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Test: First Law of Thermodynamics - 2 - Mechanical Engineering MCQ


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15 Questions MCQ Test - Test: First Law of Thermodynamics - 2

Test: First Law of Thermodynamics - 2 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: First Law of Thermodynamics - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: First Law of Thermodynamics - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: First Law of Thermodynamics - 2 below.
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Test: First Law of Thermodynamics - 2 - Question 1

According to first law of thermodynamics

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 1

Since first law of thermodynamics defined as law of conservation of energy hence total energy of a system remains constant.

Test: First Law of Thermodynamics - 2 - Question 2

Internal energy is defined by

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 2

Zeroth law of thermodynamics — concept of temperature
First law of thermodynamics — concept of internal energy
Second law of thermodynamics — concept of entropy.

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Test: First Law of Thermodynamics - 2 - Question 3

Key concept in analyzing the filling of an evacuated tank is

Test: First Law of Thermodynamics - 2 - Question 4

First law of thermodynamics is valid for

Test: First Law of Thermodynamics - 2 - Question 5

During a thermodynamic process, 84 kJ of heat flows into the system and the work done by the system is 32 kJ. The increase in internal energy of the system is

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 5

From first law of thermodynamics
δQ = dU+δW
84 = dU +32
dU = 52kJ

Test: First Law of Thermodynamics - 2 - Question 6

The specific heat at constant pressure for an ideal gas is given by 
cp = 0.9 + (2.7 x 10-4) T (kJ/kgK)
Where T is in kelvin. The change in enthalpy for this ideal gas undergoing a process in which the temperature changes from 27°C to 127°C is most nearly.

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 6

Test: First Law of Thermodynamics - 2 - Question 7

A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are +20 kJ and +50 kJ respectively. If the system is returned to state 1 and Q2-1 is - 10 kJ what is the value of work W2-1

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 7

Test: First Law of Thermodynamics - 2 - Question 8

Given that along the path 1-2-3 a system absorbs 100 kJ as heat and does 60 kJ work while along the path 1 -4-3 it does 20 kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 8

Test: First Law of Thermodynamics - 2 - Question 9

The network output for the cycle 1-2-3-4 -5-6-1 shown in figure is

Test: First Law of Thermodynamics - 2 - Question 10

In a reversible isothermal expansion process fluid expands from 10 bar and 2 m3 to 2 bar and 10 m3. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 10

Note that in case of reversible isothermal expansion change in internal energy is zero hence
δQ = δW= 100 kW

Test: First Law of Thermodynamics - 2 - Question 11

The state of an ideal gas is changed from (T1, P1) to (T2, P2) in a constant volume process. To calculate the change in enthalpy (Δh) ail of the following properties/variables are required.

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 11

Calculation of Change in Enthalpy (Δh) in Constant Volume Process

- Properties/Variables Required:
- Cp (specific heat at constant pressure)
- Initial temperature (T1)
- Final temperature (T2)

- In a constant volume process, the enthalpy change (Δh) is equal to the heat added to the system. The formula for calculating the change in enthalpy is Δh = Cp * (T2 - T1).

- Cp is the specific heat at constant pressure, which represents the amount of heat required to raise the temperature of one unit mass of a substance by one degree Celsius at constant pressure.

- We need the initial temperature (T1) and final temperature (T2) to calculate the change in enthalpy, as the enthalpy change is directly proportional to the temperature change.

- Therefore, the properties/variables required to calculate the change in enthalpy (Δh) in a constant volume process are Cp, T1, and T2.

Test: First Law of Thermodynamics - 2 - Question 12

For the two paths as shown in the figure, one reversible and one irreversible, to change the state of the system from a to b.

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 12

ΔU is a point function and it is independent of the path followed.

Test: First Law of Thermodynamics - 2 - Question 13

The net work done for the closed system shown in the given pressure-volume diagram is

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 13

Correct Answer :- B

Explanation : AC = 4 m3.

ABC = 2, ACD = 3

Work done = Area of ABC + Area of ACD

          = 1/2 * 2 * 4 + 1/2 * 3 * 4 

          = 1000 kN-m.

Test: First Law of Thermodynamics - 2 - Question 14

Two ideal heat engine cycles are represented in the given figure. Assume VQ = QR, PQ = QS and UP = PR= RT. If the work interaction for the rectangular cycle (WVRU) is 48 Nm, then the work interaction for the other cycle PST is

Test: First Law of Thermodynamics - 2 - Question 15

A mixture of gases expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is

Detailed Solution for Test: First Law of Thermodynamics - 2 - Question 15

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