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 ?
To calculate the time it would take for water to be converted into steam, we can use the equation:
q = mL
where q is the heat added, m is the mass of the substance, and L is the heat of vaporization.
First, we need to find the heat required to convert 1.50 g of water into steam. The heat of vaporization of water is approximately 40.7 J/g.
q = mL
q = (1.50 g) * (40.7 J/g)
q ≈ 61.05 J
Next, we can calculate the time using the equation:
q = Pt
where q is the heat added, P is the power, and t is the time.
In this case, the heat added is 61.05 J and the power is 24.0 W.
q = Pt
61.05 J = (24.0 W) * t
To find t, rearrange the equation:
t = q / P
t = 61.05 J / 24.0 W
Now, let's calculate t:
t ≈ 2.54 seconds
Therefore, it would take approximately 2.54 seconds for 1.50 g of water at 100.0 degrees Celsius to be completely converted into steam if heat were added at a constant rate of 24.0 watts.
To calculate the time it takes for water to be converted into steam, we need to use the equation:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the substance (in kilograms)
c = specific heat capacity (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's calculate the heat energy required to convert the water into steam:
Q = mL
Where:
m = mass of the substance (in kilograms)
L = latent heat of vaporization (in joules per kilogram)
The latent heat of vaporization for water is approximately 2,260,000 joules per kilogram.
m = 1.50 kg
L = 2,260,000 J/kg
Q = (1.50 kg) * (2,260,000 J/kg)
Q = 3,390,000 J
Next, we can substitute the values into the equation Q = mcΔT to calculate the change in temperature:
Q = mcΔT
ΔT = Q / (mc)
m = 1.50 kg
c = specific heat capacity of water (approximately 4,186 J/kg°C)
ΔT = (3,390,000 J) / (1.50 kg * 4,186 J/kg°C)
ΔT ≈ 509.8 °C
Now, let's calculate the time it would take for the water to be converted into steam, considering a constant rate of heat addition of 24.0 J/s:
Time (t) = Q / P
Where:
Q = heat energy (in joules)
P = power (in watts, which is equal to joules per second)
Power (P) = 24.0 J/s
t = (3,390,000 J) / (24.0 J/s)
t ≈ 141,250 s
So, it would take approximately 141,250 seconds for 1.50 kg 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.