How Does Pressure Affect Temperature Change in a Water Thermos?

[reynaldo85]

Im trying to figure out how to calculate the pressure necessary to create a 30 degree celsius temperature drop in water, by using a small pressurized cylinder of air within the water. The volume of the water is .00075 m^3 and the volume of the canister is .00025 m^3. I calculated the amount of energy lost by the water to be 94.05 KJ. But I am not sure how to calculate the corresponding pressure necessary in the pressurized air container to cause this temperature change in the water.

These are the values i was using:
Cair= 1.2KJ/kg*K
Mair=.0003kg
T1,air=293K
P1,air= 3bar (not sure if its enough)
Vair= 0.00025m^3

Cwater=4.18KJ/kg*K
Mwater=.75kg
T1,water=293K
T2,water=263K
P1,water=101.3Kpa
Vwater=0.00075m^3

I wasn't sure what values had to be assumed and but i assumed:
t=180seconds (for the air to escape the canister)
ρair= 1.204 kg/m^3

Any help is appreciated

 

[256bits]

If the water, as you have calculated, would 'lose' 94.05 kJ to drop in temperature by 30 degrees C, then something else has to 'gain' 94.05 kJ of energy.

You could start by asking yourself how much air ( mass ) at what temperature could cool 0.75kg of water by from 293 to 263 degrees C. Subsequently, you could calculate the pressure of that mass of air within the canister.

Also, are you considering that a release of air to atmosphere will cause the pressure of air within the canister to drop and thus lower the temperature of air within the canister? It seems that you have a T1,air=293K.
Note that if so there will be less mass of air within the canister to 'gain' 94.05kJ of energy.

 

[reynaldo85]

I used Q=mCΔT to find the mass of air required to cool the water (both at room temperature).
94.05KJ=m(1KJ/kg*k)(293K)
m~0.3kg
However the air will be escaping at a certain flow rate. Does this mean i have to assume a flow rate and time to reach P2,air = atm ?
Also, the material and surface area of the container would impact the rate of heat transfer. This means i would have to use Fourier's law but I'm not sure to incorporate it into the equation.

 

[pukb]

If you are saying that air in the cannister is escaping at some flow rate, then the figure 94.05KJ has to be associated with some time term.

(mcΔT/t)water = (mcΔT/t)air

Thereby, there will be a time in which water has to cool by 30degress and the same time must be used to calculate the flow rate of air.
There are two unknowns, the time for heat transfer and the final temp of air.

as you said assuming t = 150s

0.75*4180*30/150 = m * 0.0003 * ΔT/t

After temperature is calculated pressure can be calculted.m/t on RHS will give mass flow rate of air
It seems it has to be a numerical method with trial and error.
taking m = 0.0003kg will give very large values of ΔT which are not practical.So I think there must be some more constriants to the problem.

 

[alphy]

Hello,

Calculating the necessary pressure in the pressurized air container to cause a 30 degree Celsius temperature drop in the water is a complex problem that requires several assumptions and calculations. Here are some steps you can follow to get an approximate answer:

1. First, calculate the initial and final temperatures of the water. The initial temperature is 293K, and the final temperature is 263K (30 degrees Celsius lower). This gives a temperature change of ΔT = 30K.

2. Next, calculate the change in internal energy of the water using the specific heat capacity of water (Cwater = 4.18KJ/kg*K) and the mass of water (0.75kg). The change in internal energy (ΔU) can be calculated as follows:

ΔU = mCΔT = (0.75kg)(4.18KJ/kg*K)(30K) = 94.05KJ

This is the amount of energy that needs to be transferred from the water to the air in order to cause the temperature drop.

3. Now, we need to calculate the change in internal energy of the air. This can be calculated using the specific heat capacity of air (Cair = 1.2KJ/kg*K), the mass of air (0.0003kg), and the change in temperature (ΔT = 30K). The change in internal energy of the air (ΔUair) can be calculated as follows:

ΔUair = mCairΔT = (0.0003kg)(1.2KJ/kg*K)(30K) = 0.0108KJ

4. Next, we can use the first law of thermodynamics (ΔU = Q - W) to calculate the work done by the air (W) on the water. Since the air is expanding and doing work on the water, the work done will be negative. Therefore, the equation becomes:

ΔU = Q + |W| = 94.05KJ + |W|

Solving for |W|, we get:

|W| = 94.05KJ - ΔUair = 94.05KJ - 0.0108KJ = 94.0392KJ

5. Now, we can use the ideal gas law (PV = nRT) to calculate the pressure in the air container