a)Estimate how long it takes for the cold air to fall out when you open the door of your refrigerator. For simplicity, assume your fridge is a cube of height 1 meter, and that the cold air inside is 5% denser than the warm air outside. Imagine that you remove the top and bottom of this cube and that the cold air just falls out as a solid block, without mixing with the warm air it moves through.
b)The ideal gas law P V = N k T, relates the pressure P and the (absolute!) temperature T of an idealized gas consisting of N particles in a box of volume V (here k is a constant which will not matter for us). If air was an ideal gas, and the density difference between the air inside and outside the refrigerator is 5% (but the pressure is the same), what is the temperature difference between inside and outside?
a) How far does it have to fall? Assume one meter.
Don't forget there is a buoyancy effect. Net downward force is
[1.05*density)*V - (density)*V]*g = M*a = (1.05)*density*a
"density" is ambient density
a = (0.05/1.05) g = 0.47 m/s^2
(1/2) a t^2 = 1 m when
t^2 = (1 m)*(4.25 s^2/m)
t = 2.1 s
a) To estimate how long it takes for the cold air to fall out when you open the door of your refrigerator, we can consider the motion of the cold air as a solid block sinking due to its higher density. We need to find the time it takes for the block to fall a distance equal to the height of the fridge.
First, let's calculate the density of the cold air inside the fridge. If the warm air outside has a density of rho_warm and the cold air inside is 5% denser, then the density of the cold air (rho_cold) can be found as:
rho_cold = rho_warm * (1 + 0.05)
Next, we need to calculate the mass of the cold air block. The mass (m) can be found by multiplying the volume (V) of the cold air block by its density (rho_cold). Since the fridge is a cube of height 1 meter, the volume is equal to the base area (A) multiplied by the height (h):
V = A * h = A * 1 = A
Now, let's find the time it takes for the cold air block to fall. We can use the equation of motion for an object in free fall:
h = (1/2) * g * t^2
Solving for time (t):
2 * h = g * t^2
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
b) The ideal gas law P V = N k T relates the pressure (P), volume (V), number of particles (N), temperature (T), and a constant (k) for an ideal gas. If the air inside and outside the fridge have the same pressure and the density difference between them is 5%, we can relate their temperatures using this equation.
The density of an ideal gas can be related to its pressure and temperature using:
rho = (P / (k * T))
Since the pressure is the same inside and outside the fridge, we can equate their densities:
rho_cold = rho_warm * (1 + 0.05)
(P_cold / (k * T_cold)) = (P_warm / (k * T_warm)) * (1 + 0.05)
Simplifying and rearranging the equation:
T_cold = T_warm * (rho_warm / rho_cold)
Substituting the relation between density and temperature:
T_cold = T_warm * ((P_warm / (k * T_warm)) / (P_cold / (k * T_cold))) * (1 + 0.05)
Canceling the k terms:
T_cold = T_warm * ((P_warm * T_cold) / (P_cold * T_warm)) * (1 + 0.05)
T_cold = T_warm * ((P_warm * T_cold) / (P_cold * T_warm)) * 1.05
Simplifying:
T_cold = T_warm * 1.05
Therefore, the temperature difference between inside and outside the fridge is 5%.