Saturday, February 15, 2020

H C Verma solutions, Laws of Thermodynamics, EXERCISES, Q-1 to Q-10, Chapter-26, Concepts of Physics, Part-II

Laws of Thermodynamics

EXERCISES, Q-1 to Q-10

   1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises 17°C. The mass of the vessel is 100 g and that of the water is 200 g and. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal expansion. (a) How much heat is transferred to the liquid-vessel system? (b) How much work has been done on this system? (c) How much is the increase in the internal energy of the system?      


Answer:  (a) Since the system is insulated, no heat is transferred to the water-vessel system. The increase in the temperature is due to the energy transfer by work-done on the water-vessel system, the heat transferred to the system is zero.


     (b) The work done on the system is equal to the increase in the heat energy of the system. 

Increase in heat energy of water 

=(0.20 kg)*(4200 J/kg-K)*(17° - 15°)

=1680 J

Increase in heat energy of vessel

=(0.10 kg)*(420 J/kg-K)*(17° - 15°)

=84 J

Total = 1680+84 J =1764 J. 


       (c) ΔU = ΔQ - ΔW, Since ΔQ = 0, 

ΔU = ΔW = 1764 J. 



 

   2. Figure (26-E1) shows a paddlewheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) How much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle. 
The figure for Q-2
 


Answer: (a) The container is adiabatic, so heat transfer is zero.


     (b) The work done on the liquid =  Decrease in the potential energy of the block

=mgh

=(12 kg)*(10 m/s²)*(0.70 m)

=84 J. 


          (c) The rise in temperature will depend on the mass of the liquid which is not mentioned here. Assuming the mass of the liquid = 1 kg. Specific heat capacity of the liquid, s = 4200 J/K. If the rise in temperature =ΔT, then

Q = ms*Δt

→84 = 1*4200*Δt

→Δt = 84/4200

→Δt = 0.02°C.  


 


  3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from the frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt. (c) Find the work done in this frame by the external force holding the belt.   


Answer:  (a) m = 100 kg. u = 2.0 m/s. µ = 0.2, v = 0.

Change in internal energy ΔU =ΔQ - ΔW
Here ΔQ = 0, So
ΔU = -ΔW =-{½mv² - ½mu²}
     = -½m{v²-u²}
     = -½*100*{0²-2²}
     = 200 J.
(b) The initial speed of the block with respect to the moving frame, u = 0,
The final speed of the block, v = 2 m/s.
The increase in the kinetic energy of the block = ½mv² -½mu²
=½*100*2² - 0
=200 J.

(c) Let the force of friction between the block and belt = F. Given, µ = 0.2,
So, F = µ*R =µ*mg
→F = 0.2*100*10 N =200 N.
Acceleration, a = F/m =200/100 m/s² 
=2 m/s².
      Let the distance traveled by the block = s. From v²-u² =2as,
2² -0² =2*2*s
→s = 4/4 =1 m.
So the work done by the external force in moving the block by 1 m = Force*distance= (200 N)*(1 m) =200 J.
But the external force not only moves the block by 1 m but also gives it kinetic energy. The kinetic energy gained by the block is also due to the work done by the external force. This is equal to ½mv² =½*100*2² =200 J.
       So the total work done by the external force =200 J +200 J =400 J.    

    


 

  4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.   


Answer:  Since the container is rigid, there will be no change in volume. So, ΔV = 0. And the work done by the system ΔW =p*ΔV =0.

Given that, ΔQ = 100 J. From the first law of thermodynamics,
ΔQ = ΔU + ΔW

→100 J = ΔU +0

→ΔU = 100 J.

So the change in the internal energy of the gas =100 J. 



 

  5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied and extracted from the gas, what is the change in the internal energy of the gas?    


Answer:  (a) Let us draw the p-V curve, 
Diagram for Q-5

The work done by the gas is the area under the p-V curve. It is shaded. Since the volume decreases, the work done by the gas is negative and equal to,

= -½(50 + 10)*(200-50) J
= -30*150 J
= -4500 J
= -4.5 kJ. 

(b) The change in internal energy,
ΔU =ΔQ - ΔW
Here ΔQ = 0 and ΔW = -4.5 kJ,
so, ΔU = 4.5 kJ. 

  6. An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas? 


Answer:  Since pV = nRT

→V = nR(T/p), 

Since n, R and p/T and hence T/p are constant, the volume also remains constant from process i to f. So ΔV =0.
The work done by the gas ΔW =p*ΔV =0 (zero). 


 

  7. Figure (26-E2) shows three paths through which a gas can be taken from state A to state B. Calculate the work done by the gas in each of the three paths. 
The figure for Q-7

Answer:  The work done by the gas can be calculated in a p-V diagram by the area bounded between the p-V curve and V axis between the two ordinates of volumes at the initial and final points.

In the process AB,

Work done by the gas

=½(10 kPa+30 kPa)*(25 cc -10 cc)

=20 kPa * 15 cc

=20000 N/m² * 15x10⁻⁶ m³

=0.30 J,


In the process ADB

Work done by the gas = Area under AD + Area under DB

= (10 kPa) * (25 cc -10 cc) + 0

=10000 N/m² * 15x10⁻⁶ m³

= 0.15 J.


In the process ACB

The work done by the gas =Area under AC +Area under CB
= 0 + (30 kPa)*(25 cc -10 cc)
= 30000 N/m² * 15x10⁻⁶ m³
=0.45 J.


 

  8. When a system is taken through the process abc shown in figure (26-E3), 80 J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process?   
The figure for Q-8


Answer:  Difference of internal energies (ΔU) between states at points a and c will be the same whatever the process is adopted.

     Work done in the process abc ΔW =30 J (Given) and heat given ΔQ =80 J.

Since ΔU =ΔQ -ΔW =80 -30 J =50 J.

In the second process, adc, ΔW = 10 J.

Now ΔU =ΔQ -ΔW

→50 =ΔQ - 10 

→ΔQ = 50+10 =60 J.

Hence 60 J of heat flows in this process. 




 

  9. 50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure (26-E4). Find the quantity of heat to be supplied to take it from A to B via ADB.   
The figure for Q-9


Answer:  For the process ACB,

ΔQ = 50 cal=50*4.2 J= 210 J, 

ΔW =area between the curve ACB and V-axis 

=(50 kPa)*(400-200)cc

=(50000 N/m²)*(200x10⁻⁶ m³)

=10 J,

Now ΔU =ΔQ -ΔW =210 -10 =200 J.

The change in the internal energy between the states A and B, ΔU, will be the same in both processes. For the process ADB, (from the figure)

ΔW =(155 kPa)*(400-200)cc

     =(155x1000 N/m²)*(200x10⁻⁶ m³)

     =31 J

From ΔU =ΔQ -ΔW

→200 J =ΔQ -31 J

→ΔQ =200+31 =231 J =(231/4.2) cal

        =55 cal.

So 55 cal of heat is needed to take the system from the state A to B through the process ADC.   


 



  10. Calculate the heat absorbed by a system in going through the cyclic process shown in figure (26-E5). 
The figure for Q-10
 


Answer:  Since it is a cyclic process, the initial and final state is the same, hence the change in the internal energies ΔU =0. 

Now ΔQ -ΔW =ΔU =0

→ΔQ = ΔW

So the heat absorbed is equal to the work done. In the full cycle sometimes work is done by the system and sometimes work is done on the system. So the work done by the system is sometimes positive and some times negative. Since the work done by the system is the area under the curve and V-axis, net work done by the system is equal to the area of the circle,

=πd²/4

=π(200)²*1000*10⁻⁶/4 J

=40π/4 J
=31.4 J = Heat absorbed.

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Links to the Chapters



CHAPTER- 20 - Dispersion and Spectra


CHAPTER- 19 - Optical Instruments

CHAPTER- 18 - Geometrical Optics



CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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Click here for "The Forces" - OBJECTIVE-I


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Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


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CHAPTER- 2 - "Physics and Mathematics"

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