(a)
m=5lbR=38.7ft⋅lb/lbm⋅Rk=1.668Q=300BTU=233451lbf⋅ftT1=80Q=mcp(T2−T1)233451=5×k−1kR(T2−T1)233451=5×0.6681.668×38.7×(T2−80)T2=563.165°F
(b)
ΔU=mcv(T2−T1)=k−1mR(T2−T1)=5×0.66838.7×(563.165−80)=139958.6331lbf⋅ft
(c)
ΔH=Q=233451lbf⋅ftΔS=mcpln(T1T2)=5×0.6681.668×38.7×ln(80563.165)=942.9295
(d)
W=Q−ΔU=233451−139958.633=93492.3669lbf⋅ft
Comments