An electric kettle with 2.0kw heating element has a heat capacity of 400jk 1.0kg of water at 20 degree celcius is placed in the kettle, the kettle is switched on and it is found that 13 minutes later the mass of water in it is 0.5kg. ignoring heat losses. Calculate the value of specific latent heat vaporization water (the specific heat capacity of water=4.2*10^3jk^-1k^1
Well, well, well! Looks like we have a heating situation on our hands! Let's get crackin' with some calculations!
First things first, let's find out how much heat is required to raise the temperature of the water from 20°C to its boiling point. We know the specific heat capacity of water is 4.2 * 10^3 J/(kg·K), and we have 1.0 kg of water.
ΔQ1 = m * c * ΔT
ΔQ1 = 1.0 kg * 4.2 * 10^3 J/(kg·K) * (100°C - 20°C)
ΔQ1 = 1.0 kg * 4.2 * 10^3 J/(kg·K) * 80°C
ΔQ1 = 3.36 * 10^5 J
So, we have used up 3.36 * 10^5 J of heat to raise the temperature of the water to its boiling point. But we're not done yet!
Next, let's see how much heat is needed to vaporize the remaining 0.5 kg of water at its boiling point. We know the specific latent heat of vaporization for water is what we need to find.
ΔQ2 = m * L
ΔQ2 = 0.5 kg * L
Now, let's go back to the electric kettle. We know it has a heating element with a power of 2.0 kW, and it has been switched on for 13 minutes. We can convert this time to seconds:
t = 13 minutes * 60 seconds/minute
t = 780 seconds
The power of the kettle tells us how much energy it supplies per unit time. So, the energy supplied by the kettle is:
ΔQ = P * t
ΔQ = 2.0 kW * 780 s
ΔQ = 1.56 * 10^6 J
Putting it all together, we can set up an energy balance:
ΔQ = ΔQ1 + ΔQ2
1.56 * 10^6 J = 3.36 * 10^5 J + 0.5 kg * L
Solving for L:
L = (1.56 * 10^6 J - 3.36 * 10^5 J) / 0.5 kg
L = 1.23 * 10^6 J/kg
So, the specific latent heat of vaporization for water is approximately 1.23 * 10^6 J/kg.
Now that's some hot stuff! Just be careful not to get steamy with your calculations!
To calculate the specific latent heat of vaporization of water, we can use the formula:
Q = mcΔT + mL
Where:
Q = total heat energy supplied
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
L = specific latent heat of vaporization of water
Given:
Power of the kettle (P) = 2.0 kW = 2.0 * 10^3 W
Time (t) = 13 minutes = 13 * 60 seconds
Initial mass of water (m1) = 1.0 kg
Final mass of water (m2) = 0.5 kg
Initial temperature of water (T1) = 20°C = 20 + 273 = 293 K
Final temperature of water (T2) = boiling point of water = 100°C = 100 + 273 = 373 K
Specific heat capacity of water (c) = 4.2 * 10^3 J kg^(-1) K^(-1)
First, let's calculate the total heat energy supplied (Q):
Q = Pt
Q = (2.0 * 10^3) * (13 * 60)
Q = 156,000 J
Now, let's calculate the heat energy required to heat the water from 20°C to 100°C:
Q1 = mcΔT
Q1 = 1.0 * 4.2 * 10^3 * (373 - 293)
Q1 = 336,000 J
Next, let's calculate the heat energy required to vaporize 0.5 kg of water at 100°C:
Q2 = mL
To find L, we rearrange the equation to solve for L:
L = Q2 / m2
Q2 = Q - Q1
Q2 = 156,000 - 336,000
Q2 = -180,000 J (negative because heat is lost)
L = (-180,000) / 0.5
L = -360,000 J/kg
Since the specific latent heat of vaporization of water cannot be negative, we made a mistake somewhere in the calculations.