Exam4Sol - Principles of Physics I PHYS 2211-160 Exam #4...

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Principles of Physics I PHYS 2211-160 Exam #4 Chapter 15 and Thermodynamics Spring 2010 Name__ Solutions _______________ You might like to know that 1.00atm = 1.013x10 5 Pa , σ = 5.67x10 -8 J/(s m 2 K 4 ) , k = 1.38x10 -23 J/K , R = 8.31J/(mol K) , N A = 6.02x10 23 /mol, the density of copper is 8920kg/m 3 , the thermal conductivity of copper is 390.J/(s m o C) , the specific heat capacity of copper is 387 J/(kg o C), the density of gold is 19300kg/m 3 , the density of mercury is 13600kg/m 3 , the specific heat capacity of liquid water is 4186J/(kg o C) , the specific heat capacity of ice is 2.00x10 3 J/(kg o C), the latent heat of fusion of water is 33.5x10 4 J/kg . 1. (5 points) A water pipe has a circular intake opening with a radius of 3.00cm where the water has a speed of 2.00m/s . What is the speed of the water exiting the pipe where the radius of the pipe is 2.50cm ? a) 1.39 m/s b) 1.67 m/s From the continuity equation we have A 1 v 1 = A 2 v 2 , so c) 2.00 m/s d) 2.40 m/s v 2 = A 1 v 1 / A 2 = π (0.0300m) 2 (2.00m/s) / [ π (0.0250m) 2 ] = 2.88m/s e) 2.88 m/s 2. (5 points) What is the volume of 2.00moles of an ideal gas at 4.00atm and 30.0 o C ? a) 1.23x10 -3 m 3 From the Ideal Gas Law we have PV = nRT or b) 1.24x10 -2 m 3 c) 1.25x10 2 m 3 V = nRT / P = (2.00moles) (8.31J/ mol K) (30.0K + 273K) / d) 1.26x10 3 m 3 e) 1.27x10 8 m 3 [(4.00atm) (1.013x105Pa)] = 1.24x10
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This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.

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Exam4Sol - Principles of Physics I PHYS 2211-160 Exam #4...

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