thermodynamics Write clearly and legibly. Please turn in your sheet of notes wit

Question

thermodynamics

Write clearly and legibly. Please turn in your sheet of notes with the exam. 1. (20 points) A swimming pool contains 103 cubic meters of water at room temperature. A swimmer jumps into the pool from a 10 meter high tower. Find the change in temperature, assuming that no heat transfer occurs between the swimmer and the water, and between the pool and the surroundings. The water has density 1000 kg/m3 and heat capacity C CP 4200 J/(kg K). The earth gravitational constant is g -9.81 m/s' 2, (20 points) An air rifle shoots a projectile with mass m1 = 2 g and velocity 28 m/s. The projectile hits a wall and stops. Estimate its temperature immediately after the impact if it was initially at 25°C. The heat capacity of the projectile is Cv 0.126 J/(g K). One mol of ideal gas is compressed from an initial state of 1 bar and 25°C to a final state of 5 bar and 25°C by 3. (a) (30 points) reversible isothermal compression. Calculate the required work and heat transferred. (b) (30 points) reversible heating at constant volume and followed by reversible cooling at constant pressure Calculate the required work and heat transferred in both cases. The parameters are Cv 5R/2 and CP-7R/2, R-8.314 J/(mol K). The volume of 1 mol ideal gas at 25°C and i bar is V,- 0.02479 m3

Explanation / Answer

(1) solution:

Given:

Volume of water in swimming pool = 103 m3

Hight from which swimmer jumps (H) = 10 m

Density = 1000 kg/m3

Cp = Cv = 4200 j/kg.K

g = 9.81 m/s2

Formula used:

mswimmergH = mwaterCp (T2 – T1)

mswimmer = not given = 1Kg ( assumed to equate the unit of energy both side in above formula)

Change in temperature = (T2 – T1) = mswimmergH/ mwaterCp

                                                             = (1×9.81×10)/(1000×1000×4200)

                                                            = 2.33×10-80C

(2) solution:

Given:

Mass of projectile = 2 g

Velocity of projectile = 28 m/s

Initial temperature = 298K

Cv = 0.126 j/g.K

Mass of wall = 1g

Formula used:

(1/2)mV2 = mCv (T2 – T1)

(1/2)×2×(282) = 1×0.126×(T2 – 298)

So, T2 = 6520K

(3) solution:

Given:

P1 = 1 bar, P2 = 5 bar , T = 298.15K

Cv = 5R/2, Cp = 7R/2, R = 8.314 j/mol.K, V1 at 1 bar = 0.02479 m3

= Cp / Cv = 1.4

( a) Isothermal expansion:

Work required = - heat transferred

Q = - W = nRT ln ( P1 / P2)

           = 1×8.314×298.15× ln (1/5)

Q = - W = -3.990 KJ

(b) Reversible heating at constant volume:

V2 = V1 (P1 / P2) = 0.02479×(1/5) = 0.004958 m3

T2 = T1 (P2 / P1) = 298.15× (5/1) = 1490.75K

For this step the volume is constant, and

Q = U = Cv T = (5/2)×8.314×(1490.75-298.15) = 24788J

During the second step the air is cooled at the constant pressure of 5 bar to its final state:

Q = H = CpT = (7/2) ×8.314×(298.15-1490.75) = - 34703J

Also,

U = H-PV = -34703 – (5×105)(0.004958-0.02479) = -24788J

For the 2 step combined:

Q = 24788 – 34703 = -9915J

U = 24788 – 24788 = 0

W = U – Q = 9915J

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