Ceramic Superconductivity Experiment: Calculating Liquid Nitrogen Boiling Loss

[danield]

Homework Statement

Some ceramic materials will become superconducting if immersed in liquid nitrogent. In an experiment, a 0.150kg block piece of such material at 20C is placed in liquid nitrogent at its boiling poitn to cool in a perfectly insulated flask, whcih allows the gaseous N2 immeadiately to escape. How many liters of liquid nitrogen will be boiled away in doing this operation? (Take the specific headt of the ceramic material to be the same as that of glass, and take the density of liquid nitrogen to be 800kb/m3)

Homework Equations

Q=mc#T
L=Q/m
density=mass/volume

The Attempt at a Solution

well right now i have the data but i don't know where to go from here
I have
Mass of ceramic =.15kg
Tinitiail of ceramic= 20 C
Tinitial of Nitrogen = -196C
C for ceramic= 840
C for N= 20.8
Latent Heat of nitrogen = 2x10^5

I tried using Sum of Q=0
meaning that
MCTf - MCTi + MCTf - MCTi
But i don't hace nitrogens mass, so i am stuck
any help is appreciated

 

[anathema]

Dear forum post writer,

Thank you for sharing your problem with us. I am happy to assist you in finding a solution.

First, let's start by understanding the problem. You have a ceramic material that has a specific heat capacity (C) of 840. This means that for every 1 kg of ceramic, it takes 840 joules of energy to raise its temperature by 1 degree Celsius. You also have the mass of the ceramic (0.150 kg) and its initial temperature (20 degrees Celsius).

Next, you have liquid nitrogen, which has a density of 800 kg/m3 and a boiling point of -196 degrees Celsius. This means that when liquid nitrogen reaches its boiling point, it will start to evaporate and turn into gas. The energy required for this phase change is called the latent heat of vaporization, which in this case is 2x10^5 joules/kg.

Now, let's look at the flask in which the ceramic is placed. Since the flask is perfectly insulated, it means that no heat can enter or leave the system. This means that the total amount of energy in the system remains constant.

Using the equation Q=mcΔT, we can calculate the amount of heat required to cool the ceramic from 20 degrees Celsius to -196 degrees Celsius. This will be the same amount of heat that is released when the liquid nitrogen boils and turns into gas.

Therefore, we can set up the following equation:

Qceramic + Qnitrogen = 0

Where:

Qceramic = (0.150 kg)(840 J/kg*C)(20C - (-196C)) = 3,696 J
Qnitrogen = (m)(2x10^5 J/kg)

Since we do not know the mass of nitrogen (m), we can use the density formula to calculate it:

Density = mass/volume
800 kg/m3 = m/1 L
m = 800 kg

Now, we can substitute this value into our equation:

Qnitrogen = (800 kg)(2x10^5 J/kg) = 1.6x10^8 J

Finally, we can solve for the volume of nitrogen using the equation L=Q/m:

Volume = (1.6x10^8 J)/(800 kg) = 2x10^5 L

Therefore, the volume of nitrogen that will be boiled away is 200,000

 

[ogb p]

I would approach this problem by first calculating the heat lost by the ceramic material as it cools from 20C to -196C. This can be done using the heat capacity (C) of the ceramic material and the difference in temperature (ΔT) as follows:

Q = mcΔT

Where:
Q = heat lost by the ceramic material
m = mass of the ceramic material
c = specific heat capacity of the ceramic material
ΔT = change in temperature (20C - (-196C) = 216C)

Next, I would calculate the amount of liquid nitrogen needed to absorb this heat loss. This can be done using the latent heat of nitrogen (L) and the mass of the ceramic material as follows:

L = Q/m

Where:
L = latent heat of nitrogen
Q = heat lost by the ceramic material (calculated in the previous step)
m = mass of the ceramic material (0.150kg)

Finally, I would convert the mass of liquid nitrogen into liters using its density (ρ) as follows:

V = m/ρ

Where:
V = volume of liquid nitrogen (in liters)
m = mass of liquid nitrogen (calculated in the previous step)
ρ = density of liquid nitrogen (800kg/m^3)

By following these steps, you should be able to calculate the amount of liquid nitrogen that will be boiled away in this experiment. Keep in mind that this is a simplified calculation and there may be other factors (such as heat loss due to the insulated flask) that could affect the result.