What is the final pressure inside the thermos after dropping in a piece of iron?

[spherical23]

Homework Statement

A well insulated 2 liter thermos bottle contains 500 ml of boiling liquid Nitrogen, and the remainder of the thermos has Nitrogen gas at 1 atmosphere in equilibrium with the liquid. You drop a 200 gram piece of iron at 25 degrees Celsius inside and seal the cap. After 5 seconds, what is the pressure inside the thermos?

Homework Equations

\rho _{liquid nitrogen} = 810 kg/m³

\rho _iron = 7.87 g/cm³

PV = NkT

I think that I need more equations to solve the problem, but I don't know what they are.

The Attempt at a Solution

V _total = 2 L = 0.002 m³

V _{liquid nitrogen} = 500 ml = 500 * 10^-3 L = 0.5 L = 0.0005 m³

before dropping the iron in: V _{nitrogen gas} = 0.002 m³ - 0.0005 m³ = 0.0015 m³

V _iron: \rho = mV \rightarrow 7.87 g/cm³ = 200g(V) \rightarrow V = 0.03935 cm³ = 0.0003935 m³

after dropping the iron in: V _{nitrogen gas} = 0.0015 m³ - 0.0003935 m³ = 0.0011065 m³

T_{initial of liquid nitrogen} = boiling point at 1 atm = 77.2 K

T_{initial of iron} = 25 degrees Celsius = 298 K

P_{initial of liquid nitrogen} = P_{initial of nitrogen gas} = 1 atm = 0.013 * 10⁵ Pa

\Deltat = 5 seconds

P_final = ?

I just don't know which equation to use.

 

[mgb_phys]

You need to work out how much heat energy the iron gives up when cooled (assume final T of iron = LN2)
Assume this energy goes into boiling extra LN2.
Work out how much extra N2 this generates.
Then it's just gas laws for the pressure.

The 5 seconds is tricky, are you sure about that part? Unless you are expected to calcualte heat diffusion rate from the iron it's irrelevant.

 

[spherical23]

thanks, i will give that a try and then reply again when i come up with something. i think the 5 seconds is just enough for relaxation time so that we don't have to worry about anything during the relaxation time.