How long would it take for 1.50 of water at 100.0 to be converted completely into steam if heat were added at a constant rate of 24.0
How long would it take for 1.50mol of water at 100.0 degree celcius to be converted completely into steam if heat were added at a constant rate of 24.0 j/s?
How much heat is required?
1.5 g x heat vaporization = q.
Now if heat is added at 24.0 J/s, how many seconds will it take?
42.2 mins
To determine how long it would take for the water to be converted completely into steam, we can use the specific heat and heat of vaporization of water.
First, we need to calculate the amount of heat required to convert the water into steam. The specific heat of water is 4.18 J/g°C, and the heat of vaporization of water is 2260 J/g.
1.50 g of water at 100.0°C needs to be heated to its boiling point, which is 100.0°C. The heat required to heat the water to this temperature can be calculated using the equation:
q1 = m * C * ΔT
Where:
q1 = heat required
m = mass of water being heated (1.50 g)
C = specific heat of water (4.18 J/g°C)
ΔT = change in temperature (100.0°C - 0.0°C)
q1 = 1.50 g * 4.18 J/g°C * 100.0°C
q1 = 627 J
Next, we need to calculate the heat required to convert the water at 100.0°C into steam at 100.0°C. The equation to calculate this heat is:
q2 = m * Hv
Where:
q2 = heat required
m = mass of water being converted (1.50 g)
Hv = heat of vaporization of water (2260 J/g)
q2 = 1.50 g * 2260 J/g
q2 = 3390 J
The total heat required to convert 1.50 g of water into steam is the sum of q1 and q2:
q_total = q1 + q2
q_total = 627 J + 3390 J
q_total = 4017 J
Finally, we can use the heat applied at a constant rate of 24.0 J/s to determine the time required for the water to be converted into steam:
t = q_total / heat rate
t = 4017 J / 24.0 J/s
t ≈ 167.375 s
Therefore, it would take approximately 167.375 seconds for 1.50 g of water at 100.0°C to be converted completely into steam if heat were added at a constant rate of 24.0 J/s.