how long would it take for 1.50 mol of water at 100C to be converted completely into steam if heat were added at a constant rate of 18.0J/s?
How much energy do you need to make the conversion?
1.50 mol x 18g/mol = 27 g H2O
q = 27g H2O x 2259 J/g = about 60,999 J.
18.0 J/s x ?s = 60,999 J.
Solve for s.
To calculate the time it takes for 1.50 mol of water to be converted into steam, we need to use the concepts of heat capacity and phase change.
Here's how you can find the answer step by step:
Step 1: Determine the heat required for the phase change
To convert water from 100°C to steam, it undergoes a phase change from liquid to gas. The heat required for this phase change is given by the equation:
q = n * ΔHvap
Where:
q is the heat required (in Joules)
n is the number of moles of water
ΔHvap is the molar heat of vaporization
The molar heat of vaporization for water is approximately 40.7 kJ/mol or 40,700 J/mol.
So, for 1.50 mol of water, the heat required for the phase change is:
q = 1.50 mol * 40,700 J/mol = 61,050 J
Step 2: Determine the time taken
The rate of heat transfer is given as 18.0 J/s (Joules per second). To find the time taken, we divide the heat required by the rate of heat transfer:
t = q / rate
t = 61,050 J / 18.0 J/s ≈ 3386.11 s
Therefore, it would take approximately 3386.11 seconds (or 56.43 minutes) for 1.50 mol of water to be completely converted into steam at a constant rate of 18.0 J/s.