How many KJ are released when 75.0g of steam at 100 degree C condenses cools to 0 degree C and freezes at 0 degree C?
I set it up like this but it is wrong,
What am I doing wrong:
75g x 2260J/1g water x 1Kj/1000J =
Your setup is on the right track, but there is a mistake in your conversion.
To determine the amount of heat released when 75.0g of steam at 100°C condenses, cools to 0°C, and freezes at 0°C, we need to consider three separate processes: condensation, cooling, and freezing.
1. Condensation:
The heat released during condensation can be calculated using the heat of vaporization of water. The heat of vaporization of water is 2260 J/g. So the heat released during condensation can be calculated as:
75.0g x 2260 J/g = 169,500 J (or 169.5 kJ)
2. Cooling:
To calculate the heat released during cooling from 100°C to 0°C, you need to consider the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. Using this value, the heat released during cooling can be calculated as:
75.0g x 4.18 J/g°C x (100°C - 0°C) = 31,350 J (or 31.35 kJ)
3. Freezing:
To calculate the heat released during freezing at 0°C, you again need to consider the heat of fusion of water. The heat of fusion of water is 334 J/g. The calculation for the heat released during freezing can be done as:
75.0g x 334 J/g = 25,050 J (or 25.05 kJ)
So, to calculate the total heat released, you can simply add up the results from each step:
Total heat released = 169.5 kJ + 31.35 kJ + 25.05 kJ = 225.9 kJ
Therefore, the correct answer is 225.9 kJ.