(a)
m=5lbR=38.7ftā
lb/lbmā
Rk=1.668Q=300BTU=233451lbfā
ftT1ā=80Q=mcpā(T2āāT1ā)233451=5Ćkā1kRā(T2āāT1ā)233451=5Ć0.6681.668Ć38.7āĆ(T2āā80)T2ā=563.165°F
(b)
ĪU=mcvā(T2āāT1ā)=kā1mRā(T2āāT1ā)=5Ć0.66838.7āĆ(563.165ā80)=139958.6331lbfā
ft
(c)
ĪH=Q=233451lbfā
ftĪS=mcpāln(T1āT2āā)=5Ć0.6681.668āĆ38.7Ćln(80563.165ā)=942.9295
(d)
W=QāĪU=233451ā139958.633=93492.3669lbfā
ft