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Sample Final Exam Problems

# Sample Final Exam Problems - Sample Final Exam Problems for...

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Sample Final Exam Problems for P-40B Problem 1 (35 points): A hemispherical shaped bowl is floating on the surface of water. The mass of the bowl is 0.6 kg. Water is poured into the bowl. When 4.3 liters (1 liter = 10 -3 m 3 ) of water has been added, the bowl starts to sink. The volume of the bowl is ½ (4/3 π R 3 ). What is the radius, R, of the bowl? R Water Added Water Bouyant Force: B = " w V Bowl g (volume of hemispherical bowl just befor sinking Weight of Bowl: W Bowl = m Bowl g Weight of water in bowl: W water = w V water g Newton's 1st Law : F y # = 0 $B % W Bowl % W water = 0 (Find V Bowl ) w V Bowl g - m Bowl g % w V water g = 0$ V Bowl = m + w V water water V Bowl = 0.6 kg + (1000 kg / m 3 )(4.3 x 10 % 3 m 3 ) 1000 kg / m 3 = 0.0049 m 3 V Bowl = 2 3 R 3 $R 3 = 3(0.0049) 2 = 0.002339 m 3 R = 0.133 m This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Problem 2 (30 points): An axe head consisting of 1.8 kg of iron is left outdoors on a cold winter night where the outside temperature is 240K. The axe head is then brought inside a room where the initial temperature is 293K at one atmosphere of pressure. The volume of the room, which is well insulated, is 38 m 3 . Consider the room well-sealed so no air leaks in or out. When the axe head comes to thermal equilibrium with the air, what is the new temperature of the air in the room? Ignore the thermal interaction with the furniture, walls and so forth. Use C V = 21.0 J/mol K for the molar specific heat of air at constant volume and c Fe = 448 J/kg C o for the specific heat of iron. Calculate number of moles of air in room: n = PV RT = (1.013 x 10 5 )(38) (8.31)(293) = 1581 moles Apply Heat Balance Eq. (Conservation of Heat) Q axe + Q air = 0 " m axe c Fe ( T f # 240 K ) + nC V ( T f # 293 K ) = 0$ T f = m axe c Fe (240 K ) + nC V (293 K ) m axe c Fe + nC V = (1.8)(448)(240) + (1581)(21.0)(293) (1.8)(448) + (1581)(21.0) T f = 291.7 K
Problem 3 (35 points): A monatomic ideal gas undergoes a transformation from an initial state (P i = 1.0 atm, V i = 350 cm 3 ) to a final state (P f = 2.0 atm, V f = 175 cm 3 ). The transformation is carried out in two steps: The gas is compressed at constant pressure to the final volume and then the pressure is increased while the volume is held constant. Determine the change in internal energy of the gas, the work done on the gas and the heat flow into the gas for both steps (i.e., fill in the values in the table provided). Note that you will not need to know the actual values of n (# moles) and the temperatures to work this problem (P i = ½ P f and V i = 2 V f ). Δ E Q W Step 1 -26.6 J -44.3 J +17.7 J Step 2 +26.6 J +26.6 J 0 Step 1: Constant Pressure Work W = -P i (V f – V i ) W = -(1.013x10 5 )(175-350)(10 -6 ) W = +17.7 J Q P = nC P Δ T = n(5/2R)(T f – T i ) nRT i = P i V i nRT f = P i V f Q P = 5/2(P i V f - P i V i ) = 5/2 P i (V f – V i ) = - 5/2 W = - 44.3 J Δ E = W + Q = +17.7 J – 44.3 J = - 26.6 J Step 2: Constant Volume Work: W = 0 Q V = nC

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Sample Final Exam Problems - Sample Final Exam Problems for...

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