You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90 degrees C. the specific heat capacity of aluminum is 900J/kgK. ?
(a) how much energy did it take to heat the pot of water?
(b) the diameter of the cylindrical pot is 20cm and the height of the water is 4.8cm. How high (in mm) will the water column rise (assuming the diameter stays constant?) coefficient for volumetric expansion of water is 207e-6/deg. C.
(c) a small Air bubble is self contained trapped between the bottom of the aluminum pot and a glass stove top. One micro percent (1e-6%) of the energy that went into hitting the pot full of water (from part A) goes into the air bubble. As a result the bubble experiences of volume change of 3e-6m^3 with the pressure remain in constant. Assuming it acts like an ideal gas, find that pressure
(a) To calculate the amount of energy required to heat the pot of water, we need to use the specific heat capacity formula:
Q = m * c * ΔT
where:
Q = energy (in Joules)
m = mass (in kilograms)
c = specific heat capacity (in Joules per kilogram per Kelvin)
ΔT = change in temperature (in Kelvin)
In this case, we have:
m = 1.5 kg (mass of water)
c = specific heat capacity of water (which is approximately 4186 J/kgK)
ΔT = (90 - 5) = 85 degrees Celsius
Using the formula, we can calculate Q:
Q = 1.5 kg * 4186 J/kgK * 85 K = 533,955 J
Therefore, it took approximately 533,955 Joules of energy to heat the pot of water.
(b) To calculate how high the water column will rise, we need to consider the expansion of water due to heating. The formula for volumetric expansion is:
ΔV = V₀ * β * ΔT
where:
ΔV = change in volume
V₀ = initial volume
β = coefficient of volumetric expansion
ΔT = change in temperature
We are given that the coefficient of volumetric expansion of water is 207e-6/deg. C and the initial volume of the water is V₀ = π * (0.1 m)^2 * 0.048 m (assuming 20 cm diameter and 4.8 cm height).
ΔT = (90 - 5) = 85 degrees Celsius
Now, we can substitute the values into the formula:
ΔV = π * (0.1 m)^2 * 0.048 m * 207e-6/deg. C * 85 deg. C
Calculate:
ΔV ≈ 0.00109416 m^3
To convert this to millimeters, we multiply by 10^6:
ΔV ≈ 1094.16 mm^3
Therefore, the water column will rise approximately 1094.16 mm.
(c) To find the pressure experienced by the air bubble, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant (approximately 8.314 J/(mol·K))
T = temperature (in Kelvin)
We are given that the volume change of the air bubble is ΔV = 3e-6 m^3 and the pressure remains constant. Also, we assume the temperature to remain constant.
Using the ideal gas law, we can rearrange the equation:
P₁V₁ = P₂V₂
where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume of the air bubble.
Substituting the values into the equation:
P₁ * V₁ = P₂ * (V₁ + ΔV)
We are given that ΔV = 3e-6 m^3, so:
P₁ * V₁ = P₂ * (V₁ + 3e-6)
Since the pressure remains constant (P₁ = P₂), we can solve the equation for P₁:
P₁ = P₂ = P = P₁ * V₁ / (V₁ + 3e-6)
Therefore, the pressure experienced by the air bubble is P = P₁ * V₁ / (V₁ + 3e-6).