What is specific latent heat of vporization and specific latent heat of fusion? An electrical kettle with a 2.0kW. heating element has a heat capacity of 400JK-1. A 1.0kg of water at 20 degree celsisus is placed in the kettle. The kettle is switched on and at 13 minutes later the mass of water remaing is 0.5kg. Ignoring heat losses calculate a value for specific latent heat of vaporization of water. The specific heat capacity of water is 4200 JKg-1 K-1.
The specific latent heat of vaporization and specific latent heat of fusion are properties of a substance that represent the amount of energy required to change its state.
Specific latent heat of vaporization (Lv) is the amount of energy required to convert one unit mass of a substance from its liquid state to the vapor state at constant temperature and pressure. It is usually measured in joules per kilogram (J/kg).
Specific latent heat of fusion (Lf) is the amount of energy required to convert one unit mass of a substance from its solid state to the liquid state at constant temperature and pressure. It is also measured in joules per kilogram (J/kg).
In this problem, we are given an electrical kettle with a 2.0 kW heating element, a heat capacity of 400 J/K, a mass of water of 1.0 kg at 20 degrees Celsius, and after 13 minutes, the mass of the water remaining is 0.5 kg. We need to calculate the specific latent heat of vaporization (Lv) of water.
First, let's calculate the energy input into the water from the heating element. The power (P) of the kettle is given as 2.0 kW, which is equivalent to 2000 W. The time (t) is 13 minutes, which is equivalent to 13 * 60 = 780 seconds.
The energy input (Q) into the water can be calculated using the formula:
Q = P * t
Substituting the given values, we have:
Q = 2000 W * 780 s = 1,560,000 J
Next, let's calculate the change in temperature (ΔT) of the water. The specific heat capacity (C) of water is given as 4200 J/kg K, and the initial temperature (T1) is 20 degrees Celsius.
The change in temperature can be calculated using the formula:
ΔT = Q / (m * C)
Since the mass (m) of the water remaining is 0.5 kg, we can substitute the given values:
ΔT = 1,560,000 J / (0.5 kg * 4200 J/kg K) = 740 K
Now, let's calculate the energy required to raise the temperature of 0.5 kg of water from 20 degrees Celsius to its boiling point (100 degrees Celsius).
The energy required (Q2) can be calculated using the formula:
Q2 = m * C * ΔT
Substituting the values, we have:
Q2 = 0.5 kg * 4200 J/kg K * 80 K = 168,000 J
Finally, the energy required to vaporize the remaining 0.5 kg of water can be calculated by subtracting the energy required to raise the temperature from the total energy input:
Qvaporization = Q - Q2 = 1,560,000 J - 168,000 J = 1,392,000 J
The specific latent heat of vaporization of water (Lv) can be calculated by dividing the energy required for vaporization by the mass of the water:
Lv = Qvaporization / m
Substituting the values, we have:
Lv = 1,392,000 J / 0.5 kg = 2,784,000 J/kg
Therefore, the specific latent heat of vaporization of water is approximately 2,784,000 J/kg.