Friday, January 17, 2020

H C Verma solutions, Calorimetry, EXERCISES, Q-1 to Q-10, Chapter-25, Concepts of Physics, Part-II

CALORIMETRY

EXERCISES, Q1 to Q10

   1. An aluminum vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminum, iron, and water are 910 J/kg-K, 470 J/kg-K, and 4200 kg-K respectively.    


Answer:  Mass of aluminum vessel, m = 0.5 kg.

Mass of water, m' =0.2 kg,

Mass of iron, m" =0.2 kg.

Specific heat capacity (SHC) of aluminum, s = 910 J/kg-K.

SHC of water, s' = 4200 J/kg-K,

SHC of iron, s" = 470 J/kg-K.

The initial temperature of vessel and water, T' = 20°C =273+20 =293 K.

The initial temperature of Iron, T" =100°C =373 K.

Assume the equilibrium temperature = T.

Heat loss by the iron rod =m" s" (373-T)
=0.2*470*(373-T)
=94(373-T)
Heat gained by the vessel and water
=ms(T-293)+m's'(T-293)
=(T-293)(ms+m's')
=(T-293)(0.5*910+0.2*4200)
=1295(T-293)
In calorimetry, the heat lost = heat gained, so
94(373-T) = 1295(T-293)
→373-T =13.78(T-293)
→373-T =13.78T -13.78*293
→14.78T =373+4037 =4410
→T = 4410/14.78 =298 K
Hence the equilibrium temperature is 298-273 =25°C.

  

 


   2. A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J/kg-°C.    




Answer:  Mass of iron, m = 100 g =0.1 kg. SHC of iron, s =470 J/kg-°C. The temperature of the furnace and the iron piece = T (say), Equilibrium temperature = 60°C. So the heat lost by the iron piece,

=0.1*470*(T-60) =47(T-60).

Water equivalent of calorimeter = 10 g and water in the calorimeter = 240 g. So equivalent mass of both of the two, m' =10+240 =250 g =0.25 kg.

The initial temperature of these two, T' =20°C. Hence the heat gained by water and calorimeter =0.25*4200*(60-20)

=42000 J

Since heat loss = Heat gained

47(T-60) =42000

→T-60 =894

→T = 894+60 =954°C. 




   3. The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16 °C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?    


Answer:  Let the mass of each liquid = m. Given that,

The temperature of A =12°C 

The temperature of B = 19°C

The temperature of C = 28°C,

The temperature of the mixture of A and B = 16°C,

The temperature of the mixture of B and C =23°C.

The temperature of the mixture of A and C = T =? 

If the specific heat capacity of A, B and C are s, s' and s" respectively, then from the principle of calorimetry i.e. heat lost = heat gained, for the mixture A and B,

ms(16-12) = ms'(19-16)

→4s = 3s'

→s' = 4s/3

For the mixture B and C,

ms'(23-19) = ms"(28-23)

→4s' = 5s"

→s' = 5s"/4

→4s/3 = 5s"/4
→s" = 16s/15

For the mixture A and C,

s(T-12) = s"(28-T)

→s(T-12) = (16s/15)(28-T)
→(T-12)*15 = 16(28-T)
→15T -180 = 16*28-16T
→31T =180+448 =628
→T = 628/31 ≈ 20.3°C

  

   4. Four 2 cm x 2 cm x 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. The density of ice = 900 kg/m³, the density of drink = 1000 kg /m³, the specific heat capacity of the drink = 4200 J/kg-K, latent heat of fusion of ice = 3.4x10⁵ J/kg.     


Answer:  Mass of four ice cubes =4*0.02*0.02*0.02*900 kg

=0.0288 kg. 

Mass of the drink =200x10⁻⁶*1000 kg

=0.2 kg  

The temperature of the ice =0°C

The temperature of the drink =10°C
Latent heat of fusion of ice, L =3.4x10⁵ J/kg.
Let the equilibrium temperature = T


(a) The heat required by the ice in completely fusing =0.0288*3.4x10⁵ =9790 J {Its temperature is still 0°C}

Heat available to the drink in coming up to 0°C =0.2*4200*(10-0) =8400 J.
So since heat available is less than the required for the fusion of ice, hence all the ice will not melt. And at the equilibrium, the temperature of the ice drink mixture will be 0°C.


(b) The mass of ice melted can use only 8400 J of heat available to the drink in coming up to the temperature of 0°C. Hence this mass = 8400/3.4x10⁵ kg
=0.025 kg
=25 g. 


  


    5. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5°C. Specific heat capacity of water = 4200 J/kg-°C and latent heat of vaporization of water = 2.27x10⁶ J/kg. 


Answer:  Let the required time be t seconds. Mass of water coming out in t seconds = 0.2t/1000 kg =2x10⁻⁴t kg. The heat required to vaporize this mass of water =2x10⁻⁴t*2.27x10⁶ J

=4.54*100t J =454t J. 

This heat is taken from the water itself. 

The mass of water =10 kg

The heat released by this mass by coming down by 5°C

=10*4200*5 J

=210000 J 

equating these two,

454t = 210000 

→t =210000/454 =462.6 s =7.7 min.

    

 


   6. A cube of iron (density = 8000 kg/m³, specific heat capacity = 470 J/kg-K) is heated to a high temperature and is placed on a large block of ice at 0°C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice =900 kg/m³ and the latent heat of fusion of ice = 3.36x10⁵ J/kg.  


Answer:  Let the volume of iron cube =V. The volume of ice displaced is also V. Hence the mass of iron cube =8000V kg and the mass of ice displaced = 900V. Since the ice block is large the final temperature of the iron cube =0°C. If the initial temperature of the iron cube = T, the heat lost by the cube =8000V*470*T J

Heat gained by the displaced ice mass in fusion =900V*3.36x10⁵ J.  

Since the heat lost = heat gained,

8000V*470*T = 900V*3.36x10⁵

→T = 9*3.36*1000/8*47

→T ≈ 80°C.    




   7. 1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36x10⁵ J/kg and latent heat of vaporization of water = 2.26x10⁶ J/kg.  


Answer:  Since the latent heat of vaporization of water is much more than the latent heat of fusion of ice, the ice will melt completely. Heat available in steam till condensation at 100°C = 2.26x10⁶ J. 

The heat required by ice in melting at 0°C =3.36x10⁵ J. Heat required by this melted ice now water to reach a temperature of 100°C = 1*4200*100 J =4.2x10⁵ J. 

Hence total heat required by 1 kg of ice at 0°C to convert to water at 100°C =(3.36+4.2)x10⁵ J =7.56x10⁵ J. 

This heat will be supplied by that mass of steam which will condense into the water at 100°C. Let the mass of this steam = m. So heat supplied by m mass of steam =m*2.26x10⁶. Equating these two,

m*2.26x10⁶ = 7.56x10⁵

→m = 0.335 kg = 335 g. 

So the equilibrium temperature will be 100°C and the mass of water in it = 1.0 kg+0.335 kg =1.335 kg. The remaining mass of the steam = 1000 - 335 =665 g.        





   8. Calculate the time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 4200 J/kg-K.    


Answer:  Amount of heat required to raise the temperature of 20 kg of water from 10°C to 35°C = 20*4200*(35-10) =2.1x10⁶ J.

The rating of the immersion heater =1000 W = 1000 J/s but only 80% is used to heat the water, hence it supplies 800 J/s to the water. Thus the time needed to supply the required amount of heat,

=2.1x10⁶/800 s

=2625 s

≈ 44 min.    





   9. On a winter day, the temperature of the tap water is 20°C whereas the room temperature is 5°C. Water is stored in a tank of capacity of 0.5 m³ for household use. If it were possible to use the heat liberated by the water to lift a 10 kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take g = 10 m/s².    


Answer:  The mass of water stored in the tank = 0.5*1000 kg =500 kg.

Heat energy liberated by 500 kg of water to come at room temperature of 5°C = 500*4200*(20-5) =3.15x10⁷ J.

The energy needed by 10 kg mass to lift it by h meter vertically =mgh =10*10*h =100 h.

Equating, 100 h =3.15x10⁷

→h = 3.15x10⁵ m =3.15x100 km

→h =315 km.

   



   10. A bullet of mass 20 g enters into a fixed wooden block with a speed of 40 m/s and stops in it. Find the change in internal energy during the process.    


Answer:  The mass of the bullet, m = 20 g =0.02 kg. Speed of the bullet, v = 40 m/s. Hence the kinetic energy of the bullet =½mv²

=½*0.02*40²

=16 J.

Since the bullet stops inside the block, its final kinetic energy = 0. Thus the whole of the kinetic energy of the bullet is changed into heat energy which changes the internal energy of the block. Thus the change in the internal energy = 16 J.  

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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

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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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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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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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