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First law of thermodynamics applied to a submarineby Carbon884
Tags: thermodynamics 
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#1
Jun1311, 12:49 PM

P: 5

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(T2T1)= m(R(air)cv)*(T2T1)= 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 ^^. 


#2
Jun1311, 02:14 PM

P: 4

R(air) = Cp  Cv
=> Cp=R(air)+Cv so, Q=m(R(air)+Cv)(T2T1) 


#3
Jun1311, 08:55 PM

P: 1,970

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


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