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Class 12-science H C VERMA Solutions Physics Chapter 3 - Calorimetry

Calorimetry Exercise 47

Solution 1

Heat gain by vessel + water =Heat loss by iron block



Solution 2

Heat gained by water + Calorimeter = Heat loss by iron piece

240 × 4200 × (60 - 20) + 10 × 4200 × (60 - 20) = 100 × 470 × (T-60)


Solution 3

Let specific heat capacity of different liquids A, B and C be   respectively.

A and B are mixed

Heat gain by A= Heat loss by B



B and C are mixed

Heat gain by B=Heat loss by C



A and C are mixed, Let final temp be T

Heat gain by A= Heat loss by C




Solution 4

Mass of four ice cubes=   



Mass of drink=  


=0.2 kg

Heat required to melt ice=  


=9792 J

Heat released by drink from 10°C to 0°C =  


=8400 J

Heat required > Heat released

So ice will not melt completely.

Hence equilibrium temperature=0°C 

(b) Amount of ice melted by 8400 J



m=0.025 kg

m 25 gm

Solution 5

Rate of water coming out= 0.2gm/sec

Let time taken to cool water by 5°C be t.

Mass of water coming out=0.2 t (in gm)


Heat loss by water to cool= Heat taken to evaporate



t=462.5 sec


t=7.7 min

Solution 6

Volume of ice melted=Volume of cube

Heat loss by iron cube=Heat taken by ice




Solution 7

Heat released by steam to get converted into water=msteamL𝑣=(1)(2.26×106)


Heat required by ice to melt=   



Heat required by water to raise temperature from 0°C to 100°C 




Total heat absorbed by ice to become water at 100°C 





Excess heat released=  


Let mass converted into steam by this excess heat be m.



  gm (steam)

Mass of water at 100°C ≈ (2000-625) 

 1.335 kg (water)

Solution 8



t=2625 sec

 = 44 min

Solution 9

Heat released by water = Energy required to lift mass




h=315 Km

Solution 10

Change in internal energy= Energy loss by bullet



=16 J

Solution 11

KE of man = heat taken by water




 =15 gm

Solution 12

Thermal energy= 80% of potential energy change



=96 J


Solution 13

Rate of production of thermal energy=  



=16857 J/sec


Solution 14

Thermal energy developed = Loss in KE



=3.75 J

Solution 15

If both blocks are taken as system,




 10 m/s



= 4500 J



=1500 J

Thermal energy developed=   


=3000 J

Solution 16

Thermal energy = 40% mechanical energy lost




Solution 17

Since, block moves with constant velocity; a=0


Thermal energy gain=Mechanical Energy lost





Solution 18



Initially, spring-block is at rest andbalanced. Let stretch in spring be x.




x=0.05 m

Now, when support is broken, mechanical energy of spring-block system is transferred to both block and water due to which temperature rises by T.





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