Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Write the expression of heat for an isothermal reversible expansion of a perfect gas from initial volume V1 to final volume V2.
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- Q4: A. An ideal gas initially at 30°C undergoes an isobaric expansion at 3kpa. If the volume increases from 1.5 to 3.5 m3 and 13 kJ is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature?arrow_forwardQ4: 05kg of water at 7bar and 15¿ is contained in a der 0-3m diameter by friction-Less piston Heat supplied until the temperature of cylinder contents becomes energy is 204c, the pressure of the contents remains at 7bar. Determine: a the heat energy supplied by b. the distance moved the piston, c. the work energy, di the change in internalarrow_forwardShown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.2870 kJ/(kg·K). (The internal energy can be determined by the equation ΔU=m·cv·(T2-T1), where cv = 0.7180 kJ/(kg·K) is the specific heat at the constant volume.) Calculate the final state temperature T2.__________ (K)arrow_forward
- Shown in the figure, an insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains an indeal gas, and the other part is evacuated. The partition is then removed, and the gas expands into the entire tank. At the initial state, the mass of the gas is m= 4.00kg, initial pressure is p1 = 600.00 kPa, initial temperature is T1 = 300.00 K. The gas constant is R = 0.2870 kJ/(kg·K). (The internal energy can be determined by the equation ΔU=m·cv·(T2-T1), where cv = 0.7180 kJ/(kg·K) is the specific heat at the constant volume.) Calculate the final state pressure p2.__________ (kPa)arrow_forwardA rigid tank of volume 10 m³ initially contains saturated water vapor at a temperature of 120 °C. Steam at a pressure 1.2 MPa and a temperature of 400 °C enters the tank through a valve in steam line that is connected to the tank until the final pressure in the tank is 800 kPa, at which time the temperature is 200 °C. All kinetic and potential energy effects can be neglected. A schematic of the problem and properties at all state points except state 1 are shown in the figure below. All of the properties at state 2 and the inlet state i are provided on the figure. Initial State in Tank T₁-120 °C, Sat. vapor u₁=? kJ/kg V₁=? m³/kg Pi=1.2 MPa, Ti-400 °C hi-3261.3 kJ/kg V=10 m³ Final State in Tank T: 200 °C, P₂-800 kPa u₂= 2631.1 kJ/kg v₂=0.26088 m³/kg Qout For Question 6: The initial specific internal energy, u1, of the saturated vapor in the tank in kJ/kg isarrow_forwardA copper container of mass 0.080 kg and specific heat 387 Jkg-1 K-1 contains 0.30 kg of water and 0.040 kg of ice at 0˚C. Steam at 100˚C is passed into the water and its temperature stabilizes at 20.0˚C. Find the mass of the water left in the container. Assume the system is insulated from its environment.arrow_forward
- An ideal gas is compressed from a volume of V, = 6.00 L to a volume of V, = 3.00 L while in thermal contact with a heat reservoir at T = 295 K as in the figure below. During the compression process, the piston moves down a distance of d = 0.145 m under the action of an average external force of F = 21.0 kN. (a) Find the work done on the gas. kJ (b) Find the change in internal energy of the gas. k) (c) Find the thermal energy exchanged between the gas and the reservoir. kJ (d) If the gas is thermally insulated so no thermal energy could be exchanged, what would happen to the temperature of the gas during the compression? O The temperature would decrease. O The temperature would remain constant. O The temperature would increase.arrow_forwardInitial condition: P = 0.7 MPaT = 250oCm = 5 kgProcess: IsobaricFinal condition: vapor = 3.5 kgRequired: Heatarrow_forwardNeed in 1 hour plz do it correctly and neatly i will upvotearrow_forward
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