# First law of thermodynamics applied to a submarine

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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## Answers and Replies

Andrew Mason
Science Advisor
Homework Helper

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 cant really imagine though is what happens to the pressure during this process?

Andrew Mason
Science Advisor
Homework Helper
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 cant 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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