A heater marked 60w evaporates 6×10^-3kg of boiling water in 60secs .what is the specific latent heat of vaporization of Water in jkg^-1?
You mean 2256 kJ/kg I think.
but using the data provided:
60 watts * 60 seconds = 3600 Joules input
3600 Joules / 0.006 kg = 600,000 Joules/kg = 600 kJ/kg
I think not
To find the specific latent heat of vaporization of water, we can use the equation:
Energy (E) = Power (P) × Time (t)
Given:
Power (P) = 60 W
Time (t) = 60 s
First, we need to calculate the energy (E) used by the heater to evaporate the water. The energy used is given by:
E = Power × Time
E = 60 W × 60 s
E = 3600 J
Next, we need to find the mass (m) of the water that evaporated. The mass of the water can be calculated using the equation:
Energy (E) = mass (m) × specific latent heat of vaporization (L)
Given:
Energy (E) = 3600 J
Specific latent heat of vaporization (L) = ?
We can rearrange the equation to solve for the specific latent heat of vaporization:
L = E / m
Rearranging the equation gives:
L = E / m
L = 3600 J / (6 × 10^-3 kg)
Now, let's calculate the specific latent heat of vaporization:
L = 3600 J / (6 × 10^-3 kg)
L = (3600 J) / (0.006 kg)
L ≈ 600,000 J/kg
Therefore, the specific latent heat of vaporization of water is approximately 600,000 J/kg.
To find the specific latent heat of vaporization of water, we need to use the formula:
Q = mL
Where:
Q = heat energy absorbed or released (in joules)
m = mass of water (in kilograms)
L = specific latent heat of vaporization (in joules per kilogram, J/kg)
In this case, we know that the heater marked 60W (watts) evaporates 6×10^-3kg of water in 60 seconds (secs).
First, let's convert the power from watts to joules per second (J/s) since 1 Watt (W) = 1 Joule per second (J/s).
60W = 60J/s
Next, we need to calculate the total heat energy:
Q = Power x Time
Q = 60J/s x 60s
Q = 3600J
Now, we can calculate the specific latent heat of vaporization:
L = Q / m
Given that m = 6×10^-3kg, we can substitute the values:
L = 3600J / (6×10^-3kg)
L = 3600J / 0.006kg
L = 600,000 J/kg
Therefore, the specific latent heat of vaporization of water is 600,000 J/kg.