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

## Chapter 3 - Calorimetry Exercise 47

An aluminium 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 aluminium, iron and water are 910 J/kg-K, 470 J/kg-K and 4200 J/kg-K respectively.

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

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.

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

The temperatures of equal masses of three different liquids are 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?

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

Four
2cm × 2cm ×
2cm cubes are taken out from a
refrigerator are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when the 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. Density of ice=
900kg/, density of the drink= 1000kg/, specific heat capacity of the drink= 4200J/kg-K, latent
heat of fusion of ice= 3.4×10^{5}J/kg.

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

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 is contains 10kg of water and 0.2g
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= 2.27 × 10^{6} J/kg.

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

A
cube of iron (density=8000 kg/, 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=9000 kg/ and the latent heat
of fusion of ice=3.36 ×
10^{5} J/kg.

Volume of ice melted=Volume of cube

Heat loss by iron cube=Heat taken by ice

T=353K

1
kg of ice at 0°C is mixed
with 1 kg of steam at 100°C.
What will be the composition of the systemwhen thermal equilibrium is
reached? Latent heat of fusion of ice=3.36 × 10^{5} J/kg and latent heat of vaporization of
water= 2.26×10^{6} J/kg.

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)

Calculate
the time required to heat 20kg 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}.

t=2625 sec

= 44 min

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 0.5m^{3} for
household use. If it were possible to use the heat liberated by the water to
lift a 10 kg of mass vertically, how high can it be lifted as the water comes
to the room temperature? Take g=10 m/s^{2}.

Heat released by water = Energy required to lift mass

h=315000m

h=315 Km

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.

Change in internal energy= Energy loss by bullet

=16 J

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?

KE of man = heat taken by water

Kg

=15 gm

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 calorie.

Thermal energy= 80% of potential energy change

=96 J

A van of mass 1500 kg travelling 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.

Rate of production of thermal energy=

=16857 J/sec

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.

Thermal energy developed = Loss in KE

=3.75 J

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.

If both blocks are taken as system,

10 m/s

= 4500 J

=1500 J

Thermal energy developed=

=4500-1500

=3000 J

A ball is droppedon 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 the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J/K.

Thermal energy = 40% mechanical energy lost

A copper cube of mas 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 down through 60 cm. Specific heat capacity of copper = 420 J/kg-K.

Since, block moves with constant velocity; a=0

Thermal energy gain=Mechanical Energy lost

A metal block of density 6000 kg/ and mass 1.2 kg is suspended through a spring of spring constant 200 N/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 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. 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 is negligible.

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.

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