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Physics Case study (4 Marks)

Thermodynamics · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 5 questions.

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Case study (4 Marks)

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5 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Read the passage given below and answer the following questions from (i) to (iv). Zeroth Law of Thermodynamics states that two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other. The Zeroth Law clearly suggests that when two systems A and B, are in thermal equilibrium, there must be a physical quantity that has the same value for both. This thermodynamic variable whose value is equal for two systems in thermal equilibrium is called temperature (T). Thus, if A and B are separately in equilibrium with C, TA = TC and TB = TC. This implies that TA = TB i.e. the systems A and B are also in thermal equilibrium. Zeroth Law of Thermodynamics leads to the concept of internal energy of a system. We know that every bulk system consists of a large number of molecules. Internal energy is simply the sum of the kinetic energies and potential energies of these molecules. A certain amount of heat is supplied to the system’ or ‘a certain amount of work was done by the system its energy changes.
  1. Three thermodynamic systems are at temperature of 500 c .what can we say about them?
  1. Heat flows between them
  2. It obeys Zeroth Law of Thermodynamics
  3. Temperature of one system will increase and temperature of remaining two will decrease
  4. None of these
  1. Zeroth law of thermodynamics helped in the creation of which scale?
  1. Temperature
  2. Heat energy
  3. Pressure
  4. Internal energy
  1. State Zeroth Law of Thermodynamics:
  2. Define Internal energy of system:
Answer
  1. (b) It obeys Zeroth Law of Thermodynamics.
  2. (a) Temperature.
  3. Zeroth Law of Thermodynamics states that two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other. i.e. when two systems A and B, are in thermal equilibrium individually with system C then these two systems are also in thermal equilibrium with each other.
  4. Internal energy is the sum of the kinetic energies and potential energies of all the molecules possesses by system.
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Question 24 Marks
Read the passage given below and answer the following questions from 1 to 5. First Law of Thermodynamics The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that the energy supplied to the system goes in partly to increase the internal energy of the system and the rest in work on the environment. Mathematically, $\triangle\text{Q}=\triangle\text{U}+\triangle\text{W}$ where $\triangle\text{Q}$ is the heat supplied to the system, $\triangle\text{W}$ is the work done by the system and $\triangle\text{U}$ is the change in internal energy of the system. $\triangle\text{Q}$ and $\triangle\text{W}$ depend on the path taken to go from initial to final states, but the combination $\triangle\text{Q}​​-​​\triangle\text{W}$ is path independent.
  1. The first law of thermodynamics is concerned with conservation of:
  1. Number of molecules
  2. Number of moles
  3. Energy
  4. Temperature
  1. Which of the following is not a path function?
  1. $\triangle\text{Q}$
  2. $\triangle\text{Q}​​+\triangle\text{W}$
  3. $\triangle\text{W}$
  4. $\triangle\text{Q}​​-​​\triangle\text{W}$
  1. An electric heater supplies heat to a system at a rate of 120W. If system performs work at a rate of $80 J s^{-1}$, the rate of increase in internal energy is:
  1. $30 J s^{-1}$
  2. $40 J s^{-1}$
  3. $50 J s^{-1}$
  4. $60 J s^{-1}$
  1. Asystem goes from A to B by two different paths in the P - V diagram as shown in figure. Heat given to the system in path 1 is 1100 J, the work done by the system along path 1 is more than path 2 by 150 J. The heat exchanged by the system in path 2 is:
  1. 800 J
  2. 750 J
  3. 1050 J
  4. 950 J
  1. A certain mass of gas is carried from A to B, along three paths via ACB, ADB and AEB. Which one of the following is correct?
  1. Work done via path ACB is minimum.
  2. Work done via path ADB is maximum.
  3. Work done via path ACB is maximum.
  4. Work done via path AEB is maximum.
Answer
  1. (c) Energy
Explanation:
The first law of thermodynamics is concerned with conservation of energy.
  1. (d) $\triangle\text{Q}​​-​​\triangle\text{W}$
Explanation:
$\triangle\text{U}=\triangle\text{Q}-\triangle\text{W}$
The internal energy is independent of path.
  1. (b) $40 J s^{-1}$
Explanation:
According to first Law of thermodynamics
$\triangle\text{U}=\triangle\text{Q}+\triangle\text{W}$
$\therefore\frac{\triangle\text{Q}}{\triangle\text{t}}=120\text{W},$ $\frac{\triangle\text{W}}{\triangle\text{t}}=80\text{Js}^{-1}$
$\therefore\frac{\triangle\text{U}}{\triangle\text{t}}=120-80=40\text{Js}^{-1}$
  1. (d) 950 J
Explanation:
The change in internal energy of system will be same for both paths 1 and 2.
Along path 1, $\triangle\text{Q}_1=\triangle\text{U}+\triangle\text{W}_1\ ...(\text{i})$
Along path 2, $\triangle\text{Q}_2=\triangle\text{U}+\triangle\text{W}_2\ ...(\text{ii})$
Subtract (ii) from (i), we get

or $\triangle\text{Q}_1-\triangle\text{Q}_2=\triangle\text{W}_1-\triangle\text{W}_2$
$1100-\triangle\text{Q}_2=150$
$\triangle\text{Q}_2=1100-150=950\text{J}$
  1. Work done via path ACB is maximum.
Explanation:
Work done by a gas depends on the area enclosed between the P - V curve and the volume axis.
The area enclosed by curve ACB is maximum. Hence work done is maximum along path ACB.
The area enclosed by curve AEB is minimum.
Hence work done is minimum along path AEB.
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Question 34 Marks
Read the passage given below and answer the following questions from (i) to (v) Carnot principles are only for the cyclical devices like heat engines, which state that the efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.
  1. In a Carnot cycle, the working medium rejects heat at a ________ temperature.
    1. Higher
    2. Lower
    3. constant
    4. none of these
  2. Which of the following is NOT a state variable?
    1. work
    2. internal energy.
    3. entropy
    4. all of the above
  3. The efficiency of reversible heat engine is:
    1. $1 +(T_2/T_1)$
    2. $(T_1/T_2)+1$
    3. $(T_1 /T_2)- 1$
    4. $1 - (T_2 / T_1)$
  4. Other factors remaining constant, if the temperature of the source is increased, the efficiency of the Carnot engine will:
    1. decrease
    2. increase
    3. constant
    4. increase or decrease depending upon temperature ratio
  5. Over the complete Carnot cycle, entropy:
    1. increase
    2. decrease
    3. constant
    4. first increase and then decrease
Answer
  1. (b) lower
  2. (a) Work
  3. (d) $1- (T_2/ T_1)$
  4. (b) increase
  5. (c) constant
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Question 44 Marks
Read the passage given below and answer the following questions from (i) to (v). Kelvin-Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work. Clausius statement: No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. It can be proved that the two statements above are completely equivalent. A thermodynamic process is reversible if the process can be turned back such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. a reversible process is an idealized motion. A process is reversible only if it is quasi-static (system in equilibrium with the surroundings at every stage) and there are no dissipative effects. For example, a quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless movable piston is a reversible process. The free expansion of a gas is irreversible. The combustion reaction of a mixture of petrol and air ignited by a spark cannot be reversed. Cooking gas leaking from a gas cylinder in the kitchen diffuses to the entire room. The diffusion process will not spontaneously reverse and bring the gas back to the cylinder. The stirring of a liquid in thermal contact with a reservoir will convert the work done into heat, increasing the internal energy of the reservoir. The process cannot be reversed exactly; otherwise it would amount to conversion of heat entirely into work, violating the Second Law of Thermodynamics. Irreversibility is a rule rather an exception in nature.
  1. The diffusion process is:
  1. Reversible process
  2. Irreversible process
  1. A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless movable piston is
  1. Reversible process
  2. Irreversible process
  1. State Kelvin Planck statement.
  2. State Clausius statement.
  3. Define reversible processes and irreversible processes of thermodynamics.
Answer
  1. (b) Irreversible process
  2. (a) Reversible process
  3. Kelvin-Planck statement states that We cannot construct any device like the heat engine that operates on a cycle, absorbs the heat energy, and completely transforms this energy into an equal amount of work. Some of the heat gets released into the atmosphere. Practically no device bears 100% thermal efficiency.
  4. According to clausius It is nearly impossible for heat to move by itself from a temperature that is lower in temperature to a reservoir that is at a higher temperature. That is we can say that the transfer of heat can only occur spontaneously from high temperature to temperature. i.e No process is possible whose sole result is the transfer of heat from a colder object to a hotter object without any external work provided to do it in short we cannot construct a refrigerator that can operate without any input work.
  5. A thermodynamic process is said to be reversible if both the system and the surroundings return to their original states, with no other change anywhere else in the universe. On the other hand an irreversible process can be defined as a process in which the system and surrounding will not return to their original condition once the process is initiated.
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Question 54 Marks
Read the passage given below and answer the following questions from (i) to (v). Heat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work. It consists of a working substance-the system. For example, mixture of fuel vapors and air in a gasoline or diesel engine or steam in a steam engine are the working Substances. The working substance goes through a cycle consisting of several processes. In some of these processes, it absorbs a total amount of heat Q1 from an external reservoir at some high temperature T1. In some other processes of the cycle, the working substance releases a total amount of heat $Q_2$ to an external reservoir at some lower temperature T2. The work done (W) by the system in a cycle is transferred to the environment via some arrangement (e.g. the working substance may be in a cylinder with a moving piston that transfers mechanical energy to the wheels of a vehicle via a shaft). The basic features of a heat engine are schematically represented in Fig.

The cycle is repeated again and again to get useful work for some purpose. The discipline of Thermodynamics has its roots in the study of heat engines. A basic question relates to the efficiency of a heat engine. The efficiency ( h ) of a heat engine is defined by $n =\frac{ W }{ Q _1}$ Where Q 1 is the heat input i.e., the heat absorbed by the system in one complete cycle and $W$ is the work done on the environment in a cycle. In a cycle, a certain amount of heat $\left(Q_2\right)$ may also be rejected to the environment. Then according to the First Law of Thermodynamics, over one complete cycle. $W = Q _1-$ $Q_2 n =1-\frac{Q_2}{Q_1}$ For $Q_2=0, n=1$, i.e., the engine will have $100 \%$ efficiency in converting heat into work. Note that the First Law of Thermodynamics i.e., the energy conservation law does not rule out such an engine. But experience shows that such an ideal engine with $\eta=1$ is never possible. A refrigerator is the reverse of a heat engine. Here the working substance extracts heat $Q _2$ from the cold reservoir at temperature $T 2$, some external work $W$ is done on it and heat Q1 is released to the hot reservoir at temperature T1. The efficiency of refrigerator is expressed in terms of coefficient of performance $(\alpha)$ of a refrigerator is given by $\alpha=\frac{ Q _2}{W}$ where $Q _2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system
  1. In a heat engine the process need not be cyclic. True or False?
  1. True
  2. False
  1. Efficiency of heat engine N = 100% is it practically possible?
  1. Yes
  2. No
  1. Define efficiency of heat engine.
  2. Define coefficient of performance.
  3. Write a note on heat engine.
Answer
  1. (b) False
  2. (b) No
  3. Efficiency of heat engine is defined as ratio of work done by the engine to the heat supplied to the engine. Mathematically, $\text{n}=\frac{\text{W}}{\text{Q}_1}$ or $\text{n}=1-\frac{\text{Q}_2}{\text{Q}_1}$ where, $Q_2$ is heat rejected to sink.
  4. The efficiency of refrigerator is expressed in terms of coefficient of performance (α) of a refrigerator is defined as ratio of desired output to the work supplied to the refrigerator and given by$\alpha=\frac{\text{Q}_2}{\text{W}}$ Where $Q_2$ is the heat extracted from the cold reservoir and W is the work done on the System.
  5. Heat engine is a device in which system undergo a cyclic process that results in conversion of heat to work. It consists of, mixture of fuel vapors and air in a gasoline or diesel engine or steam in a steam engine as the working Substances. The working substance goes through a cycle consisting of several processes.
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