A 2 lbm steel *** contains 2 lbm liquid water at 60 F. It is now put on the stove where it is heated to the boiling point of the water.

Question 1

A 2 lbm steel *** contains 2 lbm liquid water at 60 F. It is now put on the stove where it is heated to the boiling point of the water. Neglect any air being heated and find the total amount of energy needed.

Question 2

Energy Eq.: U2 − U1= 1Q2 − 1W2

The steel does not change volume and the change for the liquid is minimal, so 1 W2 ≅ 0.

kratos · Answer 1 · 2020-04-16T01:24:27+0000

Energy Eq.: U2 − U1= 1Q2 − 1W2

The steel does not change volume and the change for the liquid is minimal, so 1 W2 ≅ 0.

A 2 lbm steel *** contains 2 lbm liquid water at 60 F. It is now put on the stove where it is heated to the boiling point of the water.

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A 2 lbm steel *** contains 2 lbm liquid water at 60 F. It is now put on the stove where it is heated to the boiling point of the water.

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