Here, T1=400K,T2=320KT_1= 400 K, T_2= 320 KT1=400K,T2=320K
Heat flow at constant process can be find as
H=nCvdT=1×0.718×(400−320)=57.44JH= n C_vdT = 1\times 0.718\times ( 400-320)= 57.44 JH=nCvdT=1×0.718×(400−320)=57.44J
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