Answer
Using first law of thermodynamics
ΔQ=ΔU+ΔW\Delta Q=\Delta U+\Delta WΔQ=ΔU+ΔW
ΔU=ΔW+ΔQ\Delta U=\Delta W+\Delta QΔU=ΔW+ΔQ
Heat is rejected so using positive sign here
So the change of specific internal energy
ΔU=70+42=112KJ/Kg\Delta U=70+42=112KJ/KgΔU=70+42=112KJ/Kg
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