# Thermodynamics -- total entropy generation during cooling of tank of water...

1. Oct 25, 2015

### jittapon

1. The problem statement, all variables and given/known data
A 0.2 m^3 steel container that has a mass of 30 kg is filled with liquid
water. Initially both the steel tank and the water are at 50°C. Now heat is transferred, and the entire system (i.e., steel and water) cools to the surrounding air temperature of 25°C. Determine the total entropy generation during this process.
http://postimg.org/image/43k9ywc6v/
2. Relevant equations

3. The attempt at a solution
I think that take the steel container and liquid water and air(at 25 °C) as an isolated system so from the equation Sin-Sout(Net entropy transfer by heat or mass out the isolated boundary)+Sgen=ΔSsystem so no net entropy changes across isolated system boundaries so Sgen=ΔSsystem and ΔSsystem=ΔSliquid water+ΔSsteel+ΔSair surrounding
then find the mass of water by find the average density of water at 25°C and 50°C I got Density=992.5 kg/m^3 and also mass of water=992.5*0.2=198.5 kg
ΔSliquid water=(198.5 kg)(S@25 °C - S@50 °C)=198.5*(0.3672-0.7038)=-66.8151 kJ/k
and ΔSsteel(incompressible substances)=m*C*ln(T2/T1)=30*0.5*ln((25+273)/(50+273))=-1.2084 kJ/k
and total heat transfer to an air surroundings is Qair=-Qliquid water-Qsteel
find Qliquid=m*c*(T2-T1)=198.5*4.18*(25-50)=-20743.25 kJ
Qsteel=m*c*(T2-T1)=30*0.5*(25-50)=-375 kJ
Qair=-(-375-20743.25)=21118.25 kJ
from ΔSair=Qair/T=21118.25/(25+273)=70.867 kJ/k
then from ΔSliquid water and ΔSsteel and ΔSair I can find Sgen=ΔSsystem=ΔSliquid water+ΔSsteel+ΔSair surrounding =-66.8151 kJ/k-1.2084 kJ/k+70.867 kJ/k=2.8435 kJ/k
thankyou

2. Oct 25, 2015