A thick, vertical iron pipe has an inner diameter of 0.070 m. A thin aluminum (α = 23x10-6 (C°)-1) disk, heated to a temperature of 89°C, has a diameter that is 4.4x10-5 m greater than the pipe's inner diameter. The disk is laid on top of the open upper end of the pipe, perfectly centered on it, and allowed to cool. What is the temperature of the aluminum disk when the disk falls into the pipe? Ignore the temperature change of the pipe.
To find the final temperature of the aluminum disk when it falls into the pipe, we can use the principle of thermal equilibrium. The temperature will be such that the heat gained by the aluminum disk is equal to the heat lost.
We can start by calculating the initial and final volumes of the disk.
Given:
Inner diameter of the pipe (D) = 0.070 m
Diameter of the disk (d) = Inner diameter of the pipe + 4.4x10^(-5) m = 0.070 m + 4.4x10^(-5) m
Initial volume of the disk (V_initial) = (π/4) * (d^2) * thickness
Final volume of the disk (V_final) = (π/4) * (D^2) * thickness
Since the thickness of the disk is not known, let's consider it as 't' for now.
The initial and final volumes will be equal when the disk falls into the pipe. Therefore, V_initial = V_final.
(π/4) * (d^2) * t = (π/4) * (D^2) * t
Since the thickness of the disk cancels out, we can focus on the diameters of the disk and pipe.
(d^2) = (D^2)
(d^2) = (0.070 m)^2
d = 0.070 m
So, the diameter of the disk remains the same, and it's equal to the diameter of the pipe. This implies that the disk will be just big enough to fall into the pipe when it is at the same temperature.
Therefore, the temperature of the aluminum disk when it falls into the pipe will be the same as the temperature of the environment or surroundings around the pipe. Since the temperature change of the pipe is ignored, we assume the surroundings have the same temperature as the initial temperature of the disk, which is 89°C.
Therefore, the temperature of the aluminum disk when it falls into the pipe is 89°C.
To solve this problem, we can use the concept of thermal expansion and the principle of conservation of volume.
Step 1: Find the change in diameter of the aluminum disk.
Given:
Inner diameter of the iron pipe (D) = 0.070 m
Change in diameter of the disk (ΔD) = 4.4x10^(-5) m
The new diameter of the disk (D_aluminum) = D + ΔD
D_aluminum = 0.070 m + 4.4x10^(-5) m
D_aluminum = 0.070044 m (approximated to five decimal places)
Step 2: Find the change in volume of the aluminum disk.
The change in volume (ΔV) is given by the formula:
ΔV = V * β * ΔT
Where:
V is the initial volume of the disk,
β is the thermal expansion coefficient of aluminum, and
ΔT is the change in temperature of the disk.
We can rewrite this equation as:
ΔT = ΔV / (V * β)
In this problem, we are given:
Initial temperature of the disk (T_initial) = 89°C
Thermal expansion coefficient of aluminum (β_aluminum) = 23x10^(-6) (°C)^(-1)
Inner diameter of the disk (D) = D_aluminum = 0.070044 m (approximated to five decimal places)
We need to find ΔT, which is the change in temperature of the disk.
Step 3: Find the initial volume (V_initial) of the disk.
The initial volume of the disk is given by the formula for the volume of a cylinder:
V_initial = π * r_initial^2 * h
Where:
r_initial is the initial radius of the disk, and
h is the height of the disk.
In this problem, the disk is thin, and its height can be assumed to be negligible. Therefore, h = 0.
The initial radius (r_initial) of the disk is half of the initial diameter (D_initial):
r_initial = D_initial / 2
We are given:
D_initial = D_aluminum = 0.070044 m (approximated to five decimal places)
Substituting the values, we have:
r_initial = 0.070044 m / 2
r_initial = 0.035022 m (approximated to six decimal places)
Finally, we can calculate the initial volume:
V_initial = π * (0.035022 m)^2 * 0
V_initial = 0
Since the initial volume is zero, we can skip Step 4 and directly go to Step 5.
Step 5: Find ΔT, the change in temperature of the disk.
Using the equation from Step 2:
ΔT = ΔV / (V_initial * β_aluminum)
From Step 4, we know that:
V_initial = 0
Thus, the equation becomes:
ΔT = ΔV / (0 * β_aluminum)
Since any value divided by zero is undefined, we can conclude that the change in temperature of the disk (ΔT) is undefined.
Therefore, we cannot determine the temperature of the aluminum disk when it falls into the pipe using the given information.