Thermodynamics Test

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Thermodynamics Test - 48

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Question 1
1 / -0

Three bodies $$A$$, $$B$$ and $$C$$ are in thermal equilibrium. The temperature of $$B$$ is $${45}^{o}C$$, then the temperature of $$C$$ is ____________

A
$${45}^{o}C$$

B
$${50}^{o}C$$

C
$${40}^{o}C$$

D
Any temperature

Solution

According to zeroth law of thermodynamics if two thermodynamic system are in thermal equilibrium with a third them they are in thermal equilibrium with each other.
Accordingly thermal equilibrium between system is a transitive relation. So, If $$A,B$$ and $$C$$ are in thermal equilibrium $$\&$$ temperature of $$ B$$ is $$45°C$$ then the $$ C$$ will have same $$45°C$$ temperature.
Hence, the answer is $$45°C.$$
Question 2
1 / -0

Which of the following is correct for any thermodynamic system?

A
The internal energy changes in all processes

B
Internal energy and entropy are state functions

C
The change in entropy can never be zero

D
Heat gained or lost in an adiabatic process is always zero

Solution

An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that all the change in internal energy is in the form of work done.
Hence option D is correct.
Question 3
1 / -0

Which is not true for Second Law of Thermodynamics?

A
The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process.

B
It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.

C
It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and cooling of a heat reservoir.

D
The second law of thermodynamics states that the total energy of an isolated system is constant.

Solution

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. The total entropy can remain constant in ideal cases where the system is in a steady state (equilibrium), or is undergoing a reversible process. In all spontaneous processes, the total entropy always increases and the process is irreversible. The increase in entropy accounts for the irreversibility of natural processes, and the asymmetry between future and past.

Historically, the second law was an empirical finding that was accepted as an axiom of thermodynamic theory. Statistical mechanics, classical or quantum, explains the microscopic origin of the law.
Hence option D is only incorrect option.
Question 4
1 / -0

The molar heat capacity in a process of diatomic gas if it does a work of $$\dfrac{Q}{4}$$ when a heat of Q is supplied to it is

A
$$\dfrac{2}{5}$$R

B
$$\dfrac{5}{4}$$R

C
$$\dfrac{10}{3}$$R

D
$$\dfrac{6}{4}$$R

Solution

First law of Thermodynamics we have :-

$$Q=w+ΔU$$ (1)

where,

Q = Heat supplied to the gas

w = Work done by the gas

ΔU = Change in the internal energy of the gas

Given :-

$$w=Q/4$$ (2)

From (1) & (2) we have :-

⟹$$ΔU=3Q/4$$ (3)

Now,

$$ΔU=\dfrac{nRΔT}{γ−1}$$ (4)

where,

n = No. of moles of gas

ΔT = Change in temperature

R = Universal Gas Constant

γ = Specific heat ratio

For a diatomic gas :-

$$γ=7/5$$ ( 5)

Using (3), (4) & (5) we get :-

⟹$$nΔT=\dfrac{3Q}{10R}$$ (6)

Also note that :-

$$Q=nCΔT$$ (7)

where,

C = Molar specific heat for the process

So now replace nΔT in (7) by (6) :-

⟹$$\dfrac{3QC}{10R}=Q$$

⟹$$C=\dfrac{10R}{3}$$
Question 5
1 / -0

A carnot cycle is having maximum efficiency because

A
it comprises of two adiabatic process which requires no heat in its execution.

B
it comprises of two isothermal process.

C
its every process is reversible

D
none of the above

Solution

Carnot cycle consists of 4 process. All of them are ideal and practically not possible.

1.Isothermal heat addition- Nothing can be more efficient than this heating since no finite temperature difference exists between heat source and receiver. Hence it is a reversible process and most efficient,

2.Isentropic(reversible adiabatic) expansion- Most efficient expansion, as no heat loss is taking place, full energy is utilised only for expansion and thereby doing work.

3.Isothermal heat rejection- Most efficient heat rejection to a heat sink at same temperature.

4.Isentropic compression- Compressed(without friction) with no heat lost to surrounding. Power input is used only for increasing the pressure and temperature.

In this cycle every process is reversible and hence it is most efficent.
When the cycle is completed input energy and power is used only for the purpose it was intended. Output was completely used for intended purpose
Question 6
1 / -0

In a cyclic heat engine operating between a source temperature of $$600^0C$$ and a sink temperature of $$20^0 C$$, the least rate of heat rejection per kW net output of the engine is,

A
0.505kW

B
0.490kW

C
0.333kW

D
none of the above

Solution

$$\therefore Heat\quad absorption=\dfrac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } $$

$$=\dfrac { 873-293 }{ 873 } =0.664$$

$$\therefore Heat\quad Rejected=1-heat\quad absorbed$$
$$=1-0.664$$
$$\therefore Heat\quad rejected=0.335$$
Hence the heat of rejection per kW is $$0.335$$
Question 7
1 / -0

A series combination of two Carnots engines operate between the temperatures of $$180^0C$$ and $$20^0C$$. If the engines produce equal amount of work,then what is the intermediate temperature(In $$^0C$$)?

A
80

B
90

C
100

D
110

Solution

A series combination of two Carnot engines operate between the temperatures of $$180^0C$$and $$20^0C$$.If the engines produce equal amount of work
The intermediate temperature in series combination is given by
$$T_i=\dfrac{T_1+T_2}{2}=\dfrac{180+20}{2}=100^oC$$
Question 8
1 / -0

A reversible engine operates between temperatures 900 K & $$T_2$$($$T_2$$ < 900 K), & another reversible engine between $$T_2$$ & 400 K ($$T_2$$ > 400 K) in series. What is the value of $$T_2$$ if work outputs of both the engines are equal?

A
600K

B
625K

C
650K

D
675K

Solution

Output work $$W = \Delta T$$
Thus for equal work output, temperature difference should be equal.
$$900-T_2=T_2-400$$
Or $$2T_2 = 1300$$
$$\implies$$ $$T_2=650 \ K$$
Question 9
1 / -0

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at $$27^0C$$, then what is the source temperature (in $$^0C$$)?

A
477

B
346

C
564

D
none of the above

Solution

$$\eta =\dfrac { 60 }{ 100 } $$

$${ T }_{ 2 }={ 27 }^{ o }C$$
$${ T }_{ 2 }={ 300 }^{ o }K$$
$${ T }_{ 1 }=?$$

$$\eta =\dfrac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } $$

Here, $${ T }_{ 1 }=Source\quad temperature$$
$${ T }_{ 2 }=sink\quad temperature$$
$$\eta =Efficiency$$
$$\eta =100-rejecion$$

$$\therefore \dfrac { 60 }{ 100 } =\dfrac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } $$

$$\therefore 0.6=\dfrac { { T }_{ 1 }-300 }{ { T }_{ 1 } } $$

$$\therefore { T }_{ 1 }\left( 1-0.6 \right) =300$$

$${ T }_{ 1 }=\dfrac { 300 }{ 0.4 } =\dfrac { 3000 }{ 4 } ={ 750 }^{ o }K$$

$$\therefore { T }_{ 1 }={ \left( 750-273 \right) }^{ o }C$$
$$\therefore { T }_{ 1 }={ 477 }^{ o }C$$
Question 10
1 / -0

One reversible heat engine operates between 1600 K and $$T_2$$ K, and another reversible heat engine operates between $$T_2$$ K and 400 K. If both the engines have the same heat input and output, then the temperature $$T_2$$ must be equal to:

A
$$600$$

B
$$800$$

C
$$650$$

D
$$675$$

Solution

$${ T }_{ 1 }=1600K$$

$${ T }_{ 2 }={ T }_{ 2 }K$$

$${ T }_{ 1 }^{ 1 }={ T }_{ 2 }K$$

$${ T }_{ 2 }^{ 1 }=400K$$

Same input and same output.Then the efficiency is same.
$$\eta =\dfrac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } $$

$$\therefore \dfrac { 1600-{ T }_{ 2 } }{ 1600 } =\dfrac { { T }_{ 2 }-400 }{ { T }_{ 2 } } $$

$$=1600{ T }_{ 2 }-{ T }_{ 2 }^{ 2 }=1600{ T }_{ 2 }-640000$$

$$\therefore { T }_{ 2 }^{ 2 }=640000$$
$$\therefore { T }_{ 2 }=800K$$

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Thermodynamics Test - 48

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Thermodynamics Test - 48

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Question 1

$${45}^{o}C$$

$${50}^{o}C$$

$${40}^{o}C$$

Any temperature

Solution

Question 2

The internal energy changes in all processes

Internal energy and entropy are state functions

The change in entropy can never be zero

Heat gained or lost in an adiabatic process is always zero

Solution

Question 3

The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process.

It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.

It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and cooling of a heat reservoir.

The second law of thermodynamics states that the total energy of an isolated system is constant.

Solution

Question 4

$$\dfrac{2}{5}$$R

$$\dfrac{5}{4}$$R

$$\dfrac{10}{3}$$R

$$\dfrac{6}{4}$$R

Solution

Question 5

it comprises of two adiabatic process which requires no heat in its execution.

it comprises of two isothermal process.

its every process is reversible

none of the above

Solution

Question 6

0.505kW

0.490kW

0.333kW

none of the above

Solution

Question 7

80

90

100

110

Solution

Question 8

600K

625K

650K

675K

Solution

Question 9

477

346

564

none of the above

Solution

Question 10

$$600$$

$$800$$

$$650$$

$$675$$

Solution

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