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
(0.5)(910)(T-20)+(0.2)(4200)(T-20)=(0.2)(470)(100-T)
T=25°C
Solution 2
Heat gained by water + Calorimeter = Heat loss by iron piece
240 × 4200 × (60 - 20) + 10 × 4200 × (60 - 20) = 100 × 470 × (T-60)
T=950°C
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
T=20.3°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 =
=(0.2)(4200)(10)
=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
Q=ml
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)
So,
Heat loss by water to cool= Heat taken to evaporate
t=462.5 sec
min
t=7.7 min
Solution 6
Volume of ice melted=Volume of cube
Heat loss by iron cube=Heat taken by ice
T=353K
Solution 7
Heat released by steam to get converted into water=msteamL𝑣=(1)(2.26×106)
J
Heat required by ice to melt= 
J
Heat required by water to raise temperature from 0°C to 100°C
J
Total heat absorbed by ice to become water at 100°C
J
Since,
Excess heat released=
J
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=315000m
h=315 Km
Solution 10
Change in internal energy= Energy loss by bullet
=16 J
Solution 11
KE of man = heat taken by water
Kg
=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= 
=4500-1500
=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.
So,
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.
