The tank B (3 bar, 50°C) has more stored energy.
The availability of the system is
"a=m((u_1-u_0)-T_0(s_1-s_0))=m(c_v(T_1-T_0)-T_0\\Delta s)"
"\\Delta s_A=c_p\\ln\\frac{T_1}{T_0}-R\\ln\\frac{p_1}{p_0}=1.005\\ln\\frac{50+273}{15+273}-0.287\\ln\\frac{1}{1}=0.115 \\frac{kJ}{kg\\cdot K}"
"a_A=m(c_v(T_1-T_0)-T_0\\Delta s_A)=1(0.718(323-288)-288\\cdot0.115)=-7.99kJ"
"\\Delta s_B=c_p\\ln\\frac{T_2}{T_0}-R\\ln\\frac{p_2}{p_0}=1.005\\ln\\frac{50+273}{15+273}-0.287\\ln\\frac{3}{1}= -0.2\\frac{kJ}{kg\\cdot K}"
"a_B=m(c_v(T_2-T_0)-T_0\\Delta s_B)=1(0.718(323-288)-288\\cdot(-0.2))=82.73kJ"
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