How long would it take for 1.50mol of water at 100.0 degrees to be converted completely into steam if heat were added at a constant rate of 23.0J/s?
Heatrate*time= masswater*Heatvaporization.
Watch units, solve for time.
To find out how long it would take for 1.50 mol of water to be completely converted into steam, we need to consider the process of vaporization and use the concept of specific heat and heat of vaporization. Here's how you can calculate it step by step:
1. Calculate the heat needed to convert all the water into steam.
The heat needed is given by the equation:
q = n * ΔHvap
where q is the heat, n is the number of moles, and ΔHvap is the heat of vaporization.
In this case, n = 1.50 mol and ΔHvap for water is approximately 40.7 kJ/mol (or 40,700 J/mol).
So, the heat needed is:
q = 1.50 mol * 40,700 J/mol = 61,050 J
2. Calculate the time it takes to convert all the water into steam.
Time (t) can be calculated using the equation:
t = q / P
where P is the power (rate of heat transfer).
In this case, P is given as 23.0 J/s.
t = 61,050 J / 23.0 J/s
t ≈ 2655.65 s
3. Convert the time to a more convenient unit.
The time is currently in seconds, so let's convert it to minutes or hours for easier understanding.
There are 60 seconds in a minute, so:
t = 2655.65 s / 60 s/min
t ≈ 44.26 min
Therefore, it would take approximately 44.26 minutes for 1.50 mol of water at 100.0 degrees Celsius to be completely converted into steam if heat were added at a constant rate of 23.0 J/s.