1) Compute the change in the internal energy of a thermodynamic system under the

Question

1) Compute the change in the internal energy of a thermodynamic system under the following scenarios: (i) 50 J of heat added and 40 J work extracted, (ii) 50 J of heat added and 75 J work extracted, (iii) 50 J of heat removed and 40 J work extracted, (iv) 50 J of heat removed and 75 J work input. 2) By how much does the internal energy of 0. 25 kg of liquid water change when it is heated from 20 C to 25 C? 3) By how much does the internal energy of 0.25 kg of ice (solid water) change when it is heated from ?10 C to ?5 C? 4) By how much does the internal energy of 0 .25 kg of ice at 0 C change when melted to liquid water (also at 0 C)? 5)By how much does the internal energy of 0.25 kg of ice at ?5 C change when it is converted entirely to liquid water at 5 C?

Explanation / Answer

1)U1=50-40=10J U2=50-75=-25J U3=-50-40=-90J U4=-50+75=25J U=U1+U2+U3+U4=10-25-90+25=-80J 2)U=msdT=0.25*4.186*(5) U=5.2325kJ 3)U=msdT=0.25*4.18(5)=5.2325kJ 4)U=.25*Lf=0.25*335=83.75kJ 5)U=83.75kJ+5.2325kJ U=89kJ

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