Tuesday, September 1, 2026

H C Verma solutions, Calorimetry, Chapter-25, Concepts of Physics, Part-II

CALORIMETRY

QUESTIONS FOR SHORT ANSWER


   1. Is heat a conserved quantity?  


Answer:  Since heat is a form of energy, it is conserved. If different bodies at different temperatures are put in closed surroundings in contact, heat is exchanged among them till all are at the same temperature. But the total amount of heat before contact is the same as at the final temperature.       



   2. The calorie is defined as 1 cal = 4.186 joules. Why not as 1 cal = 4 J to make the conversions easy?   


Answer:  Both one calorie of heat and 4.186 joules of mechanical work raise the temperature of 1 g of water by 1°C. Therefore, 1 cal is defined as equal to 4.186 Joules; it is not arbitrary.    



   3. A calorimeter is kept in a wooden box to insulate it thermally from the surroundings. Why is it necessary?  


Answer:  The principle of calorimetry states that the total heat given by the hot objects equals the total heat received by the cold objects. Therefore, the calorimeter is kept in a wooden box to insulate it thermally from the surroundings so that the heat of hot bodies is not lost to the surroundings.     



   4. In a calorimeter, the heat given by the hot object is assumed to be equal to the heat taken by the cold object. Does it mean that the heat of the two objects taken together remains constant?  


Answer:  Yes, heat is a form of energy, so it is conserved. The heat of the two objects taken together remains constant.   



   5. In Renault's apparatus for measuring the specific heat capacity of a solid, there is an inlet and an outlet in the steam chamber. The inlet is near the top, and the outlet is near the bottom. Why is it better than the opposite choice where the inlet is near the bottom, and the outlet is near the top?  


Answer:  The steam, when it enters the chamber, starts losing heat and goes down. That is why the inlet of steam is kept on the upper side to make the solid evenly hot. The opposite arrangement is not suitable because it will not heat the solid correctly.     



   6. When a solid melts or a liquid boils, the temperature does not increase even when the heat is supplied. Where does the energy go?  


Answer:  The heat supplied is used to make the bonds between the molecules weaker so that the molecules move more freely than in the solid state in a melting solid and than in the liquid state in a boiling liquid. It is the reason the temperature does not increase even when the heat is supplied.   



   7. What is the specific heat capacity of (a) melting ice (b) boiling water?  


Answer:  The specific heat capacity is defined as the amount of heat needed to raise the temperature of the unit mass of a substance by 1°C or 1K. Since the temperature of melting ice or the boiling water does not increase, whatever amount of heat is supplied, both of them will have theoretically infinite specific heat capacity.    



   8. A person's skin is more severely burnt when put in contact with 1 g of steam at 100°C than when put in contact with 1 g of water at 100°C. Explain.  


Answer:  Due to the latent heat of vaporization, 1 g of steam at 100°C has greater energy than 1 g of water at 100°C. Therefore, a person's skin is more severely burnt when put in contact with 1 g of steam at 100°C than water at 100°C.   



   9. The atmospheric temperature in the cities on the sea coast changes very little. Explain.  


Answer:  The specific heat capacity of water is much higher than that of land. It means the seawater warms or cools very slowly compared to the land. When the coastal land warms, the air in contact rises up, and the air over the sea (also called the sea breeze) rushes towards the land to fill the space. Thus the temperature above the land does not increase much. Similarly, when the land cools quickly due to the low temperature, the cool land air (land breeze) rushes towards the sea, and the warmer sea air fills the gap, traveling over the land breeze. So the temperature near the sea-coast changes very little. 



   10. Should a thermometer bulb have large heat capacity or small heat capacity?  


Answer:  The thermometer bulb should have a small heat capacity. Otherwise, it will take more heat and pass it on to the mercury. In calorimetry, we do not take into account the heat transferred to the thermometer. The large heat capacity of the bulb will make the calculations faulty.    



OBJECTIVE-I


   1. The specific heat capacity of a body depends on

(a) the heat given
(b) the temperature raised
(c) the mass of the body
(d) the material of the body.


Answer:  (d)   


Explanation: Specific heat capacity is a property of the material due to which the same masses of different materials need different amounts of heat to get the same temperature difference. Hence, the option (d). 


 

   2. Water equivalent of a body is measured in

(a) kg

(b) calorie

(c) kelvin

(d) m³ 


Answer:  (a)   

Explanation: The mass of water that has the same heat capacity as the given body is called the water equivalent of that body. So this mass is measured in kg. Option (a). 

 

   3. When a hot liquid is mixed with a cold liquid, the temperature of the mixture

(a) first decreases then becomes constant
(b) first increases then becomes constant
(c) continuously increases
(d) is undefined for some time and then becomes nearly constant


Answer:  (d)   


Explanation: When hot and cold liquids are mixed, the molecules of both liquids take some time to spread evenly and to transfer their energies. During this time, the temperature is undefined. After this process is complete, the temperature is nearly constant. Option (d). 



   4. Which of the following pairs represent units of the same physical quantity? 

(a) kelvin and joule

(b) kelvin and calorie

(c) newton and calorie

(d) joule and calorie.  


Answer:  (d)   


Explanation: Joule and calorie both are units of energy. Traditionally, heat was measured in calories. When it was established that heat is also a form of energy and mechanical work can also produce heat and raise the temperature, the relationship between calorie and joule was established through experiments. 1 Calorie = 4.186 joules. Hence option (d). 



   5. Which of the following pairs of physical quantities may be represented in the same unit?

(a) heat and temperature

(b) temperature and mole

(c) heat and work

(d) specific heat and heat.  


Answer:  (c)   


Explanation: Heat is a form of energy, so both can be represented in the same unit. But energy is the capacity to do work, and both are represented in the same unit. Hence, heat and work can be represented in the same unit. Option (c). 

 

   6. Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved?

(a) the sum of the temperatures of the two bodies 
(b) total heat of the two bodies
(c) total internal energy of the two bodies
(d) the internal energy of each body. 


Answer:  (c)   


Explanation: When two bodies at different temperatures are mixed in a calorimeter, energy flows from the body at a higher temperature to the body at a lower temperature. This energy in transit is called heat. Once both bodies reach the same temperature and the energy transfer is complete, it becomes the internal energy of the receiving body. Since in a calorimeter heat neither can enter nor can escape, the total internal energy of the two bodies remains constant. Option (c).

     

   7. The mechanical equivalent of heat

(a) has the same dimension as heat

(b) has the same dimension as work

(c) has the same dimension as energy

(d) is dimensionless. 


Answer:  (d)   


Explanation: If mechanical work W produces the same temperature Change as heat H, it is written

W = JH, where J is called the mechanical equivalent of heat.

So, J = W/H.

Since W and H have the same dimensions, J will be dimensionless. The option (d).     


OBJECTIVE-II


   1. The heat capacity of a body depends on

(a) the heat given

(b) the temperature raised

(c) the mass of the body

(d) the material of the body.   


Answer:  (c), (d).  


Explanation:  The heat capacity of a body is the amount of heat required by it to raise the temperature by a unit degree. But this amount of heat is proportional to the mass and specific heat of the material of the body. Hence options (c) and (d).



     2. The ratio of specific heat capacity to the molar heat capacity of a body

(a) is a universal constant

(b) depends on the mass of the body

(c) depends on the molecular weight of the body

(d) is dimensionless.   


Answer:  (c)  


Explanation: Specific heat capacity, s=Q/mT, and

Molecular heat capacity, C =Q/nT

where Q = heat energy, m = mass, T = temperature difference, n = number of moles. Hence,

s/C = n/m =n/nM =1/M

M = molecular weight. So the ratio depends on the molecular weight of the body. The option (c).  



     3. If heat is supplied to a solid, its temperature

(a) must increase

(b) may increase

(c) may remain constant

(d) may decrease.    


Answer:  (b), (c).  


Explanation: If heat is supplied to a solid, then depending upon the conditions, its temperature may increase or remain constant. If the solid is at its melting point, then the temperature will remain constant if heat is supplied; otherwise, its temperature will rise. Its temperature will never decrease if heat is supplied. Hence options (b) and (c).  



     4. The temperature of a solid object is observed to be constant during a period. In this period 

(a) heat may have been supplied to the body

(b) heat may have been extracted from the body

(c) no heat is supplied to the body

(d) no heat is extracted from the body.    


Answer:  (a), (b).  


Explanation: When heat is supplied to a solid at its melting point, the temperature does not increase till all the solid is melted. This heat is used in changing the state of the solid and is called "Latent heat". Hence option (a).

   Similarly, midway during the melting of the solid, if heat is extracted from it, the surrounding liquid starts freezing, and the temperature of the solid does not fall till all the liquid is frozen. Hence, the option (b).

     There is another possibility for these two options to be true. If heat is supplied to the solid and by some arrangement, this heat is totally used to do work by the solid, then the temperature will not change. Similarly, if heat is extracted from the solid and simultaneously the same amount of work is done on it, then also the temperature will remain constant.

       Since no condition is given about the solid, only that it is observed for a period, there is a possibility that no heat is supplied or extracted during this period. In my opinion, options (c) and (d) are also true.     



     5. The temperature of an object is observed to rise in a period. In this period

(a) heat is certainly supplied to it.

(b) heat is certainly not supplied to it

(c) heat may have been supplied to it

(d) work may have been done on it.   


Answer:  (c), (d).  


Explanation: The temperature of an object may rise either due to the heat supplied or due to work being done on it. Hence, the options (c) and (d).



     6. Heat and work are equivalent. This means

(a) when we supply heat to a body, we do work on it.

(b) when we do work on a body, we supply heat to it

(c) the temperature of a body can be increased by doing work on it

(d) a body kept at rest may be set into motion along a line by supplying heat to it.   


Answer:  (c)  


Explanation: Supplying heat to a body does not mean we do work on it; nor can a body at rest be set in motion by supplying heat alone. The options (a) and (d) are not true. Also, if we do work on a body, its temperature may increase, just like supplying heat {option (c) is correct, and means equivalence of heat and work}, but it does not mean that we supply heat to it. Option (b) is also incorrect.  


EXERCISES


   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 J/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 the vessel and water, T' = 20°C =273+20 =293 K.

The initial temperature of the 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 capacities 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 the heat available is less than the required for the fusion of ice, 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 to come 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 the iron cube =V. The volume of ice displaced is also V. Hence the mass of the 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 a 10 kg mass to lift it by h meters 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. 




   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 to 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

= 0.20 mg J

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

{Given that the 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 a 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 of 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 the 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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Part-II

Solutions - "Concepts of Physics" Part-II, by H C Verma

CHAPTER- 23 - Heat and Temperature

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34


CHAPTER- 24 - Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30


EXERCISES - Q-31 to Q-40

EXERCISES - Q-41 to Q-50

EXERCISES - Q-51 to Q-62

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