MCQ in Thermodynamics Part 8 | ECE Board Exam

MCQ Thermodynamics Part 8 | ECE Board Exam

This is the Multiple Choice Questions Part 8 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.

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Continue Practice Exam Test Questions Part 8 of the Series

MCQ in Thermodynamics Part 7 | ECE Board Exam

Choose the letter of the best answer in each question.

351. A bayabas falls from a branch 5 m above the ground with what speed in meter per second does it strike the ground assume g = 10m/s².

A. 11 m/s

B. 12 m/s

C. 13 m/s

D. 10 m/s

▲KE = mV2/2gc

View Answer:

Answer: Option D

Explanation:

352. While swimming at a depth of 13 m in a freshwater lake a fish emits an air bubble of volume 2 mm² atmospheric pressure is 100 kpa what is the original pressure of the bubble.

A. 217.17 kpa

B. 119 kpa

C. 326.15 kpa

D. 210 kap

Pabs = Pg + Patm

View Answer:

Answer: Option A

Explanation:

353. Oxygen at 15°C and 10.3 Mpa gauge pressure occupies 600 L. What is the occupied by the oxygen at 8.28 Mpa gauge pressure and 35°C?

A. 789.32 L

B. 796.32 L

C. 699 L

D. 588 L

V2= P1V1/T1P2

View Answer:

Answer: Option B

Explanation:

354. Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is the volume flow rate of water in ft³/sec?

A. 7.85

B. 6.85

C. 8.85

D. 5.85

V = Aν

View Answer:

Answer: Option A

Explanation:

355. A certain fluid is flowing in a 0.5 m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.

A. 267 kg/s

B. 378 kg/s

C. 375 kg/s

D. 456.5 kg/s

m = Aν/V

View Answer:

Answer: Option C

Explanation:

356. A gas having a volume of100 ft³ at 27°C is expanded to 120 ft³ by heated at constant pressure to what temperature has it been heated to have this new volume? 

A. 87°C

B. 85°C

C. 76°C

D. 97°C

t2= T2–T1

View Answer:

Answer: Option A

Explanation:

357. Water flow to a terminal 3 mm diameter and has an average speed of 2 m/s. What is the rate of flow in cubic meter/mm?

A. 0.0001m³/min

B. 0.076 m³/min

C. 0.085 m³/min

D. 0.097 m³/min

View Answer:

Answer: Option C

Explanation:

358. Water flowing at a 6m/s through a 60 mm pipe is suddenly channeled into a 30 mm pipe. What is the velocity in the small pipe?

A. 34 m/s

B. 24 m/s

C. 15 m/s

D. 27 m/s

View Answer:

Answer: Option B

Explanation:

359. A vertical column of water will be supported to what height by standard atmospheric pressure.

A. 33.9 ft

B. 45 ft

C. 67 ft

D. 25.46 ft

ho= Po/Yo

View Answer:

Answer: Option A

Explanation:

360. A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1 ft², V1 = 100 fpm, volume1 of 4 ft³/lB. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.

A. 625 fpm

B. 567 fpm

C. 356 fpm

D. None of the above

View Answer:

Answer: Option A

Explanation:

361. The weight of an object is 50 lB. What is its mass at standard condition?

A. 50 lbm

B. 60 lbm

C. 70 lbm

D. 80 lbm

formula: m = Fgk /g

View Answer:

Answer: Option A

Explanation:

362. A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4 lb/ft3 po = 14.7 psi.

A. 44.9 ft

B. 33.9 ft

C. 22.9 ft

D. 55.9 ft

formula: ho= po/Yw

View Answer:

Answer: Option B

Explanation:

363. For a certain gas R = 320 J/kg.K and cv= 0.84 kJ/kg.K. Find k?

A. 1.36

B. 1.37

C. 1.38

D. 1.39

formula: k= R / cv+1

View Answer:

Answer: Option C

Explanation:

364. Ten cu. ft of air at 300 psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?

A.-120 Btu

B. -220 Btu

C.-320 Btu

D. -420 Btu

formula: Q= mcv(T2-T1)

View Answer:

Answer: Option D

Explanation:

365. Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a 12 hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m3.

A. 7.9°C

B. 8.9°C

C. 9.9°C

D. 10.9°C

formula: ΔQ = m c ΔT

View Answer:

Answer: Option A

Explanation:

366. The concrete roof of a house is 10 m by 8 m and 10 cm thick (4″). Estimate the total heat the roof would absorb over the 12 day?

A. 1.3 x 108 J

b 2.3 x 108 J

C. 3.3 x 108 J

D. 4.3 x 108 J

formula: ΔQ = ΔQ/Δtx Δt

View Answer:

Answer: Option A

Explanation:

367. The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?

A. -540 J

B. -640 J

C. -740 J

D. -840 J

formula: ΔU = q +w

View Answer:

Answer: Option A

Explanation:

368. When the pressure on a 1 kg liquid is increased isothermally from 1 bar to 3000 bar the Gibbs free energy increases by 360 kJ. Estimate the density of the liquid.

A. 0.66 kg liter-1

B. 0.77 kg liter-1

C. 0.88 kg liter-1

D. 0.99 kg liter-1

View Answer:

Answer: Option B

Explanation:

369. A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.

A. 11,120 N

B. 11,320 N

C. 11,420 N

D. 11520 N

formula: F= ma / k

View Answer:

Answer: Option A

Explanation:

370. An ideal gas of volume 1liter and pressure 10 bar undergoes a quasi static adiabatic expansion until the pressure drops to 1 bar. Assume γ to be 1.4 what is the final volume?

A. 3.18 l

B. 4.18 l

C. 5.18 l

D. 6.18 l

View Answer:

Answer: Option C

Explanation:

371. Two masses, one of the 10 kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2. The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.

A. 20 kg

B. 19 kg

C. 18 kg

D. 17 kg

formula: m=Fg k / g

View Answer:

Answer: Option C

Explanation:

372. The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lB. Determine the pressure at this point.

A. 222 psi

B. 333 psi

C. 444 psi

D. 111 psi

formula: Ef= pV

View Answer:

Answer: Option D

Explanation:

373. Find и for steam at 100 psia and 600°F. If h = 1329.6 and v = 6.216

A. 1214 Btu/lb

B. 1234 Btu/lb

C. 1342 Btu/lb

D. 1324 Btu/lb

formula: и = h– pv/ J

View Answer:

Answer: Option A

Explanation:

374. What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.

A. 194 lb

B. 214 lb

C. 394 lb

D. 413 lb

formula: m=pV /RT

View Answer:

Answer: Option C

Explanation:

375. A rotary compressor receives 6 m3/ min of a gas(R=410J/ kgK, cp=1.03kJ /kgK,k= 1.67) at 105 k/Paa, 27°C and delivers it at 630 kPaa: ΔP = 0, ΔK= 0. Find the work if the process is isentropic?

A. –1664 kJ/min

B. –1774 kJ/min

C. –1884 kJ/min

D. –1994 kJ/min

formula: WSF = Q- ΔH m=p1V1/RT1 T2= T1(p2/p1)(k-1)/k

View Answer:

Answer: Option A

Explanation:

376. A carnot power cycle operates on 2 lb of air between the limits of 71°F and 500°F. The pressure at the beginning of isothermal expansion is 400 psia and at the end of isothermal expansion is 185 psig. Determine the volume at the end of isothermal compression.

A. 7.849 ft3

B. 7.850 ft3

C. 7.851 ft3

D. 7852 ft3

formula: V= mRT/ P P3= P2[T3/ T2]

View Answer:

Answer: Option A

Explanation:

377. During a polytropic process,10lb of an ideal gas, whose R= 40 ft.lb/lB.R and cp = 0.25 Btu/lB.R, changes state from 20 psia and 40°F to 120psia and 340°F. Determine n?

A. 1.234

B. 1.345

C. 1.456

D. 1.356

formula: [ p2/p1]n-1 / n = T2/T1

View Answer:

Answer: Option D

Explanation:

378. A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kfg of this gas at constant pressure when the initial temp is 32.2°C? Find T2.

A. 339.4 K

B. 449.4 K

C. 559.4K

D. 669.4K

formula: cp = kR/ k-1 Q= mcp(T2-T1)

View Answer:

Answer: Option A

Explanation:

379. A certain gas, with cp = 0.529 Btu/ lB. °Rand 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.

A.1520°R

B. 1620°R

C. 1720°R

D. 1820°R

formula: T2= T1V2/V1

View Answer:

Answer: Option B

Explanation:

380. A System has a temperature of 250°F. Convert this Value to °R?

A. 740°R

B.730°R

C. 720°R

D. 710°R

formula: °R= °F + 460

View Answer:

Answer: Option D

Explanation:

381. Steam with a specific volume of 0.09596 m³/kg undergoes a constant pressure process at 1.70 MPa until the specific volume becomes 0.13796 m³/kg. What are (A. the final temperature, (B. ∆u, (C. W, (D.∆s, and (e) Q?

A. 265.4°C, 430.7kJ/kg, 71.4kJ/kg, 1.0327kJ/(kg)(K),502.1 kJ/kg

B. 204.2°C, -703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ/(kg)(K), -787.4 kJ/kg

C. 304.2°C, -803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ/(kg)(K), -987.4 kJ/kg

D. 279.4°C, 439.7kJ/kg, 79.4kJ/kg, 3.0327kJ/(kg)(K),602.1 kJ/kg

View Answer:

Answer: Option A

Explanation:

382. Steam with an enthalpy of 2843.5 kJ/kg undergoes a constant pressure process at 0.9 MPa until the enthalpy becomes 2056.1 kJ/kg. What are (A. the initial temperature or quality, (B. ∆u, (C.W, (D. ∆s, and(e) Q?

A. 265.4°C, 430.7kJ/kg, 71.4kJ/kg, 1.0327kJ/(kg)(K),502.1 kJ/kg

B. 204.2°C, -703.2 kJ/kg, -84.15 kJ/kg, -1.7505 kJ/(kg)(K),-787.4 kJ/kg

C. 304.2°C, -803.2 kJ/kg, -89.15 kJ/kg, -2.7505 kJ/(kg)(K), -987.4 kJ/kg

D. 279.4°C, 439.7kJ/kg, 79.4kJ/kg, 3.0327kJ/(kg)(K), 602.1 kJ/kg

Formula of #1and #2: ∆u = u2 –u1, W = p(v2-v1), ∆s =s2-s1, Q = h2 –h1

View Answer:

Answer: Option B

Explanation:

383. At throttling calorimeter receives steam from a boiler drum at0.11 MPa and is superheated by 10 degrees. If the boiler drum pressure is 1.55 MPa, what is the quality of the steam generated by the boiler?

A. 95.20%

B. 70.10%

C. 65.60%

D. 95.56%

Formula: h1 = hf1 + x1hfg1

View Answer:

Answer: Option A

Explanation:

384. A steam calorimeter receives steam from a pipe at 0.1 MPa and 20°SH. For a pipe steam pressure of 2 MPa, what is the quality of the steam?

A. 95.56%

B. 70.10%

C. 95.20%

D. 85.10%

Formula: h1 = hf1 + x1hfg1

View Answer:

Answer: Option A

Explanation:

385. A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.

A. 1378.7 kJ

B. 1348.5 kJ

C. 1278,7 kJ

D. 1246,5 kJ

Formula: Q = (h2 – p2v2) –(h1 –p1v1)

View Answer:

Answer: Option A

Explanation:

386. Water vapor at 100 KPa and 150°C is compressed isothermally until half the vapor has condensed. How much work must be performed on the steam in this compression process per kilogram?

A. -1384.7 kJ

B. 1384.7 kJ

C. -2384.7 kJ

D. 2384.7 kJ

View Answer:

Answer: Option A

Explanation:

387. Wet steam at 1 MPa flowing through a pipe is throttled to a pressure of 0.1 MPA. If the throttling temperature is110°C, What is the quality of the steam in the pipe?

A. 96%

B. 86%

C. 76%

D. 66%

View Answer:

Answer: Option A

Explanation:

388. Steam is throttled to 0.1 MPa with 20 degrees of superheat. (A. What is the quality of throttled steam if its pressure is 0.75 MPa (B. What is the enthalpy of the process?

A. 97.6%,2713 kJ/kg

B. -97.6%, 2713 kJ/kg

C. 87.6%,3713 kJ/kg

D. -87.6%, 3713 kJ/kg

View Answer:

Answer: Option A

Explanation:

389. The pressure gauge on a 2000 m³ tank of oxygen gas reads 600 kPA. How much volumes will the oxygen occupied at pressure of the outside air 100 kPa?

A. 14026.5 m³

B. 15026.5 m³

C. 13026.5 m³

D. 16026.5 m³

Formula: P1V1/T1 =P2V2/T2

View Answer:

Answer: Option A

Explanation:

390. Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.

A. 130m³

B. 136m³

C. 120m³

D. 126m³

Formula: P1V1/T1 =P2V2/T2

View Answer:

Answer: Option D

Explanation:

391. A Gas tank registers1000 kPA. After some gas has been used, the gauge registers 500 kPA. What percent of the gas remains in the tank?

A. 64.40%

B. 74.60%

C. 58.40%

D. 54.60%

Formula: Pabs = Patm + Pgage & %= P2/P1 * 100%

View Answer:

Answer: Option D

Explanation:

392. The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?

A. 180 in³

B. 170 in³

C. 160 in³

D. 190 in³

Formula: P2V2 = P1V1

View Answer:

Answer: Option D

Explanation:

393. While swimming at depth of 120 m in a fresh water lake, A fish emits an air bubbles of volume 2.0 mm³ atmospheric pressure is 100 kPA. What is the pressure of the bubble?

A. 217.7 kPa

B. 317.7 kPa

C. 417.7 kPa

D. 517.7 kPa

Formula: P= δh

View Answer:

Answer: Option A

Explanation:

394. How many joules of work is the equivalent of 15000 cal of heat?

A. 62850 joules

B. 3579.95 joules

C. 14995.81 joules

D. 15004.19 joules

Formula: J =Work/Heat

J = mechanical equivalent of heat whose value is 4.19 joules/calorie

View Answer:

Answer: Option A

Explanation:

395. Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?

A. 4,190,000 joules

B. 8,390,000 joules

C. 839,000 joules

D. 419 000 joules

Formula: J =Work/Heat

J = mechanical equivalent of heat whose value is 4.19 joules/calorie

View Answer:

Answer: Option D

Explanation:

396. 3 horsepower (hp) = _____________watts?

A. 1492 watts

B. 2238 watts

C. 746 watts

D. 2238 kilowatts

Formula: 1hp= 746 watts

View Answer:

Answer: Option B

Explanation:

397. How many Newton’s (N) in 900,000 dynes?

A. 8 Newton’s

B. 9 Newton’s

C. 7 Newton’s

D. 6 Newton’s

Formula: 1Newton (N)=100,000dynes

View Answer:

Answer: Option B

Explanation:

398. Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.

A. 744.8 hp

B. 0.998 hp

C. 746 hp

D. 1.998 hp

Formula: Power = Fd/t = mgh/t

F = W = mg

d = h

View Answer:

Answer: Option B

Explanation:

399. How many calories of heat will be needed to raise the temperature of 200 g of iron from 27°C to 80°C? (c = 0.11 cal/g. °C.

A. 1.16 kcal

B. 2166 cal

C. 3.16 kcal

D. 4166 cal

Formula: H = mc∆T

View Answer:

Answer: Option A

Explanation:

400. 100g of iron was heated to 100°C and mixed with 22g of water at 40°C. The final temperature of the mixture was 60°C. Show that the heat given off by the iron equals the heat absorbed by the water.

A. 440 cal

B. 540 cal

C. 340 cal

D. 640 cal

Formula: H (given off by iron) = H (absorbed by water),

mc∆T(iron)= mc∆T(water)

View Answer:

Answer: Option A

Explanation:

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