How much heat must be added to turn 3 grams of water at 100°C into a vapor and what will be the final temperature after this much heat is added
To find out how much heat is required to turn 3 grams of water at 100°C into vapor, we can use the specific heat formula and the heat of vaporization formula.
1. First, we need to calculate the heat required to raise the temperature of 3 grams of water from 100°C to its boiling point, which is 100°C.
The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, the heat required is given by:
Heat = mass × specific heat capacity × change in temperature
Heat = 3 g × 4.18 J/g°C × (100°C - 100°C) = 0 J
So, no heat is required to raise the temperature of water at its boiling point.
2. Next, we need to calculate the heat required for the phase change from liquid to vapor. The heat of vaporization of water is approximately 2260 J/g.
Heat = mass × heat of vaporization
Heat = 3 g × 2260 J/g = 6780 J
So, 6780 J of heat must be added to turn 3 grams of water at 100°C into vapor.
Now, let's calculate the final temperature after this much heat is added.
The final temperature will depend on the pressure because boiling point varies with pressure. Assuming standard atmospheric pressure at sea level, the boiling point of water is 100°C.
However, once all the water has vaporized, any additional heat added will not change the temperature. At this point, the water has reached its boiling point, and any extra heat is used only for the phase change (from liquid to vapor), not to increase temperature further.
So, the final temperature after adding heat to vaporize 3 grams of water at 100°C will still be 100°C.