MCQ in Thermodynamics Part 7 | ECE Board Exam

This is the Multiple Choice Questions Part 7 of the Series in Thermodynamics as one of the General Engineering and Applied Sciences (GEAS) topics. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize yourself with each and every question compiled here taken from various sources including past Board Questions in General Engineering and Applied Sciences (GEAS) field, Thermodynamics Books, Journals, and other Thermodynamics References.

Continue Practice Exam Test Questions Part 7 of the Series

⇐ MCQ in Thermodynamics Part 6 | ECE Board Exam

Choose the letter of the best answer in each question.

301. If air is at pressure, p, of 3200 lbf/ft2, and at a temperature, T, of 800°R, what is the specific volume, v? (R=5303 ft-lbf/lbm-°R, and air can be modeled as an ideal gas.)

A. 9.8 ft^3/lbm

B. 11.2 ft^3/lbm

C. 13.33 ft^3/lbm

D. 14.2 ft^3/lbm

Formula: pv = RT v = RT / p

302. Steam at 1000 lbf/ft^2 pressure and 300°R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.

A. 2500 lbf-ft/lbm

B. 3300 lbf-ft/lbm

C. 5400 lbf-ft/lbm

D. 6900 lbf-ft/lbm

Formula: h= u+ pV u= h– pV

303. 3.0 lbm of air are contained at 25 psia and 100 °F. Given that R_air = 53.35 ft-lbf/lbm- °F, what is the volume of the container?

A. 10.7 ft^3

B. 14.7 ft^3

C. 15 ft^3

D. 24.9 ft^3

Formula: use the ideal gas law

pV = mRT

T = (100 +460) °R

V = mRT/p

304. The compressibility factor, x, is used for predicting the behavior of non-ideal gases. How is the compressibility ty factor defined relative to an ideal gas? (subscript “c”refers to critical value)

A. z = P / Pc

B. z = pV/ RT

C. z = T /Tc

D. z = RT / pV

Hint: for an real gases the compressibility factor, x, is an dimensionless constant given by pV= zRT. Therefore z = pV / RT

305. From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and 45.8°C

A. 1.79 kJ/ kg-°C

B. 10.28 kJ/ kg-°C

C. 30.57 kJ/ kg-°C

D. 100.1 kJ/ kg-°C

Formula: ∆h = c∆T

From the steam table

At 47.7 °C h= 2588.1 kJ/ kg

At 43.8 °C h= 2581.1 kJ/ kg

306. A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPA. Calculate the internal energy of the system using the steam table.

A. 5 x 10^5 kJ

B. 8 x 10^5 kJ

C. 1 x 10^6 kJ

D. 2 x 10^6 kJ

Formula: from the steam table

vƒ = 0.001043 m^3 / kg

vg = 1.6940 m^3 / kg

u ƒ= 417.3 kJ/kg ug= 2506kJ/kg

formula: Mvap = V vap/vg

M liq = Vliq/ vƒ

u =uƒM liq + ug M vap

307. A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPA. The liquid water has mass of 1 kg. Using the steam table, calculate the mass of the water vapor.

A. 0.99 kg

B. 1.57 kg

C. 2.54 kg

D. 3.16 kg

Formula: from the steam table at 600 kPa

vƒ = 0.001101 m^3 / kg

vg = 0.3157 m^3 / kg

Vtot = mƒ vƒ + mg vg

mg = (tot-mƒ vƒ) / vg

308. Calculate the entropy of steam at 60 psia with a quality of 0.8

A. 0.4274 BTU/lbm-°R

B. 0.7303 BTU/lbm-°R

C. 1.1577 BTU/lbm-°R

D. 1.2172 BTU/lbm-°R

Formula: from the steam table at 60 psia:

sƒ = 0.4274 BTU/lbm-°R

sƒg = 1.2172 BTU/lbm-°R)

s = sƒ + x sƒg where x = is the quality

309. Find the change in internal energy of 5 lB. of oxygen gas when the temperature changes from 100°F to 120°F. CV = 0.157 BTU/lbm-°R

A. 14.7 BTU

B. 15.7 BTU

C. 16.8 BTU

D. 15.9 BTU

Formula: ∆U= mcv∆T

310. Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1 kg of the water?

A. 0.043 K/s

B. 0.179 K/s

C. 0.357 K/s

D. 1.50 K/s

Formula: Q = mcv (∆T)

311. A system weighing 2 kN. Determine the force that accelerate if to 12 m/s^2.

A. vertically upward when g = 9.7 m/s^2

A. 4474.23 N

B. 5484.23 N

C. 4495.23 N

D. 5488.23 N

Formula: F = m/k (a +g)

312. Refer to problem #311. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.

A. 2474.23 N

B. 2574.23 N

C. 3474.23 N

D. 2374.23 N

Formula :

M = wk / g

F = ma /k

313. A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.

A. 42.41 ft^3

B. 44.35 ft^3

C. 45.63 ft^3

D. 41.23 ft^3

Formula: Vf = (pi d^2 h) / 4

314. What is the resulting pressure when one pound of air at 15 psia and 200°F is heated at constant volume to 800°F?

A. 15 psia

B. 28.6 psia

C. 36.4 psiA.

D. 52.1 psia

Formula :

T1/p1 = T2/p2

p2= p1T2 / T1

315. What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?

A. 28 hp

B. 108 hp

C. 256 hp

D. 13900 hp

Formula: W= p1V1 ln (p1/p2)

Power = dW / dt

316. What is the equation for the work done by a constant temperature system?

A. W = mRTln(V2-V1)

B. W = mR( T2-T1 ) ln( V2/V1)

C. W = mRTln (V2/V1)

D. W = RT ln (V2/V1)

Formula : W=∫ pdV lim1,2

р = mRT / V

317. Twenty grams of oxygen gas are compressed at a constant temperature of 30°C to 5% of their original volume. What work is done on the system.

A. 824 cal

B. 924 cal

C. 944 cal

D. 1124 cal

Formula:

W = -mRTln (V2/V1)

Where R = (1.98 cal/gmole·K) (32 g/gmole)

318. Helium ( R= 0.4698 BTU/lbm-°R ) is compressed isothermally from 14.7 psia and 68°F. The compression ratio is 1:4. Calculate the work done by the gas.

A. –1454 BTU/lbm

B. -364 BTU/lbm

C. -187BTU/lbm

D. 46.7 BTU/lbm

Formula: W = RT ln (V2/V1)

319. Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200 kPA. Calculate the work done by the system.

A. 8 kJ

B. 10 kJ

C. 12 kJ

D. 14 kJ

Formula: W = p(V2-V1)

320. refer to problem no.313. Determine the specific volume.

A. 0.02 ft^3/lbm

B. 0.05 ft^3/lbm

C. 1.0 ft^3/lbm

D. 1.2 ft^3/lbm

Formula :

Vf = ( pi d^2 h) / 4

Pf = mf / vf

Specific volume= Vf /mf

321. What is the weight of a66-kgm man at standard condition?

(Formula: Fg= mg / k)

A. 66 kgf

B. 66 kgm

C. 66 lbm

D. 66 gf

322. What is the specific weight of water at standard condition?

(Formula: γ = ρg / k)

A. 1000 kgm/m³

B. 9.8066 m/s²

C. 1000 kgf/m³

D. None of the above

323. 746°R = ______ °F

A. 254

B. 345

C. 286

D. None of the above

324. A 30-m vertical column of fluid (density 1878 kg/m³ ) is located where g= 9.65 mps². Find the pressure at the base of the column.

(Formula: pg= gρhg/k )

A. 543680 N/m²

B. 543.68 kPa (gauge)

C. Both a & b

D. None of the above

325. Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure?

(formula: p2 = p1T2/T1)

A. 0

B. 209 psia

C. – 420 psia

D. None of the above

326. 876°R = _____ °F

A. 335

B. 416

C. 400

D. None of the above

327. There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPA. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2.

(Formula: T2= T1p2/ p1)

A. 999 K

B. 888 K

C. 456 K

D. One of the above

328. 5 atm = ____mmHg

A. 8300

B. 3800

C. 3080

D. None of the above

329. A certain gas, with cp = 0.529 Btu/lB. °R and R = 96.2 ft.lb/lB. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psiA. Compute for T2.

(Formula: T2= T1V2/V1)

A. 460°R

B. 270°R

C. 1620°R

D. None of the above

330. In the above problem, compute for the mass.

(Formula: m = p1V1 / RT1)

A. 0.2148 lb

B. 0.2134 lb

C. 0.1248 lb

D. None of the above

331. 710°R= ______ °C

A. 214

B. 121

C. 213

D. None of the above

332. 212 °F = _____ °C

A. 200

B. 150

C. 100

D. None of the above

333. Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find ∆E.

(Formula: ∆E = Q- W)

A. 125 J

B. 123 J

C. 126 J

D. None of the above

334. A pressure gage registers 50 psig in a region where the barometer is 14.25 psiA. Find absolute pressure in psia, PA.

(Formula; p = patm+ pg)

A. 433 kPa

B. 443 kPa

C. 343 kPa

D. None of the above

335. A mass of 5 kg is 100m above a given datum where local g = 9.75 m/s². Find the gravitational force in newtons.

(Formula: Fg= mg/k )

A. 48.75 N

B. 50 N

C. 45 N

D. None of the above

336. In the above problem, find the potential energy of the mass with respect to datum.

(Formula: P = mgz/k )

A. 4875 j

B. 0.51 j

C. 0.46 j

D. None of the above

337. The combined mass of car and passengers travelling at 72 km/hr is 1500 kg. Find the kinetic energy of this combined mass.

(Formula: K =mv2/ 2k )

A. 300 kJ

B. 200 kJ

C. 500 kJ

D. None of the above

338. 14.696 psia = _____ mmHg

A. 760

B. 1

C. 350

D. None of the above

339. 212°C = _____ K

A. 485

B. 435

C. 498

D. None of the above

340. 212 °F = _____R

A. 567

B. 672

C. 700

D. None of the above

341. An automobile tire has a gauge pressure of 200 kpa at 0°C assuming no air leaks and no change of volume of the tire, what is the gauge pressure at 35ºC.

A. 298.645

B. 398.109

C. 291.167

D. 281.333

Pg = Pabs – Patm

342. An ideal gas at 45 psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?

A. 65.10 psi

B. 65.11 psi

C. 65.23 psi

D. 61.16 psi

P1V1/T1= P2V2/T2;V = Constant

343. A wall of the firebrick has an inside temperature of 313ºF and an outside temperature of 73ºF. What is the difference in the surface temperature in Rankin?

A. 70

B. 68

C. 72

D. 94

ºR = ºF + 460

344. What is the force required to accelerate amass of 30 kg at a rate of 15 m/s².

A. 460 N

B. 380 N

C. 560 N

D. 450 N

F = ma

345. How much does an object having the mass of 100 kg weight in newton.

A. 981 N

B. 991 N

C. 981.6 N

D. 980.1N

F = ma

346. The volume of the gas held at constant pressure increases 4 cm² at 0°C to 5 cm². What is the final pressure?

A. 68.65ºC

B. 68.25ºC

C. 70.01°C

D. 79.1ºC

t2= T2–T1

347. A certain gas with cp = 0.529 Btu/lb°R and R = 96.2 ft/lbºR expands from 5 ft and 80ºF to 15 ft while the pressure remains constant at 15.5 psiA.

A. T2=1.620ºR, ▲H = 122.83 Btu

B. T2 = 2°R, ▲H = 122.83 Btu

C. T2 = 2.620ºR, ▲H = 122.83 Btu

D. T2 = 1°R, ▲H = 122.83 Btu

T2= V2(t2)/V1 and ▲H = mcp (T2-T1)

348. A vacuum is connected to a tank reads 3 kpa at a location with the barometric pressure reading is 75mmhg. Determined the P absolute in the tank

A. 70.658 kpa

B. 68 kpa

C. 58.78 kap

D. None of the above

Pabs = Patm – Pvacuum

349. Calculate:

A. Mass flow rate in lb/hr.

B. The velocity at section 2 in fps

A. 800,000lb/hr;625ft/s

B. 900,000lb/hr;625 ft/s

C. 888,000lb/hr;269 ft/s

D. 700,000lb/hr;269 ft/s

m = A1V!/V1

350. A 600kg hammer of a pile driver is lilted 2m the pilling head. What is the change of potential energy? If the hammer is realest. What will be its velocity and the instant if it sticks the pilling?

A. 10,772 N-m and 5.26 m/s

B. 13,200 N-m and 5.26 m/s

C. 11,772 N-m and 6.26 m/s

D. 11,77 2N-m and 5.26 m/s

▲PE = mgo(▲Z)/gc

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