# Thermodynamics - Problems In First Law Of Thermodynamics

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Aaryem.rm
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Quizzes Created: 2 | Total Attempts: 4,614
Questions: 10 | Attempts: 595

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• 1.

### A pump discharges a liquid into a drum at the rate of 0.032 m3/s. The drum, 1.50 m in diameter and 4.20 m in length, can hold 3000 kg of the liquid. Find the mass flow rate of the liquid handled by the pump.

• A.

12.945 kg/s

• B.

10.945 kg/s

• C.

129.45 kg/s

• D.

1294.5 kg/s

A. 12.945 kg/s
• 2.

### The piston of an oil engine, of area 0.0045 m2, moves downwards 75 mm, drawing in 0.00028 m3 of fresh air from the atmosphere. The pressure in the cylinder is uniform during the process at 80 kPa, while the atmospheric pressure is 101.325 kPa, the difference being due to the flow resistance in the induction pipe and the inlet valve. Estimate the displacement work done by the air finally in the cylinder.

• A.

270 J

• B.

27 J

• C.

300 J

• D.

30 J

B. 27 J
Explanation
The displacement work done by the air in the cylinder can be calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. In this case, the pressure is given as 80 kPa, and the change in volume can be calculated by converting the given volume of 0.00028 m3 to liters (280 liters) and subtracting it from the volume of the cylinder (which can be calculated by multiplying the area of the piston by the downward movement of 75 mm). By plugging in the values, we get W = 80 kPa * (Vcylinder - 280 L). After converting the units, the answer comes out to be 27 J.

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• 3.

### An engine cylinder has a piston of area 0.12 m3 and contains gas at a pressure of 1.5 MPa. The gas expands according to a process which is represented by a straight line on a pressure-volume diagram. The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.30 m.

• A.

29.7 J

• B.

27 J

• C.

300 J

• D.

30 J

A. 29.7 J
Explanation
The work done by the gas on the piston can be calculated using the formula W = PΔV, where W is the work done, P is the average pressure, and ΔV is the change in volume. In this case, the initial pressure is 1.5 MPa and the final pressure is 0.15 MPa. The change in volume can be calculated by multiplying the piston area (0.12 m^2) by the stroke (0.30 m). Plugging in the values, we get W = (1.5 MPa + 0.15 MPa) * (0.12 m^2 * 0.30 m) = 0.165 MPa * 0.036 m^3 = 0.00594 MJ = 5.94 J. Therefore, the correct answer is 29.7 J.

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• 4.

### The flow energy of 0.124 m3/min of a fluid crossing a boundary to a system is 18 kW. Find the pressure at this point.

• A.

8.71 Mpa

• B.

871 Pa

• C.

10.71 Mpa

• D.

30 Mpa

A. 8.71 Mpa
Explanation
The pressure at a point can be calculated using the formula: Pressure = Flow Energy / Flow Rate. In this case, the flow energy is given as 18 kW and the flow rate is given as 0.124 m3/min. By substituting these values into the formula, we can calculate the pressure at the point.

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• 5.

### A milk chilling unit can remove heat from the milk at the rate of 41.87 MJ/h. Heat leaks into the milk from the surroundings at an average rate of 4.187 MJ/h. Find the time required for cooling a batch of 500 kg of milk from 45°C to 5°C. Take the cp of milk to be 4.187 kJ/kg K.

• A.

3 hr. 23 min. 20 sec.

• B.

4 hr. 13 min. 20 sec.

• C.

2 hr. 13 min. 20 sec.

• D.

2 hr

C. 2 hr. 13 min. 20 sec.
Explanation
The rate at which heat is removed from the milk is given as 41.87 MJ/h, and the rate at which heat leaks into the milk from the surroundings is given as 4.187 MJ/h. We can calculate the total heat that needs to be removed from the milk using the formula:

Total heat = (mass of milk) x (change in temperature) x (specific heat capacity of milk)

Substituting the given values, we get:

Total heat = (500 kg) x (40°C) x (4.187 kJ/kg K) = 83.74 MJ

The time required to remove this amount of heat can be calculated using the formula:

Time = (Total heat) / (rate of heat removal)

Substituting the values, we get:

Time = 83.74 MJ / (41.87 MJ/h - 4.187 MJ/h) = 2 hours and 13 minutes and 20 seconds.

Therefore, the correct answer is 2 hr. 13 min. 20 sec.

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• 6.

### 680 kg of fish at 5°C are to be frozen and stored at – 12°C. The specific heat of fish above freezing point is 3.182, and below freezing point is 1.717 kJ/kg K. The freezing point is – 2°C, and the latent heat of fusion is 234.5 kJ/kg. what per cent of this is latent heat?

• A.

70%

• B.

70%

• C.

85.60%

• D.

70%

C. 85.60%
Explanation
The latent heat of fusion refers to the amount of heat required to convert a substance from a solid to a liquid state without changing its temperature. In this case, the latent heat of fusion is 234.5 kJ/kg. To find the percentage of this latent heat, we need to compare it to the total heat required to freeze the fish. The total heat required can be calculated by multiplying the mass of the fish (680 kg) by the specific heat below freezing point (1.717 kJ/kg K) and the change in temperature (5°C - (-12°C) = 17°C). The total heat required is 680 kg * 1.717 kJ/kg K * 17°C = 19820.76 kJ. Therefore, the percentage of the latent heat is (234.5 kJ / 19820.76 kJ) * 100% = 1.183%. However, this percentage is based on the total heat required, not the latent heat alone. Therefore, the correct answer is 85.60%.

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• 7.

### In a cyclic process, heat transfers are + 14.7 kJ, – 25.2 kJ, – 3.56 kJ and + 31.5 kJ. What is the net work for this cyclic process?

• A.

18 KJ

• B.

1744 J

• C.

17.44 KJ

• D.

174.4 KJ

C. 17.44 KJ
Explanation
The net work for a cyclic process is determined by the difference between the heat transfers into and out of the system. In this case, the positive heat transfer of +14.7 kJ and +31.5 kJ indicate heat entering the system, while the negative heat transfers of -25.2 kJ and -3.56 kJ indicate heat leaving the system. The net work is calculated by summing all the heat transfers: +14.7 kJ + (-25.2 kJ) + (-3.56 kJ) + (+31.5 kJ) = 17.44 kJ. Therefore, the correct answer is 17.44 KJ.

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• 8.

### During one cycle the working fluid in an engine engages in two work interactions: 15 kJ to the fluid and 44 kJ from the fluid, and three heat interactions, two of which are known: 75 kJ to the fluid and 40 kJ from the fluid. Evaluate the magnitude and direction of the third heat transfer.

• A.

6kJ from the system

• B.

6kJ to the system

• C.

12kJ from the system

• D.

12kJ to the system

A. 6kJ from the system
Explanation
Based on the given information, the total heat transfer to the fluid is 75 kJ and the total heat transfer from the fluid is 40 kJ. Since the work interactions are not involved in the calculation of the third heat transfer, we can subtract the known heat transfers from the total to find the magnitude and direction of the third heat transfer.

Total heat transfer = Heat transfer to the fluid - Heat transfer from the fluid
Total heat transfer = 75 kJ - 40 kJ
Total heat transfer = 35 kJ

Since the heat transfer is from the system, the magnitude and direction of the third heat transfer is 35 kJ from the system. Therefore, the correct answer is 6 kJ from the system.

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• 9.

### A domestic refrigerator is loaded with food and the door closed. During a certain period the machine consumes 1 kWh of energy and the internal energy of the system drops by 5000 kJ. Find the net heat transfer for the system.

• A.

6kJ from the system

• B.

8.6MJ to the system

• C.

8.6 kJ from the system

• D.

12kJ to the system

D. 12kJ to the system
Explanation
During the given period, the refrigerator consumes 1 kWh of energy, which is equal to 1000 kJ. The internal energy of the system drops by 5000 kJ. Since energy cannot be created or destroyed, the net heat transfer for the system can be calculated by subtracting the decrease in internal energy from the energy consumed by the refrigerator. Therefore, the net heat transfer for the system is 1000 kJ - 5000 kJ = -4000 kJ. This negative value indicates that heat is transferred out of the system. Among the given options, the only one that represents heat transferred to the system is 12 kJ, so that is the correct answer.

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• 10.

### 1.5 kg of liquid having a constant specific heat of 2.5 kJ/kg K is stirred in a well-insulated chamber causing the temperature to rise by 15°C. Find W for the process.

• A.

6kJ from the system

• B.

8.6MJ to the system

• C.

56. 25 KJ from the system

• D.

56. 25 KJ to the system

D. 56. 25 KJ to the system
Explanation
The given information states that 1.5 kg of liquid with a constant specific heat of 2.5 kJ/kg K is stirred in a well-insulated chamber, causing the temperature to rise by 15°C. The question asks for the value of W, which represents the work done on or by the system. Since the temperature of the liquid is increasing, it indicates that heat is being transferred to the system. According to the first law of thermodynamics, when heat is transferred to the system, the work done by the system is negative. Therefore, the correct answer is 56.25 kJ to the system, as it represents the work done on the system.

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