Solution.
lb=0.15m;
la=0.1m;
d=0.03m;
tb=50oC;
ta=25oC;
ρb=8500kg/m3;
ρa=2700kg/m3;
cb=380J/kgoC;
ca=920J/kgoC;
mbcbΔTb=macaΔTa;
mb=ρbSblb;ma=ρaSala;
ρblbcbΔTb=ρalacaΔTa;
ΔTaΔTb=ρblbcbρalaca;
ΔTaΔTb=8500⋅0.15⋅3802700⋅0.1⋅920=0.5;
tf−tatb−tf=0.5⟹tf=1.5tb+0.5ta;
tf=1.550+12.5=41.7oC;
Answer: tf=41.7oC.
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