First law of thermodynamics applied to a submarine

AI Thread Summary
The discussion focuses on applying the first law of thermodynamics to a submarine with a constant volume of air at specific initial conditions. A heat flow of 60 MJ/h and dissipative work of 21 kW are introduced, prompting questions about the average temperature after one hour. The calculations initially led to an incorrect conclusion about the final temperature, highlighting the need to correctly determine the specific heat capacities. It is clarified that during this process, as temperature increases due to heat flow, pressure will also increase since the volume and amount of gas remain constant. The discussion emphasizes the importance of accurately applying thermodynamic principles to understand the behavior of the submarine's air.
Carbon884
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First law of thermodynamics

Hallo,

i hope someone can help me with the following question:

A submarine contaiins 1000m^3 of air and has a temperature and pressure of 15°C and 0.1MPa respectively. Due to the cold seawater a heatflow of 60 MJ/h occurs. The machines on the otherhand add disspiative Workof 21 kW to the system. The specific heat capacity of air is c(p,air) = 1.005 kJ/Kg*K.

what average Temperature will the air have after one hour of diving?

V=constant=1000m^3
P=constant?
T1=288.15K => T2=?

Firstly i found the density and mass of the gas:

R(air)= 286.9 J/K*Kg => density= P/T*R(air)=1.209 Kg/m^3 => m= V*density= 1209.63 Kg

Secondly i found the total Energy:

Q/h=Wdiss - Q/h = 15600 KJ/h

Then i used the first law:

dW=0 because V=constant

dQ=dU

Q=mcp(T2-T1)= m(R(air)-cv)*(T2-T1)= 288.19K ... which can't possibly be right because that is essentially my T1 temerature. So what did I do wrong?

Thanks for your help in advance ^^.
 
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Carbon884 said:
Q=mcp(T2-T1)= m(R(air)-cv)*(T2-T1)= 288.19K ... which can't possibly be right because that is essentially my T1 temerature. So what did I do wrong?
I am not sure I follow what you have done here. Since it is a constant volume process you have to find Cv. You are given Cp. So what is Cv for the air?

AM
 
R(air)= cp-cv => cv=R(air)+cp...i accidentally wrote a minus sign but i meant plus -.-.

What I can't really imagine though is what happens to the pressure during this process?
 
Carbon884 said:
R(air)= cp-cv => cv=R(air)+cp...i accidentally wrote a minus sign but i meant plus -.-.
I think you meant: Cv = Cp-R.

What I can't really imagine though is what happens to the pressure during this process?
If the submarine is air-tight and rigid, the volume is constant and the quantity of gas does not change. Since heat flow is into the submarine, temperature goes up. Apply the ideal gas law:

PV = nRT

If V and n are constant, what must happen to P if T increases?

AM
 
*nods* so the pressure will increase with temperature.
 
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