# 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=m_{steam}L_{𝑣}=(1)(2.26×10^{6})

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.