Question
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?
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?
Answers
drwls
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
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