If we instead used 90 g of steam, how much of the 200 g of water would have turned into 100° C steam in the end?
To find out how much of the 200 g of water would have turned into 100°C steam if we used 90 g of steam instead, we need to calculate the amount of water that evaporated.
First, let's determine the amount of steam formed when 90 g of steam is used instead.
Since water and steam have the same density (at the same temperature and pressure), we can assume that 90 g of steam has the same volume as 90 mL of water.
Next, we need to consider that the initial 200 g of water and the 90 g of steam will be mixed together.
To calculate the total mass of the mixture, we add the mass of water and steam: 200 g + 90 g = 290 g.
Now, assuming no loss of heat or energy during the process, the total volume of the mixture will be equal to the sum of the volumes of water and steam. Since the volume of water is equal to its mass (200 mL), and the volume of steam is equal to its mass (90 mL), the total volume of the mixture is 290 mL.
Finally, since the mixture is at 100°C, which is the boiling point of water, the entire mixture will be in the form of steam. Therefore, none of the 200 g of water would have turned into steam.
In conclusion, if we used 90 g of steam instead, none of the 200 g of water would have turned into 100°C steam in the end.