First law of thermodynamics applied to a submarine

Carbon884

Hallo,

I hope someone can help me with the following question:

A submarine conatins 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 adds disspiative Work of 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 ^^.

fahim_naruto

R(air) = Cp - Cv
=> Cp=R(air)+Cv

so, Q=m(R(air)+Cv)(T2-T1)

nasu

The specific heat at constant pressure is given in the problem.
If you assume a constant-volume process, you can calculate cv=cp-R.
Either way the temperature increase is of the order of 10K (around 14K if you consider constant volume).
Review your last step.