A pot of water is boiling under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In two minutes, the mass of water boiled away is m = 2.5 kg. The radius of the pot bottom is R = 8.5 cm and the thickness is L = 2.0 mm. What is the temperature of the heating element in contact with the pot?
The specific latent heat of evaporation of water at 100 C is:
L= 2260 kJ/kg
The amount of heat that entered the pot in two minutes is thus:
Q = 2.5 kg * 2260 kJ/kg = 5650 kJ
Dividing this by the time gives you the power:
Q/(2 minutes) = 5650 kJ/(120 s) =
47.1 kW
If you divide this by the surface area, you get the heat flux:
47.1 kW/[pi (8.5 cm)^2] =
2.07*10^6 W/m^2
This heat flux is delivered to the water via thermal conduction, it is therefore equal to minus the gradient of the temperature in the copper inbetween the heating element and the water, times the heat conduction coefficient of copper.
If we take the heat conduction coefficient of copper to be
lambda = 385 W/(m K), then we have:
lambda (T_heating element - 100 C)/(2 mm) = 2.07*10^6 W/m^2 ------->
T_heating element = 110.8 C
Thank you. Your answer is correct, but I get a different answer when I try to calculate that last equation myself. I must be in kelvin or something.
Well, let's dive into the physics of boiling water. But fear not, I'll try to keep it entertaining!
First, let's acknowledge that water boils at 100°C under standard atmospheric pressure. So, if the water is boiling, it means it's already reached that temperature.
Now, we turn our attention to the heating element. Since the heat enters the pot only through its bottom, we can assume that the temperature of the heating element is the same as the boiling water, which is 100°C.
But hold your laughter, because it's time for some calculations!
We have the mass of water boiled away in 2 minutes, which is 2.5 kg. To find the heat required to boil this amount of water, we can use the equation:
Q = m * L_v
Where Q is the heat, m is the mass, and L_v is the latent heat of vaporization. For water, the latent heat of vaporization is approximately 2260 kJ/kg.
Plugging in the values, we get:
Q = 2.5 kg * 2260 kJ/kg ≈ 5650 kJ
Since we are assuming the heat enters the pot only through the bottom, we can assume all the heat is transferred to the water. And since the heating element is in contact with the pot, we can say that the temperature of the heating element is also 100°C.
So, the temperature of the heating element in contact with the pot is a cracking 100°C!
To find the temperature of the heating element in contact with the pot, we can use the formula for the amount of heat transferred through conduction:
Q = (k * A * ΔT * t) / L
Where:
Q = Heat transferred
k = Thermal conductivity of copper
A = Area of the pot bottom
ΔT = Change in temperature
t = Time
L = Thickness of the pot bottom
First, we need to find the area of the pot bottom. The area of a circle is given by:
A = π * R^2
Given that R = 8.5 cm, we can calculate the area:
A = π * (8.5 cm)^2
Next, we rearrange the formula to solve for the change in temperature:
ΔT = (Q * L) / (k * A * t)
Substituting the given values:
ΔT = (m * g * L) / (k * A * t)
Where:
m = Mass of water boiled away
g = Acceleration due to gravity
Given that m = 2.5 kg, g = 9.8 m/s^2, L = 2.0 mm, and t = 2 minutes = 120 seconds, we can calculate the change in temperature:
ΔT = (2.5 kg * 9.8 m/s^2 * 0.002 m) / (k * π * (0.085 m)^2 * 120 s)
Now, we need the thermal conductivity of copper, which is approximately 401 W/(m*K). Substituting this value and calculating:
ΔT ≈ (2.5 kg * 9.8 m/s^2 * 0.002 m) / (401 W/(m*K) * 3.14 * (0.085 m)^2 * 120 s)
Finally, we can evaluate this expression to find the change in temperature, ΔT.
To determine the temperature of the heating element in contact with the pot, we can use the principles of heat transfer and energy conservation.
1. Step one is to calculate the amount of heat required to boil the water. This can be done using the formula:
Q = mL
where Q is the heat required, m is the mass of water boiled away, and L is the latent heat of vaporization.
In this case, m = 2.5 kg and the latent heat of vaporization of water is approximately 2.26 x 10^6 J/kg. Therefore, we can calculate:
Q = (2.5 kg) x (2.26 x 10^6 J/kg)
= 5.65 x 10^6 J
2. The next step is to determine the rate of heat transfer through conduction from the heating element to the pot bottom.
The rate of heat transfer through conduction is given by:
Q = (kAΔT) / L
where Q is the rate of heat transfer, k is the thermal conductivity of copper, A is the surface area of the pot bottom, ΔT is the temperature difference between the heating element and the boiling water, and L is the thickness of the pot bottom.
We need to solve this equation for ΔT, so we rearrange it as:
ΔT = (QL) / (kA)
3. Now, we need to determine the thermal conductivity of copper. The thermal conductivity of copper is typically around 400 W/(m·K).
4. We can now calculate the surface area of the pot bottom. The surface area of a cylinder is given by:
A = 2πR^2 + 2πRL
where R is the radius of the pot bottom and L is the thickness of the pot bottom.
Plugging in the values, we have:
A = 2π(0.085 m)^2 + 2π(0.085 m)(0.002 m)
≈ 0.0453 m^2
5. Finally, we can substitute the known values into the equation derived in step 2 to calculate ΔT:
ΔT = (5.65 x 10^6 J) / (400 W/(m·K) × 0.0453 m^2)
≈ 314.5 K
Therefore, the temperature of the heating element in contact with the pot is approximately 314.5 Kelvin.