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Thermodynamics - 2 - Mechanical Engineering MCQ


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20 Questions MCQ Test GATE Mechanical (ME) Mock Test Series 2025 - Thermodynamics - 2

Thermodynamics - 2 for Mechanical Engineering 2024 is part of GATE Mechanical (ME) Mock Test Series 2025 preparation. The Thermodynamics - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Thermodynamics - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics - 2 below.
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Thermodynamics - 2 - Question 1

A new temperature scale in degrees N is to be defined. The boiling and freezing points of water on this scale are 400° N and 100° N, respectively. What will be the reading on new scale corresponding to 60° C?

Detailed Solution for Thermodynamics - 2 - Question 1

Thermodynamics - 2 - Question 2

Measurement of temperature is based on which law of thermodynamics?

Detailed Solution for Thermodynamics - 2 - Question 2

Zeroth law of thermodynamics states that if a body A is in thermal equilibrium with body B and body C, then the body B and body C are also in thermal equilibrium with each other. This law is specifically applied in temperature measurements.
 

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Thermodynamics - 2 - Question 3

A steam turbine receives steam steadily at 10 bar with an enthalpy of 3000 kj/kg and discharges at 1 bar with an enthalpy of 2700 kj/kg. The work output is 250 kJ/kg. The changes in kinetic and potential energies are negligible. The heat transfer from the turbine casing to the surroundings is equal to

Detailed Solution for Thermodynamics - 2 - Question 3

Given that
h1= 3000 kJ/kg

h2 = 2700 kJ/kg

W = 350 kJ/kg
Using steady state energy equation, h1 = h+ W + Q

3000 = 2700 + 250 + Q

Q = 50 kJ

Thermodynamics - 2 - Question 4

A system undergoes a change of state during which 80 kJ of heat is transferred to it and it does 60 kj of work. The system is brought back to its original state through a process during which 100 kj of heat is transferred to it. The work done by the system is

Detailed Solution for Thermodynamics - 2 - Question 4

For  the first process

Thermodynamics - 2 - Question 5

The internal energy of a certain system is a function of temperature alone and is given by the formula E = 25 + 0.25t kJ
If this system executes a process for which the work done by it per degree temperature increase is

 the heat interaction per degree temperature increase in kJ is

Detailed Solution for Thermodynamics - 2 - Question 5

Given that

Thermodynamics - 2 - Question 6

The air with enthalpy of 100 kJ/kg is compressed by an air compressor to a pressure and temperature at which its enthalpy becomes 200 kJ/kg. The loss of heat is 40 kJ/kg from the compressor as the air passes through it. Neglecting the change in kinetic and potential energies, the power required for an air mass flow of 0.5 kg/s is

Detailed Solution for Thermodynamics - 2 - Question 6

For given condition of air compressor steady flow of the energy equation is

Thermodynamics - 2 - Question 7

The cyclic integral of (δQ−δW) for a process is

Detailed Solution for Thermodynamics - 2 - Question 7

The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.
When heat energy is supplied to a gas, two things may occur:
• The internal energy of the gas may change
• The gas may do some external work by expanding
According to the first law of Thermodynamics:
δQ = δW + ΔU
When a process is executed by a system, the change in stored energy of the system is numerically equal to the net heat interaction minus the net work interaction during the process:
ΔU = δQ−δW where U is the internal energy which is introduced by this law.
For the cyclic process: ΔU = 0
∮ δQ =  δW
∮ (dQ - dW) = 0
The first law of thermodynamics states that for a cyclic process, the cyclic integral of heat is equal to the cyclic integral of work.

Thermodynamics - 2 - Question 8

Suppose 0.70 kg/s of air enters the compressor with a specific enthalpy of 290 kJ/kg and leaves it with specific enthalpy 450 kJ/kg. Velocities at inlet and exit are 6 m/s and 2 m/s, respectively. Assuming adiabatic process, what is power input to the compressor?

Detailed Solution for Thermodynamics - 2 - Question 8

Thermodynamics - 2 - Question 9

In a throttling process, which one of the following parameters remain constant?

Detailed Solution for Thermodynamics - 2 - Question 9

The enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling. The drop in pressure is compensated by corresponding change in density or internal energy of the flowing fluid.
 

Thermodynamics - 2 - Question 10

A refrigerating machine working on reversed Carnot cycle takes out 2 kW per minute of heat from the system while between temperature limits of 300 K and 200 K. COP and power consumption of the cycle will be respectively

Detailed Solution for Thermodynamics - 2 - Question 10

Given that

Thermodynamics - 2 - Question 11

A reversible engine operates between temperatures T1 and T2. The energy rejected by this engine is received by a second reversible engine at temperature T2 and rejected to a reservoir at temperature T3. If the efficiencies of the engines are the same then the relationship between T1, T2 and T3 is given by

Detailed Solution for Thermodynamics - 2 - Question 11

Thermodynamics - 2 - Question 12

A Carnot engine operates between 327° C and 27° C. If the engine produces 300 kJ of work, what is the entiopy change during head addition?

Detailed Solution for Thermodynamics - 2 - Question 12

Given that

Thermodynamics - 2 - Question 13

The relation

ds = dQ/T

where s represents entropy, Q represents heat and T represents temperature (absolute), holds good in which one of the following processes?

Thermodynamics - 2 - Question 14

An inventor says that his new concept of an engine, while working between temperature limits of 27° C and 327° C rejects 45% of heat absorbed from the source. His engine is then equivalent to which one of the following engines?

Detailed Solution for Thermodynamics - 2 - Question 14

The efficiency of any reversible engine cannot be more than that of Carnot cycle. For the given case, the Carnot cycle efficiency is


Thermodynamics - 2 - Question 15

Three engines A, B and C operating on Carnot cycle use working substances as argon, oxygen and air, respectively. Which will have higher efficiency?

Detailed Solution for Thermodynamics - 2 - Question 15

The efficiency of reversible engine is independent of the working fluid

Thermodynamics - 2 - Question 16

A series of combination of two Carnot engines operate between the temperatures of 180° C and 20°C. If the engine produces equal amount of work, then what is the intermediate temperature?

Detailed Solution for Thermodynamics - 2 - Question 16

For equal amount of work, the intermediate temperature will be given as 

Thermodynamics - 2 - Question 17

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at 27°C, then what is the source temperature?

Detailed Solution for Thermodynamics - 2 - Question 17

Thermodynamics - 2 - Question 18

A reversible heat engine rejects 50% of the heat supplied during a cycle of operation. If this engine is reversed and operates as a heat pump, then what is its coefficient of performance?

Detailed Solution for Thermodynamics - 2 - Question 18

The efficiency of the given reversible heat engine is η = 0.5. The COP of the heat pump from this reversible cycle will be

Thermodynamics - 2 - Question 19

When the lower temperature is fixed, COP of a refrigerating machine can be improved by

Detailed Solution for Thermodynamics - 2 - Question 19

 The performance parameter for a refrigerator cycle, called the coefficient of performance (COP), is defined as the ratio of desired thermal effect and work input to the system. It is expressed in terms of lower and higher temperatures as

Hence, COP can be increased by reducing higher temparature

 

Thermodynamics - 2 - Question 20

The heat absorbed or rejected during a polytropic process is equal to 

Detailed Solution for Thermodynamics - 2 - Question 20

Heat transfer Q in a polytropic process can be represented in terms of work transfer W as

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