FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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A 300-lb iron casting, initially at 1050°F, is quenched in a tank filled with 2121 lb of oil, initially at 80°F. The iron casting and oil can be
modeled as incompressible with specific heats 0.10 Btu/lb · °R, and 0.45 Btu/lb · °R, respectively.
(a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F.
Ignore heat transfer between the system and its surroundings.
Tf =
i
°F
(b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R.
Ignore heat transfer between the system and its surroundings.
O =
i
Btu/°R
Touthoolk ond Medie
Three-tenths kmol of carbon monoxide (CO) in a piston– cylinder assembly undergoes a process from p1 = 150 kPa, T1 = 300 K to p2 = 500 kPa, T2 = 420 K. For the process, W = -300 kJ.Employing the ideal gas model, determine:(a) the heat transfer, in kJ.(b) the change in entropy, in kJ/K.
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- A vessel of volume 0.04 m3 contains a mixture of saturated water and steam at a temperatureof 250°C. The mass of the liquid present is 9 kg. Find the pressure, mass, specific volume,enthalpy, entropy and internal energy.arrow_forwardcan water ever acts counter to entropy to perhaps stop a reaction from moving forward?arrow_forwardfor steam, the specific ideal gas constant = 461.5 J/kg K A closed system is comprised of pure water substance initially at a temperature of 500 oC and a pressure of 20 MPa (state 1). The system undergoes an isochoric process whereby its pressure drops to 0.1 Mpa (state 2). Evaluate the specific entropy, specific volume and temperature of the system at state 2.arrow_forward
- 6. A certain quantity of gas occupies 0.56 m³ at 400° C and 28 bar. Determine the gain in entropy if the gas expands isothermally to a final volume of 2.8 m². R= 287 J/kg K. the following creas Ans. 3.746 kJ/KIarrow_forwardThree-tenths kmol of carbon monoxide (CO) in a piston- cylinder assembly undergoes a process from p1 = 150 kPa, T1 = 300 K to p2 = 500 kPa, T2 = 370 K. For the process, W = -300 kJ. Employing the ideal gas model, determine: (a) the heat transfer, in kJ. (b) the change in entropy, in kJ/K. Part A Employing the ideal gas model, determine the heat transfer, in kJ. kJ Save for Later Attempts: 0 of 1 used Submit Answer Part B The parts of this question must be completed in order. This part will be available when you complete the part above.arrow_forwardA pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero (the third law of thermodynamics).arrow_forward
- The entropy change between two specified states is the same whether the process is reversible or irreversible.arrow_forward1. thermodynamicsarrow_forwardA 300-lb iron casting, initially at 1500°F, is quenched in a tank filled with 2121 lb of oil, initially at 80°F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lb · °R, and 0.45 Btu/lb · °R, respectively. (a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F. Ignore heat transfer between the system and its surroundings. Tf = i °F (b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R. Ignore heat transfer between the system and its surroundings. O = i Btu/°Rarrow_forward
- One kg of an ideal gas (gas constant R = 287 J/kg.K) undergoes an irreversible process from state-1 (1 bar, 300 K) to state -2 (2 bar, 300 K). The change in specific entropy (52 - s1) of the gas (in J/kg. K) in the process isarrow_forwardA container of 1.5 Kg of gas is at a temperature and pressure of 293 K and 1 bar respectively. The gas is adiabatically compressed until its temperature and pressure are 450 K, 4.49 bars. Adiabatic processes are processes with no heat transfer. The properties of this gas are cv = 10.3 KJ/(Kg K) and R = 4.158 KJ/(Kg K). Neglect kinetic and potential energy terms. Use the first law to determine the work into the system. Calculate the entropy production for this process. Is this a reversible process?arrow_forwardfind the changes in h as appropriate. The initial state pressure is p1 = 0.5 MPa. the final state is 2. a. constant volume : v1 = 0.3 m3/kg, p2 = 0.3 MPa; b. constant entropy : s1 = 6.3 kJ/kg K, p2 = 0.15 MPa; c. constant volume : h1 = 2500 kJ/kg, p2 = 0.2 MPa; d. constant enthalpy : s1 = 6.4 kJ/kg K, p2 = 0.2 MPa;arrow_forward
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