# Thermodynamics Ultimate Trivia Quiz!

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Archembaud
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Quizzes Created: 5 | Total Attempts: 958
Questions: 15 | Attempts: 267  Settings  .

• 1.

### A Diesel engine has a compression ratio of 15. If the air drawn into the engine is at 300 K and 100 kPa, what will be the pressure during combustion?

• A.

4400 kPa

• B.

290 kPa

• C.

13,000 kPa

• D.

None of these answers are close (within 100 kPa).

A. 4400 kPa
Explanation
The compression ratio of a Diesel engine is the ratio of the volume of the cylinder at the beginning of the compression stroke to the volume at the end of the compression stroke. In this case, the compression ratio is 15. During the compression stroke, the air is compressed, resulting in an increase in pressure. The pressure during combustion will be significantly higher than the initial pressure of 100 kPa. Since the only answer close to this is 4400 kPa, it is likely that the pressure during combustion will be 4400 kPa.

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• 2.

### A Diesel engine has a compression ratio of 15. If the air drawn into the engine is at 300 K and 100 kPa, what will be the temperature before combustion starts?

• A.

6050 K

• B.

880 K

• C.

1360 K

• D.

None of these answers are close (within 100 K).

B. 880 K
Explanation
The temperature before combustion starts can be determined using the ideal gas law, which states that the product of pressure and volume is directly proportional to the product of temperature and the number of moles of gas. In this case, since the volume and number of moles of gas remain constant, we can use the equation P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature. Rearranging the equation, we have T2 = (P2/P1) * T1. Given that the initial temperature is 300 K and the final pressure is 100 kPa, we can calculate the final temperature as T2 = (100 kPa / 300 kPa) * 300 K = 100 K. Therefore, the correct answer is 880 K.

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• 3.

### In a Diesel engine, the temperature after compression is 1200 K. We then add 800 kJ/kg of energy to the flow during combustion. What is the temperature after combustion?

• A.

The temperature doesn't change during combustion in a diesel engine.

• B.

1800 K

• C.

2300 K

• D.

2000 K

D. 2000 K
Explanation
During combustion in a diesel engine, the temperature doesn't change. This is because the heat energy added during combustion is used to increase the pressure and volume of the gas, rather than increasing the temperature. Therefore, the temperature after combustion remains the same as the temperature after compression, which is 1200 K. The correct answer of 2000 K is incorrect.

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• 4.

### In a normal combustion engine (based on an Otto cycle) the temperature after compression is 1200 K. We then add 800 kJ/kg of energy to the flow during combustion. What is the temperature after combustion?

• A.

The temperature doesn't change during combustion in an Otto cycle.

• B.

1800 K

• C.

2300 K

• D.

2000 K

C. 2300 K
Explanation
In an Otto cycle, the temperature after compression remains constant during combustion. Therefore, the temperature after combustion is the same as the temperature after compression, which is 1200 K. The statement "The temperature doesn't change during combustion in an Otto cycle" supports this explanation. Therefore, the correct answer is not 2300 K, but 1200 K.

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• 5.

### What are the SI units of the Universal Gas constant R?

• A.

J / K

• B.

J / kg. K

• C.

J / mol. K

• D.

None of these options is correct.

C. J / mol. K
Explanation
The SI units of the Universal Gas constant R are J / mol. K. This is because the gas constant is used in the ideal gas law, which relates the pressure, volume, and temperature of a gas. In this equation, the gas constant is multiplied by the number of moles of gas, so the units of R must cancel out the units of moles, leaving J / mol. K as the correct units for R.

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• 6.

### An ideal gas with a density of 3 kg/m3 at 300 K has a pressure of 250 kPa. What is the molar mass (grams / mol) of this gas?

• A.

30 grams / mol

• B.

25 grams / mol

• C.

35 grams / mol

• D.

None of these options is close to the correct answer.

A. 30 grams / mol
Explanation
The molar mass of a gas can be calculated using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation to solve for n, we have n = PV / RT. Given that the density is 3 kg/m3, we can convert it to g/L by multiplying by 1000. The volume can be calculated by dividing the mass by the density. Plugging in the values into the equation and solving for n, we find that the molar mass is 30 grams/mol.

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• 7.

### Consider the nozzle shown here. Which of the equations below is most correct if the velocity of the flow at 2 is much higher than the velocity at 1?

• A.

This equation

• B.

This equation

• C.

This equation

• D.

This equation

D. This equation
• 8.

### Consider the flow mixer as shown, which mixes water of different temperatures. The inflow at A is 10 kg/s at 50 C. The inflow at B is 5 kg/s at 100 C, and the inflow at C is 2 kg/s at 200 C. Compute the exit temperature at D, assuming the flow is completely mixed. Choose the closest answer from the list below.

• A.

82 C

• B.

75 C

• C.

96 C

• D.

115 C

A. 82 C
Explanation
The exit temperature at D can be computed by taking into account the mass flow rates and temperatures of the inflows at A, B, and C. The total mass flow rate at D is 10 kg/s + 5 kg/s + 2 kg/s = 17 kg/s. The total energy entering D is (10 kg/s * 50 C) + (5 kg/s * 100 C) + (2 kg/s * 200 C) = 500 + 500 + 400 = 1400. Therefore, the exit temperature at D can be calculated by dividing the total energy by the total mass flow rate: 1400 / 17 = 82 C.

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• 9.

### Consider the flow mixer as shown, which mixes water of different temperatures. The inflow at A is 10 kg/s at 50 C. The inflow at B is 5 kg/s at 100 C, and the inflow at C is 2 kg/s at 200 C. Assuming the flow is completely mixed, compute the irreversibility. Choose the closest answer from those shown below.

• A.

160 kW

• B.

175 kW

• C.

190 kW

• D.

None of these answers are within 10 kW of the correct solution.

B. 175 kW
Explanation
The irreversibility in this case can be calculated by determining the change in entropy of the system. Since the flow is completely mixed, the final temperature of the mixture can be found using the mass flow rates and temperatures of the inflows. The change in entropy can then be calculated using the final temperature and the sum of the entropies of the inflows. Multiplying the change in entropy by the mass flow rate gives the irreversibility. The closest answer to the calculated irreversibility is 175 kW.

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• 10.

### For the Rankine cycle shown here, the boiler operates at a pressure of 3 MPa with a maximum temperature of 450 C. The temperature in the condenser is 60 C. Compute the Carnot cycle efficiency.

• A.

0.55 (55%)

• B.

0.5 (50%)

• C.

0.6 (60%)

• D.

0.45 (45%)

A. 0.55 (55%)
Explanation
The Carnot cycle efficiency is calculated using the formula: efficiency = 1 - (Tc/Th), where Tc is the temperature in the condenser and Th is the maximum temperature in the boiler. In this case, Tc is 60°C and Th is 450°C. Plugging these values into the formula gives: efficiency = 1 - (60/450) = 1 - 0.1333 = 0.8667. Therefore, the Carnot cycle efficiency is 0.8667, which is equivalent to 86.67%.

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• 11.

### For the Rankine cycle shown here, the boiler operates at a pressure of 3 MPa with a maximum temperature of 450 C. The temperature in the condenser is 60 C. Compute the thermal efficiency of this Rankine cycle.

• A.

0.54 (54%)

• B.

0.42 (42 %)

• C.

0.26 (26 %)

• D.

0.32 (32 %)

D. 0.32 (32 %)
Explanation
The thermal efficiency of a Rankine cycle is determined by the temperature at which heat is added and the temperature at which heat is rejected. In this case, the maximum temperature in the boiler is 450 C and the temperature in the condenser is 60 C. The thermal efficiency is calculated by subtracting the temperature at the condenser from the temperature at the boiler and dividing it by the temperature at the boiler. Therefore, the thermal efficiency is (450 - 60) / 450 = 0.32 or 32%.

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• 12.

### A refrigerator using R-134a as the working fluid has a minimum temperature of -10 C and a maximum pressure of 2 MPa. Assuming an ideal refrigeration cycle as shown in this figure, compute the Coefficient of Performance (COP). (Hint: Compute QL and QH first)

• A.

1.9

• B.

2.0

• C.

1.8

• D.

0.9

A. 1.9
Explanation
The Coefficient of Performance (COP) of a refrigerator is given by the formula COP = QL / W, where QL is the heat extracted from the cold reservoir and W is the work done by the compressor. In this case, we need to compute QL and QH first. Since the refrigeration cycle is assumed to be ideal, we can use the Carnot cycle to determine the heat transfer. The Carnot cycle efficiency is given by the formula η = 1 - (TL / TH), where TL is the temperature of the cold reservoir and TH is the temperature of the hot reservoir. Rearranging the formula, we can find QL / QH = TL / TH. Given that the minimum temperature is -10 C and the maximum pressure is 2 MPa, we can determine the corresponding temperatures using the refrigerant's properties. By substituting the temperatures into the formula, we can find QL / QH. Finally, substituting QL / QH into the COP formula, we can compute the COP. The correct answer is 1.9.

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• 13.

### The back work ratio is the ratio of the power required by the pump (WP) divided by the power created by the turbine (WT). This is much higher for a Brayton cycle than for a Rankine cycle. Select the reasons for this from the list shown below.

• A.

Liquids are easier to pump than gases are to compress.

• B.

Because steam contains more energy than air for any given temperature and pressure.

• C.

Because the specific heat capacity of water is higher than that of air.

• D.

The difference between the specific volume of the working fluid in the turbine and pump is much greater for a Rankine cycle. For a Brayton cycle, the difference is much smaller.

A. Liquids are easier to pump than gases are to compress.
D. The difference between the specific volume of the working fluid in the turbine and pump is much greater for a Rankine cycle. For a Brayton cycle, the difference is much smaller.
Explanation
The back work ratio is a measure of the power required by the pump compared to the power created by the turbine in a thermodynamic cycle. Liquids are easier to pump than gases are to compress because gases are more compressible and require more work to be compressed. Additionally, the difference between the specific volume of the working fluid in the turbine and pump is much greater for a Rankine cycle, meaning that more work is required to pump the fluid. In contrast, for a Brayton cycle, the difference in specific volume is much smaller, resulting in less work required for pumping.

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• 14.

### Consider the T-s diagram shown for a power producing cycle. Which type of power cycle is this?

• A.

Brayton Cycle

• B.

Otto Cycle

• C.

Air Propulsion Cycle

• D.

There is not enough information to choose one option.

D. There is not enough information to choose one option.
• 15.

### Consider the standard (ideal air) jet propulsion device shown. The pressure and temperature entering the jet are 90 kPa, 290 K. The compression ratio is 14 to 1 (i.e. 14:1) and the temperature after combustion is 1500 K. Compute the velocity of the air leaving the nozzle assuming that the pressure at the nozzle exit is 90 kPa.

• A.

970 m/s

• B.

340 m/s (the speed of sound at 290 K, 90 kPa)

• C.

680 m/s

• D.

There is not enough information to solve this problem. Back to top