Q: The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

\${Q}_{1}={T}_{0}{S}_{0}+\cfrac{1}{2}{T}_{0}{S}_{0}=\cfrac{3}{2}{T}_{0}{S}_{0}\$\${Q}_{2}={T}_{0}(2{S}_{0}-{S}_{0})={T}_{0}{S}_{0}\$ and \${Q}_{3}=0\$\$\eta=\cfrac{W}{{Q}_{1}}=\cfrac{{Q}_{1}-{Q}_{2}}{{Q}_{1}}\$\$=1-\cfrac{{Q}_{2}}{{Q}_{1}}=1-\cfrac { { T }_{ 0 }{ S }_{ 0 } }{ \cfrac { 3 }{ 2 } { T }_{ 0 }{ S }_{ 0 } } =\cfrac { 1 }{ 3 } \$

Q: 5 mole of an ideal gas expands reversibly from a volume of \$8d\$ to \$80d\$ at a temperature of \$27\$. The change in entropy is:-(A) \$41.57{ JK }^{ -1 }\$(B) \$-95.73{ JK }^{ -1 }\$(C) \$95.73{ JK }^{ -1 }\$(D) \$-41.57{ JK }^{ -1 }\$

The correct option is CWe have,\$n=5,v_1=8dm^3, v_2=80dm^3, T=300k\$ConstantThus the change in entropy is:\$ \Delta s=nR \$ in \$\dfrac{v_2}{v_1}\$\$=5\times8.314\times \dfrac{80}{8}\$\$=95.73jk^{-1}\$

Q: Calculate the change in entropy when \$1\ g\$ of ice at \$0^oC\$ is heated to form water at \$40^oC\$

The change in entropy would be given as \$\Delta S=\Delta S_1+\Delta S_2\$ \$\Delta S_1\$ is to melt the ice and \$\Delta S_2\$ is to rise the temperature. Thus we get \$\Delta S=\displaystyle\dfrac{mL}{T}+m\int^{T_2}_{T_1}\dfrac{d T}{T}\$\$\Delta S=\displaystyle\dfrac{1\times 80}{273}+1\times\$\$\ln \dfrac{313}{273}=0.28+0.13=0.41\ cal/^oC\$

Q: The amount of heat energy required to convert \$1\, kg\$ of ice at \$-10^o C\$ to water at \$100^o C\$ is \$7,77,000\, J\$. Calculate the specific latent heat of ice. Specific heat capacity if ice = \$ 2100 J kg^{-1} K^{-1}\$, Specific heat capacity of water = \$ 4200 J kg^{-1} K^{-1}\$

Amount of heat energy gained by 1kg of ice at −10°C to raise its temperature to 0°C = 1 × 2100 × 10 = 21000 JAmount of heat energy gained by 1kg of ice at 0°C to convert into water at 0°C = LAmount of heat energy gained when temperature of 1kg of water at 0°C rises to 100°C = 1 × 4200 × 100 = 420000 JTotal amount of heat energy gained = 21000 + 420000 + L = 441000 + L.Given that total amount of heat gained is = 777000J. So,\$441000 + L = 777000.\$\$L = 777000 − 441000.\$\$L = 336000JKg^{-1}\$

Question 1

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

Accepted Answer

${Q}_{1}={T}_{0}{S}_{0}+\cfrac{1}{2}{T}_{0}{S}_{0}=\cfrac{3}{2}{T}_{0}{S}_{0}$${Q}_{2}={T}_{0}(2{S}_{0}-{S}_{0})={T}_{0}{S}_{0}$ and ${Q}_{3}=0$$\eta=\cfrac{W}{{Q}_{1}}=\cfrac{{Q}_{1}-{Q}_{2}}{{Q}_{1}}$$=1-\cfrac{{Q}_{2}}{{Q}_{1}}=1-\cfrac { { T }_{ 0 }{ S }_{ 0 } }{ \cfrac { 3 }{ 2 } { T }_{ 0 }{ S }_{ 0 } } =\cfrac { 1 }{ 3 } $

Question 2

5 mole of an ideal gas expands reversibly from a volume of $8 d$ to $80 d$ at a temperature of $27$ . The change in entropy is:-
(A) $41.57 J K^{- 1}$
(B) $- 95.73 J K^{- 1}$
(C) $95.73 J K^{- 1}$
(D) $- 41.57 J K^{- 1}$

Accepted Answer

The correct option is CWe have,$n=5,v_1=8dm^3, v_2=80dm^3, T=300k$ConstantThus the change in entropy is:$ \Delta s=nR $ in $\dfrac{v_2}{v_1}$$=5\times8.314\times \dfrac{80}{8}$$=95.73jk^{-1}$

Question 3

Define entropy. Give its units.

Accepted Answer

Entropy is a measure of randomness or disorder of the system.The greater the randomness, the higher the entropy.It is state function and extensive property.Its unit is $JK^{-1}mol^{-1}$.

Question 4

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

Accepted Answer

From Kelvin Plank statement and Claussius's statement we can say that heat can not be flow form lower temperature to higher temperature until and unless we are adding any external source like heat pump, which justifies ${SECOND \  LAW \  OF  \ THERMODYNAMICS}$

Question 5

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

Accepted Answer

Heat is a form of energy that gets transferred from one medium or object to another medium/object. Suppose if we a colder body and a hotter body and if it brought in contact, the heat gets transferred from the hotter body to colder body until the system of hotter & colder attains equilibrium.When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then The heat lost by the hotter body will be equal to the heat gained by the colder body.option D is correct

Question 6

Calculate the change in entropy when $1\ g$ of ice at $0^oC$ is heated to form water at $40^oC$

Accepted Answer

The change in entropy would be given as $\Delta S=\Delta S_1+\Delta S_2$ $\Delta S_1$ is to melt the ice and $\Delta S_2$ is to rise the temperature. Thus we get $\Delta S=\displaystyle\dfrac{mL}{T}+m\int^{T_2}_{T_1}\dfrac{d T}{T}$$\Delta S=\displaystyle\dfrac{1\times 80}{273}+1\times$$\ln \dfrac{313}{273}=0.28+0.13=0.41\  cal/^oC$

Question 7

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

Accepted Answer

The second law of Thermodynamic states that the total entropy of an isolated system increases over time, or it remains same in ideal cases (where the system underdoes a reversible process or when it is on a steady state). The first law of Thermodynamic tells the basic definition of internal energy and states the law of conservation of energy. The second law of Thermodynamic deals with the direction of natural process. it states that a natural process is not reversible. For e.g., heat flows hot to cold region, and reverse, unless some external work is done the system.In Terms of EntropyIn a reversible process, a very small increase in the entropy $(dS)$ of a system is defined to result from a small heat transfer to a closed divided by the common temperature $(T$ of the system and surrounding that supply of the heat:$dS=\dfrac {dQ}{T}$ (closed system)Various Statements of the Planck's Law:1. Statement of Kelvin: According to this statements it is not possible to make an engine in which the working substance can take heat from the source and change it completely into, work and remain unchanged itself in one complete cycle. It is concluded from this statement that it is not possible to get work continuously from heat source. Actually the sink in necessary and it necessary to release. Some heat to the sink. This statements is based on the principle of carnot heat engine.2. Statement of Clausius: According to this statements, it is an impossible process that in cyclic process the working substance transfer heat from low temperature object to high temperature object directly in the absence of external work.It is concluded from this statement that transfer of heat from a low temperature object high temperature object on its own is not possible. For this it is important that external work be done on the working substance. This statements is based on the principle of refrigerator.Equivalence of Clausius and Kelvin Statements:Let there be an engine violating the Kelvin statement: i.e., on that drains heat and converts it completely into work in a cyclic manner without any other results. Now, pairing it with a reverse Carnot engine.The newly created engine (consists of two engines) transfers heat $\Delta Q=Q\left(\dfrac {1}{\eta}-1\right)$ from the colder region (sink) to the hotter region (source), which violates the statements of Clausius. Thus, violation of Kelvin statement means violation of Clausius statement, i.e., Clausius statement implies the Kelvin statement. Similarly, we can prove that the Kelvin Statement implies the Clausius statement and hence both the statements are equivalent.

Question 8

The amount of heat energy required to convert $1\, kg$ of ice at  $-10^o C$ to water at $100^o C$ is $7,77,000\, J$. Calculate the specific latent heat of ice. Specific heat capacity if ice = $ 2100 J kg^{-1} K^{-1}$, Specific heat capacity of water = $ 4200 J kg^{-1} K^{-1}$

Accepted Answer

Amount of heat energy gained by 1kg of ice at &#8722;10&#176;C to raise its temperature to 0&#176;C = 1 &#215; 2100  &#215; 10 = 21000 JAmount of heat energy gained by 1kg of ice at 0&#176;C to convert into water at 0&#176;C = LAmount of heat energy gained when temperature of 1kg of water at 0&#176;C rises to 100&#176;C = 1 &#215;  4200 &#215; 100 = 420000 JTotal amount of heat energy gained = 21000 + 420000 + L = 441000 + L.Given that total amount of heat gained is = 777000J. So,$441000 + L = 777000.$$L = 777000 &#8722; 441000.$$L = 336000JKg^{-1}$

Question 9

Entropy is a measure of

Accepted Answer

Entropy is a measure of disorder and randomness

Question 10

All the naturally occurring processes proceed spontaneously in a direction which leads to

Accepted Answer

The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve toward thermodynamic equilibrium the state of maximum entropy.Therefore, all the naturally occurring processes proceed spontaneously in a direction which leads to increase in entropy of universe.

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Other statements of Second Law of Thermodynamics

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Second Law of Thermodynamics

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

5 mole of an ideal gas expands reversibly from a volume of $8 d$ to $80 d$ at a temperature of $27$ . The change in entropy is:-
(A) $41.57 J K^{- 1}$
(B) $- 95.73 J K^{- 1}$
(C) $95.73 J K^{- 1}$
(D) $- 41.57 J K^{- 1}$

Define entropy. Give its units.

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

Calculate the change in entropy when $1 g$ of ice at $0^{o} C$ is heated to form water at $4 0^{o} C$

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

The amount of heat energy required to convert $1 k g$ of ice at $- 1 0^{o} C$ to water at $10 0^{o} C$ is $7, 77, 000 J$ . Calculate the specific latent heat of ice. Specific heat capacity if ice = $2100 J k g^{- 1} K^{- 1}$ , Specific heat capacity of water = $4200 J k g^{- 1} K^{- 1}$

Entropy is a measure of

All the naturally occurring processes proceed spontaneously in a direction which leads to

Revise with Concepts

Other statements of Second Law of Thermodynamics

Learn with Videos

Quick Summary With Stories

Second Law of Thermodynamics

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

5 mole of an ideal gas expands reversibly from a volume of 8d to 80d at a temperature of 27. The change in entropy is:- (A) 41.57JK−1 (B) −95.73JK−1 (C) 95.73JK−1 (D) −41.57JK−1

Define entropy. Give its units.

Heat cannot by itself flow from a body at lower temperature to a body at higher temperature is a statement of consequence of

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at a higher temperature until they both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, then:

Calculate the change in entropy when 1 g of ice at 0oC is heated to form water at 40oC

Write the statements of Kelvin-Plank and Clausius of second law of Thermodynamics and slow the equivalence of the two statements.

The amount of heat energy required to convert 1kg of ice at −10oC to water at 100oC is 7,77,000J. Calculate the specific latent heat of ice. Specific heat capacity if ice = 2100Jkg−1K−1, Specific heat capacity of water = 4200Jkg−1K−1

Entropy is a measure of

All the naturally occurring processes proceed spontaneously in a direction which leads to

5 mole of an ideal gas expands reversibly from a volume of $8 d$ to $80 d$ at a temperature of $27$ . The change in entropy is:-
(A) $41.57 J K^{- 1}$
(B) $- 95.73 J K^{- 1}$
(C) $95.73 J K^{- 1}$
(D) $- 41.57 J K^{- 1}$

Calculate the change in entropy when $1 g$ of ice at $0^{o} C$ is heated to form water at $4 0^{o} C$

The amount of heat energy required to convert $1 k g$ of ice at $- 1 0^{o} C$ to water at $10 0^{o} C$ is $7, 77, 000 J$ . Calculate the specific latent heat of ice. Specific heat capacity if ice = $2100 J k g^{- 1} K^{- 1}$ , Specific heat capacity of water = $4200 J k g^{- 1} K^{- 1}$