T1=600oF,T2=200oFT_1= 600 ^oF, T_2= 200 ^o FT1=600oF,T2=200oF
Internal Energy = nCvdT= 0.1×0.17128×(200−600)=−68.512(BTU)0.1\times 0.17128 \times (200-600)= -68.512 (BTU)0.1×0.17128×(200−600)=−68.512(BTU)
for enthalpy of air we can use below formula for finding
dH=nCpdT=0.1×0.24×(−400)=−96dH=n CpdT=0.1\times 0.24\times (-400)= - 96dH=nCpdT=0.1×0.24×(−400)=−96 BTU
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