FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its maximum thermal efficiency?

• a.0.25
• b.0.50✓
• c.2.0
• d.0.75

Carnot efficiency eta = 1 - T_cold/T_hot = 1 - 300/600 = 0.50, or 50%. Temperatures must be absolute (kelvin). No real engine between the same reservoirs can exceed this limit.

How much heat is required to raise the temperature of 2 kg of water by 10 degC (specific heat 4,186 J/kg·degC)?

• a.837 kJ
• b.8.37 kJ
• c.41.9 kJ
• d.83.7 kJ✓

Sensible heat Q = m·c·(delta T) = 2 x 4,186 x 10 = 83,720 J = 83.7 kJ. Water's high specific heat is why it is an effective coolant and thermal-storage medium.

For an ideal gas held at constant temperature, if the absolute pressure is doubled, the volume becomes:

• a.Doubled
• b.Unchanged
• c.Halved✓
• d.Quadrupled

Boyle's law (a special case of the ideal gas law at constant temperature) states P·V = constant, so pressure and volume are inversely proportional. Doubling pressure halves the volume.

A Carnot engine operates between a hot reservoir at 800 K and a cold reservoir at 320 K. What is its maximum thermal efficiency?

• a.0.60✓
• b.1.67
• c.0.40
• d.0.50

Carnot efficiency eta = 1 - T_cold/T_hot = 1 - 320/800 = 1 - 0.40 = 0.60 (60%). Reservoir temperatures must be absolute (kelvin); 0.40 is the temperature ratio T_c/T_h itself.

A reversible (Carnot) refrigerator operates between a cold space at 250 K and surroundings at 300 K. What is its coefficient of performance?

• a.6
• b.1.2
• c.0.2
• d.5✓

For a Carnot refrigerator COP = T_cold/(T_hot - T_cold) = 250/(300 - 250) = 250/50 = 5. Using T_hot in the numerator (300/50 = 6) gives the heat-pump COP by mistake.

A reversible (Carnot) heat pump maintains a house at 300 K while drawing heat from outdoor air at 270 K. What is its coefficient of performance?

• a.10✓
• b.0.1
• c.9
• d.11

For a Carnot heat pump COP = T_hot/(T_hot - T_cold) = 300/(300 - 270) = 300/30 = 10. Note COP(heat pump) = COP(refrigerator) + 1, so the corresponding refrigerator COP would be 9.

A 2 kg mass of air (R = 0.287 kJ/kg·K) is at 300 K and 100 kPa. What volume does it occupy?

• a.1.5 m^3
• b.3.444 m^3
• c.1.722 m^3✓
• d.0.861 m^3

The ideal gas law PV = mRT gives V = mRT/P = (2)(0.287)(300)/100 = 172.2/100 = 1.722 m^3. With R in kJ/kg·K and P in kPa the result is directly in cubic meters.

Air (k = 1.4) at 300 K is compressed isentropically from 100 kPa to 200 kPa. What is the final temperature?

• a.366 K✓
• b.396 K
• c.246 K
• d.600 K

For an isentropic process T2 = T1(P2/P1)^((k-1)/k) = 300(2)^(0.4/1.4) = 300(2)^0.2857 = 300(1.219) = 366 K. The exponent (k-1)/k = 0.2857; scaling temperature linearly with pressure (600 K) is the common error.

Heat conducts through a plane wall of area 10 m^2, thickness 0.1 m, and thermal conductivity 0.5 W/m·K with a 20 K temperature difference across it. What is the heat-transfer rate?

• a.10 W
• b.1,000 W✓
• c.2,000 W
• d.100 W

Fourier's law for a plane wall: Q = kA(delta T)/L = (0.5)(10)(20)/0.1 = 100/0.1 = 1,000 W. Heat flow is proportional to conductivity and area and inversely proportional to thickness.

What is the conductive thermal resistance of a plane wall with thickness 0.2 m, thermal conductivity 0.04 W/m·K, and area 5 m^2?

• a.25 K/W
• b.0.04 K/W
• c.0.008 K/W
• d.1.0 K/W✓

Conduction resistance R = L/(kA) = 0.2/((0.04)(5)) = 0.2/0.2 = 1.0 K/W. Thermal resistances add in series just like electrical resistances, with Q = (delta T)/R.

A surface at 80 degC loses heat by convection to air at 20 degC. If the convection coefficient is 25 W/m^2·K over an area of 2 m^2, what is the heat-transfer rate?

• a.5,000 W
• b.1,500 W
• c.3,000 W✓
• d.300 W

Newton's law of cooling: Q = hA(delta T) = (25)(2)(80 - 20) = (25)(2)(60) = 3,000 W. The driving potential is the surface-to-fluid temperature difference of 60 degC.

What is the emissive power of a blackbody at 500 K? (Stefan-Boltzmann constant = 5.67x10^-8 W/m^2·K^4)

• a.3,544 W/m^2✓
• b.1,772 W/m^2
• c.7,087 W/m^2
• d.28.35 W/m^2

The Stefan-Boltzmann law gives Eb = sigma·T^4 = (5.67x10^-8)(500)^4 = (5.67x10^-8)(6.25x10^10) = 3,544 W/m^2. Emissive power scales with the fourth power of absolute temperature.

A closed system absorbs 100 kJ of heat while doing 30 kJ of work on its surroundings. What is the change in its internal energy?

• a.100 kJ
• b.130 kJ
• c.-70 kJ
• d.70 kJ✓

The first law for a closed system is delta U = Q - W = 100 - 30 = 70 kJ, using the sign convention that heat added is positive and work done BY the system is positive.

One mole of an ideal gas expands isothermally and reversibly at 300 K to twice its initial volume. How much work does it do? (R = 8.314 J/mol·K)

• a.1,729 J✓
• b.865 J
• c.3,457 J
• d.2,494 J

Isothermal reversible work W = nRT·ln(V2/V1) = (1)(8.314)(300)·ln(2) = 2494.2(0.693) = 1,729 J. For an isothermal ideal-gas process Q = W since internal energy does not change.

For air (cp = 1.005 kJ/kg·K) heated at constant pressure through a 50 K rise, what is the change in specific enthalpy?

• a.15 kJ/kg
• b.50.25 kJ/kg✓
• c.36 kJ/kg
• d.100 kJ/kg

For an ideal gas delta h = cp·(delta T) = (1.005)(50) = 50.25 kJ/kg. Enthalpy change of an ideal gas depends only on temperature and uses the constant-pressure specific heat.

For air (cv = 0.718 kJ/kg·K) undergoing a 100 K temperature rise, what is the change in specific internal energy?

• a.50.25 kJ/kg
• b.100.5 kJ/kg
• c.71.8 kJ/kg✓
• d.35.9 kJ/kg

For an ideal gas delta u = cv·(delta T) = (0.718)(100) = 71.8 kJ/kg. Internal-energy change of an ideal gas depends only on temperature and uses the constant-volume specific heat.

For an ideal gas with cp = 1.005 kJ/kg·K and gas constant R = 0.287 kJ/kg·K, what is the constant-volume specific heat cv?

• a.0.718 kJ/kg·K✓
• b.1.292 kJ/kg·K
• c.1.005 kJ/kg·K
• d.3.5 kJ/kg·K

Mayer's relation for an ideal gas is cp - cv = R, so cv = cp - R = 1.005 - 0.287 = 0.718 kJ/kg·K. Adding rather than subtracting R (1.292) is the common error.

An air-standard Otto cycle has a compression ratio of 8 and k = 1.4. What is its thermal efficiency?

• a.0.50
• b.0.75
• c.0.565✓
• d.0.435

Otto-cycle efficiency eta = 1 - 1/r^(k-1) = 1 - 1/8^0.4 = 1 - 1/2.297 = 1 - 0.435 = 0.565 (56.5%). Efficiency rises with compression ratio; 0.435 is the reciprocal factor itself.

How much heat is required to melt 0.5 kg of ice at 0 degC? (latent heat of fusion = 334 kJ/kg)

• a.83.5 kJ
• b.334 kJ
• c.1,130 kJ
• d.167 kJ✓

Latent heat Q = m·h_fusion = (0.5)(334) = 167 kJ. During a phase change temperature stays constant while latent heat is absorbed; no specific-heat term applies.

How much heat is required to vaporize 2 kg of saturated water at 100 degC? (latent heat of vaporization = 2,257 kJ/kg)

• a.4,514 kJ✓
• b.9,028 kJ
• c.2,257 kJ
• d.1,128 kJ

Latent heat Q = m·h_fg = (2)(2,257) = 4,514 kJ. Vaporization absorbs far more energy than melting because intermolecular bonds are fully broken as liquid becomes vapor.

1 kg of water at 80 degC is mixed with 1 kg of water at 20 degC in an insulated container. What is the final equilibrium temperature?

• a.50 degC✓
• b.100 degC
• c.60 degC
• d.40 degC

Energy balance m·c·(T - 80) + m·c·(T - 20) = 0. With equal masses and the same specific heat, the final temperature is the simple average (80 + 20)/2 = 50 degC.

A power plant delivers 400 MW of net work while receiving 1,000 MW of heat from its boiler. What is its thermal efficiency?

• a.2.5
• b.0.40✓
• c.0.25
• d.0.60

Thermal efficiency eta = W_net/Q_in = 400/1,000 = 0.40 (40%). The rejected heat is Q_out = Q_in - W_net = 600 MW, which gives the 0.60 distractor.

A heat engine receives 500 kJ from a high-temperature source and rejects 300 kJ to a sink each cycle. What is its net work output per cycle?

• a.800 kJ
• b.200 kJ✓
• c.300 kJ
• d.500 kJ

For a cycle the first law gives W_net = Q_in - Q_out = 500 - 300 = 200 kJ, since internal energy returns to its starting value over a complete cycle.

A refrigerator removes 200 kJ from the cold space while consuming 50 kJ of work input. What is its coefficient of performance?

• a.5
• b.4✓
• c.1.25
• d.0.25

Refrigerator COP = Q_cold/W_in = 200/50 = 4. Unlike an efficiency, a COP can exceed 1. Using the heat rejected (250 kJ) would wrongly give 5.

A heat pump delivers 600 kJ of heat to a house while consuming 120 kJ of work input. What is its coefficient of performance?

• a.6
• b.0.2
• c.5✓
• d.4

Heat-pump COP = Q_hot/W_in = 600/120 = 5. The heat absorbed from outside is Q_cold = 600 - 120 = 480 kJ, giving a refrigerator COP of 4 (one less than the heat-pump value).

In a counterflow heat exchanger the terminal temperature differences are 50 degC and 20 degC. What is the log-mean temperature difference?

• a.35 degC
• b.25 degC
• c.30 degC
• d.32.7 degC✓

LMTD = (delta T1 - delta T2)/ln(delta T1/delta T2) = (50 - 20)/ln(50/20) = 30/ln(2.5) = 30/0.916 = 32.7 degC. The LMTD is always below the arithmetic mean of 35 degC.

Two wall layers in series have thermal resistances of 0.5 K/W and 1.5 K/W. If the overall temperature difference is 100 K, what is the heat-transfer rate?

• a.200 W
• b.25 W
• c.100 W
• d.50 W✓

Series resistances add: R_total = 0.5 + 1.5 = 2.0 K/W. Then Q = (delta T)/R_total = 100/2.0 = 50 W, the thermal analog of Ohm's law.

An ideal gas at constant pressure occupies 1 m^3 at 300 K. If it is heated to 600 K, what volume does it occupy?

• a.2 m^3✓
• b.4 m^3
• c.0.5 m^3
• d.1 m^3

Charles's law at constant pressure gives V2 = V1·(T2/T1) = 1·(600/300) = 2 m^3. Volume is directly proportional to absolute temperature when pressure is held constant.

Steam enters a nozzle with negligible velocity and accelerates adiabatically as its specific enthalpy drops by 50 kJ/kg. What is the approximate exit velocity?

• a.224 m/s
• b.100 m/s
• c.316 m/s✓
• d.158 m/s

The steady-flow energy balance for an adiabatic nozzle gives V = sqrt(2·delta h) = sqrt(2 x 50,000 J/kg) = sqrt(100,000) = 316 m/s. Enthalpy drop converts directly to kinetic energy.

An ideal gas (R = 0.287 kJ/kg·K) expands isothermally and reversibly to three times its initial volume. What is its specific entropy change?

• a.0.315 kJ/kg·K✓
• b.0.861 kJ/kg·K
• c.0.2 kJ/kg·K
• d.0.63 kJ/kg·K

For an isothermal ideal-gas process delta s = R·ln(V2/V1) = 0.287·ln(3) = 0.287(1.099) = 0.315 kJ/kg·K. Entropy rises as the gas expands into a larger volume.

What is the specific volume of air (R = 0.287 kJ/kg·K) at 350 K and 200 kPa, treating it as an ideal gas?

• a.0.251 m^3/kg
• b.0.502 m^3/kg✓
• c.0.2 m^3/kg
• d.1.004 m^3/kg

For an ideal gas v = RT/P = (0.287)(350)/200 = 100.45/200 = 0.502 m^3/kg. Specific volume rises with temperature and falls with pressure.

An aluminum plate (k = 200 W/m·K) is 0.02 m thick with a 100 K temperature difference across it. What is the conduction heat flux?

• a.1,000,000 W/m^2✓
• b.10,000 W/m^2
• c.400,000 W/m^2
• d.500,000 W/m^2

The one-dimensional conduction heat flux is q = k(delta T)/L = (200)(100)/0.02 = 20,000/0.02 = 1,000,000 W/m^2 (1 MW/m^2). Flux is heat transfer per unit area.

What is the maximum possible thermal efficiency of any heat engine operating between reservoirs at 500 K and 300 K?

• a.0.55
• b.0.60
• c.0.40✓
• d.0.45

The Carnot efficiency sets the upper bound: eta_max = 1 - T_cold/T_hot = 1 - 300/500 = 0.40 (40%). Any claim of a higher efficiency between the same reservoirs violates the second law.

A rigid sealed tank of gas is at 100 kPa and 300 K. If it is heated to 450 K, what is the new pressure?

• a.300 kPa
• b.200 kPa
• c.150 kPa✓
• d.66.7 kPa

At constant volume Gay-Lussac's law gives P2 = P1·(T2/T1) = 100·(450/300) = 150 kPa. Pressure is directly proportional to absolute temperature in a rigid container.

During an adiabatic compression, 40 kJ of work is done ON a closed system of gas. What is the change in its internal energy?

• a.+80 kJ
• b.+40 kJ✓
• c.0 kJ
• d.-40 kJ

For an adiabatic process Q = 0, so delta U = -W. Work done ON the system is W = -40 kJ, giving delta U = -(-40) = +40 kJ. Compression raises internal energy and temperature.

A heat exchanger surface has an overall heat-transfer coefficient of 10 W/m^2·K, an area of 20 m^2, and a mean temperature difference of 30 K. What is the heat-transfer rate?

• a.600 W
• b.6,000 W✓
• c.60 W
• d.3,000 W

The overall rate equation is Q = U·A·(delta T) = (10)(20)(30) = 6,000 W. The coefficient U bundles all series conduction and convection resistances into one value.

For air with cp = 1.005 kJ/kg·K and cv = 0.718 kJ/kg·K, what is the specific-heat ratio k?

• a.1.29
• b.1.4✓
• c.2.4
• d.0.71

The specific-heat ratio is k = cp/cv = 1.005/0.718 = 1.40. This ratio (about 1.4 for diatomic gases like air) governs isentropic processes and the speed of sound.

What is the density of air (R = 0.287 kJ/kg·K) at 101.325 kPa and 300 K?

• a.0.85 kg/m^3
• b.1.18 kg/m^3✓
• c.1.5 kg/m^3
• d.2.36 kg/m^3

From the ideal gas law density rho = P/(RT) = 101.325/((0.287)(300)) = 101.325/86.1 = 1.18 kg/m^3. Density is the reciprocal of specific volume.