Monday, January 20, 2020

H C Verma solutions, Calorimetry, EXERCISES, Q-11 to Q-18, Chapter-25, Concepts of Physics, Part-II

CALORIMETRY

EXERCISES, Q11 to Q18

   11. A 50 kg man is running at a speed of 18 km/h. If all the kinetic energy of the man can be used to increase the temperature of water from 20°C to 30°C, how much water can be heated with this energy?     


Answer:  Mass of the man, M = 50 kg, Speed of the man, v = 18 km/h =18000/3600 m/s = 5 m/s. Hence

The kinetic energy of the man, E =½Mv².

=½*50*5² J =625 J.   

Now, this energy is used to heat the water from 20°C to 30°C. If the mass of water is m, then

625 = m*4200*(30-20)

{specific heat capacity of water =4200 J/kg-K}

→m =625/42000 kg =625/42 g ≈15 g. 




 

    12. A brick weighing 4.0 kg is dropped into a 1.0 m deep river from a height of 2.0 m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find this thermal energy in calories.     


Answer:  Mass of the brick, m =4.0 kg,

Total height from the bed of the river, h =2.0 + 1.0 = 3.0 m. Taking g = 10 m/s². The gravitational potential energy of the brick = mgh

=4.0*10*3.0 =120 J.

Conversion in thermal energy =80% of 120 J

=0.8*120 J 

=96 J 

=96/4.186 cal 

=23 cal.  




 

   13. A van of mass 1500 kg traveling at a speed of 54 km/h is stopped in 10 s. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy in cal/s.     


Answer:  Mass of the van, m =1500 kg,

Speed of the van, v =54 km/h 

=54000/3600 m/s

=15 m/s

The kinetic energy, E =½mv²

=½*1500*15² J

=168750 J

This energy is produced in 10 s, hence the rate of production of thermal energy

=16875 J/s

=16875/4.186 cal/s

≈4000 cal/s.  




 

   14. A block of mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10 m/s to 5 m/s, find the thermal energy developed in the process.     


Answer:  Mass of the block, m =100 g =0.10 kg, Initial speed, v = 10 m/s, final speed, v' =5 m/s. 

The thermal energy developed = loss in kinetic energy =½m(v²-v'²)

=½*0.10*(10²-5²) J

=½*0.10*75 J

=7.5/2 J

=3.75 J.  



 

   15. Two blocks of masses 10 kg and 20 kg moving at speeds of 10 m/s and 20 m/s respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.     


Answer:  Mass of 1st block, m =10 kg, the mass of the 2nd block, m' =20 kg, speed of the first block, v =10 m/s, speed of the second block, v' =20 m/s. let the speed of the combined block after the collision = V. From the conservation of linear momentum,

m'v'-mv=(m+m')V

→20*20 - 10*10 =(10+20)V

→30V = 300

→V =10 m/s.

The total initial kinetic energy of the system =½mv²+½m'v'²

=½*10*10²+½*20*20² J

=500+4000 J

=4500 J.   

The final kinetic energy of the system

=½(m+m')V²

=½*30*10² J

=30*50 J

=1500 J.

The thermal energy developed in the process = loss in kinetic energy

=Initial kinetic energy-final kinetic energy

=4500 - 1500 J

=3000 J.

 



 

   16. A ball is dropped on a floor from a height of 2.0 m. After the collision, it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into a ball. Calculate the rise in the temperature of the ball in the collision. The heat capacity of the ball is 800 J/K.     


Answer:  Let the mass of the ball = m,

Initial height, h = 2.0 m, final height after the collision, h' =1.5 m. hence the potential energy lost =mg(h-h')

=mg(2.0-1.5)

=0.5mg

Thermal energy in the ball 

=40% of 0.5mg J

=0.40*0.5mg J

=0.20mg J

If the rise in temperature of the ball =T, then,
msT =0.20mg

{Given that heat capacity of the ball =800 J/K.

Since the mass of the ball is not given, we assume that the given heat capacity is actually specific heat capacity, s =800 J/kg-K. }

→T = 0.20g/s
→T =0.20*10/800 =0.0025°C 
→T =2.5x10⁻³°C.

    


 

   17. A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides through 60 cm. Specific heat capacity of copper =420 J/kg-K.     


Answer:  Mass of the copper cube, m =200 g =0.20 kg. Length of the slide, L =60 cm =0.60 m. Since the speed is constant, there is no loss of kinetic energy. The only loss of mechanical energy is in the potential energy of the cube, which is converted into heat energy due to the force of friction on the surface.
Diagram for Q-17

      Drop in vertical height in sliding through 60 cm, =L*sin37°

=0.60*(3/5) m

=0.36 m

Loss in mechanical energy =mgh

=0.20*10*0.36 J

=0.72 J

This energy goes into the cube as thermal energy. If the rise in temperature =T. Then,

msT = 0.72, (given s=420 J/kg-K)

→T =0.72/(420*0.20)

→T =0.0086 °C =8.6x10⁻³°C.

      


 

   18. A metal block of density 6000 kg/m³ and mass 1.2 kg is suspended through spring of spring constant 200N/m. The spring block system is dipped in water kept in a vessel. The water has a mass of 260 g and the block is at a height 40 cm above the bottom of the vessel. If the support to the spring is broken, what will be the rise in the temperature of the water? The specific heat capacity of the block is 250 J/kg-K and that of water is 4200 J/kg-K. Heat capacities of the vessel and the spring are negligible. 


Answer:  Mass of the block, m =1.2 kg, density = 6000 kg/m³. The volume of the block, v = 1.2/6000 m³ 

=2x10⁻⁴ m³.

The apparent weight of the block in the water =(m-ρv)g

=(1.2-1000*2x10⁻⁴)10 N

=10 N

Extension of the spring, x =W/k 

=10/200 m

=0.05 m

Energy stored in the spring =½kx²

=½*200*0.05² J

=0.25 J

When the support breaks, this stored energy is released and the potential energy of the block is lost in a fall of 40 cm.

Loss of P.E. =(m-ρv)g*h =10*0.40 J

=4 J.

Total energy lost by the system =0.25+4 J

=4.25 J

If the rise in the temperature of block and water is T, then

msT+m's'T =4.25 

→1.2*250T + 0.26*4200T =4.25

{m'=0.26 kg, s' =4200 J/kg-K, s=250 J/kg-K}
→1392 T =4.25
→T =4.25/1392 =0.003°C.
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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

Click here for → Questions for Short Answer 

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)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

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

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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