Asked by Miko
                The internal energy change in a system that has absorbed 4 Kcals of heat and 1000 J of work is:
            
            
        Answers
                    Answered by
            Elena
            
    ΔQ =4KCal=16747.2 J
ΔQ=ΔU+W
ΔU=ΔQ-W =16747.2-1000=15747.2 J
    
ΔQ=ΔU+W
ΔU=ΔQ-W =16747.2-1000=15747.2 J
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