
Calculating:
calculatorturn0calculator1
c) Heat removed from the water at 100°C to reach 0°C:
The water is cooled from 100°C to 0°C. The heat removed is:
[ Q = m \times c_{\text{water}} \times \Delta T ]
Where:
- ( c_{\text{water}} = 1 \, \text{kcal/kg·°C} )
- ( \Delta T = 100°C – 0°C = 100°C )
Calculating:
calculatorturn0calculator2
d) Heat removed to freeze the water into ice at 0°C:
The water freezes at 0°C, releasing heat equal to its latent heat of fusion:
[ Q = m \times L_{\text{fusion}} ]
Where:
- ( L_{\text{fusion}} = 80 \, \text{kcal/kg} )
Calculating:
calculatorturn0calculator3
Total heat removed:
Summing the heat removed in all steps:
[ Q_{\text{total}} = Q_a + Q_b + Q_c + Q_d ]
Calculating:
Therefore, the total heat removed is 2433.6 kcal.
This calculation accounts for the cooling of steam from 113°C to 100°C, its condensation at 100°C, the cooling of water from 100°C to 0°C, and its freezing at 0°C. Each phase change and temperature change involves specific amounts of heat energy, which are summed to determine the total heat removed.