2300_hw_2_sol

2300_hw_2_sol - Homework #2 Solutions Thermodynamics 1-56...

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Homework #2 Solutions Thermodynamics 1-56 The pressure in chamber 2 of the two-piston cylinder shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives 1 1 2 1 3 2 2 1 3 2 ) ( A P A A P A P F F F = - + = + which when solved for P 2 gives ˜ ˜ ˆ - - = 1 2 1 3 2 1 1 2 A A P A A P P since the areas of the piston faces are given by 4 / 2 D A p = the above equation becomes kPa 3625 = - ˜ ˆ - ˜ ˆ = - ˜ ˜ ˆ - ˜ ˜ ˆ = 1 4 10 kPa) 500 ( 4 10 kPa) 1000 ( 1 2 2 2 2 1 3 2 2 1 1 2 D D P D D P P 2-1C Initially, the rock possesses potential energy relative to the bottom of the sea. As the rock falls, this potential energy is converted into kinetic energy. Part of this kinetic energy is converted to thermal energy as a result of frictional heating due to air resistance, which is transferred to the air and the rock. Same thing happens in water. Assuming the impact velocity of the rock at the sea bottom is negligible, the entire potential energy of the rock is converted to thermal energy in water and air.

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This note was uploaded on 09/04/2009 for the course CHEM 2300 taught by Professor Thermo during the Summer '08 term at University of Utah.

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2300_hw_2_sol - Homework #2 Solutions Thermodynamics 1-56...

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