What would the final volume of water be if 80.0 kJ were added to a 100.0 mL sample at 100.0 degrees Celsius?
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The heat of vaporization of water is 2257 J/gm. The addition of 80,000 J to liquid water that is already 100 C will result in the evaporation of 35.5 g, That will leave you with 64.5 g of water, occupying 64.5 mL.
The evaporated water will expand into whatever volume is available.
To determine the final volume of water after adding energy to a sample, we need to consider the change in temperature and the expansion of water.
First, let's calculate the change in temperature (∆T) using the initial and final temperatures. You have not provided the final temperature, so we'll assume it remains constant at 100.0 degrees Celsius.
∆T = Final Temperature - Initial Temperature
∆T = 100.0 ˚C - 100.0 ˚C
∆T = 0 ˚C
Since there is no change in temperature (∆T = 0 ˚C), there will be no change in volume due to thermal expansion. Therefore, we can calculate the final volume of water by adding the initial volume to the volume change (∆V).
∆V = Energy Added / Density of Water
The density of water at 100.0 degrees Celsius is approximately 0.9584 g/mL. However, you have provided the energy added in kilojoules (kJ) and the sample volume in milliliters (mL). So, we need to convert the energy from kilojoules to joules and the sample volume from milliliters to liters.
1 kJ = 1000 J
1 mL = 0.001 L
Let's perform the necessary conversions:
Energy Added (J) = 80.0 kJ x 1000 J/kJ = 80,000 J
Sample Volume (L) = 100.0 mL x 0.001 L/mL = 0.1 L
Now, we can calculate the volume change (∆V):
∆V = 80,000 J / (0.1 L x 0.9584 g/mL)
To complete the calculation, we need to know the density of water at the final temperature, but since the temperature remains constant at 100.0 degrees Celsius, we can use the density at that temperature. The calculation will give us the volume change in liters.
Finally, we add the initial volume (100.0 mL) to the volume change (∆V) to get the final volume of water.