Part A) Two identical pieces of machinery are lifted from the sidewalk to the roof of a...

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Chemistry

Part A) Twoidentical pieces of machinery are lifted from the sidewalk to theroof of a 100.0 m tall building. One is lifted directly to thebuilding's roof and has a change in internal energy of 1059 kJ. Theother is lifted to twice the height of the building and thenlowered to the roof. What is the change in internal energy of thesecond piece of machinery once it has reached the roof?
Part B)As a reaction occurs, the system loses 1150 J of heat to thesurroundings. A piston moves downward as the gases react to form asolid. As the volume of the gas decreases under the constantpressure of the atmosphere, the surroundings do 480 J of work onthe system. What is the change in the internal energy of thesystem?â€‹
Part C) Calculate the change in the internalenergy of the system for a process in which the system absorbs 140J of heat from the surroundings and does 85 J of work on thesurroundings.â€‹

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4.1 Ratings (763 Votes)

Part A )

the internal energy becomes twice.

Part B)

system loses heat Q = 1150 J

work done on the system = 480 J

change in internal energy U = Q + W

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = - 1150 + 480

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = - 670 J

change in internal energy U = - 670 J

Part C)

heat absorbed Q = 140 J

work done by the system = 85 J

internal energy change U = q + W

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 140 - 85

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 55 J

internal energy change U = 55 J

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Part A) Two identical pieces of machinery are lifted from the sidewalk to the roof of a...

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