HC Verma Concepts Of Physics Exercise Solutions Of Chapter 25 (Calorimetry)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 25 (Calorimetry):

EXERCISE

1. 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.

Sol:

Given: Mass of aluminium m_a = 0.5kg; Mass of water m_w = 0.2 kg; Mass of Iron m_i = 0.2 kg; Temperature of aluminium and water = 20°C = 297 K; Temperature of Iron = 100°C = 373 K; Specific heat of aluminium = 910 J/kg-k; Specific heat of Iron = 470 J/kg-k; Specific heat of water = 4200J/kg-k.

Assumption: heat interaction outside of the boundary is zero.

∴ Heat loss by iron block = heat gain by aluminium vessel and water

Or, 0.2 * 470 * (373 – T) = (T – 293) (0.5 * 910 + 0.2 * 4200)

Or, 94 * (373 – T) = (T – 293) (455 + 840)

Or, 373 – T = (T – 293) * 13.8

Or, T = 298 K = 25 ⁰C.

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.

Sol:

Given: mass of iron, m_iron = 100 g = 0.1 kg; mass of water, m_w = 240 g = 0.24 kg; water equivalent of calorimeter, m_eq = 10 g = 0.01 kg; final temp of mixture, T_f = 60⁰ C; temp of water, T_w = 20⁰ C; Specific heat capacity of iron, C_iron = 470 J/kg-°C; C_w = 4184 J/kg-°C.

Let the furnace temperature be T.

Now, heat loss by the iron = heat gain by the water and calorimeter

Or, m_iron * C_iron * (T – T_f) = (m_w + m_eq) * C_w * (T_f – T_w)

Or, 0.1 * 470 * (T – 60) = (0.24 +0.01) * 4184 * (60 – 20)

Or, T = 950⁰ C.

So, the temperature of the furnace is 950⁰ 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?

Sol:

Given: T_A = 12⁰C; T_B = 19⁰C; T_C = 28⁰C; m_A = m_B = m_C = m; temperature of mixture A and B is 16⁰C; temperature of mixture B and C is 23⁰C.

For mixture A and B: temperature of mixture, T = 16⁰C.

→Heat gain by A = heat loss by B

Or, m_A C_A (T – T_A) = m_B C_B (T_B – T)

Or, m C_A (16 – 12) = m C_B (19 – 16)

Or, 4 C_A = 3 C_B

For mixture B and C: temperature of mixture, T = 23⁰C.

→Heat gain by B = heat loss by C

Or, m_B C_B (T – T_B) = m_C C_C (T_C – T)

Or, m C_B (23 – 19) = m C_C (28 – 23)

Or, 4 C_B = 5 C_C

For mixture A and C: temperature of mixture, T.

→Heat gain by A = heat loss by C

Or, m_A C_A (T – T_A) = m_C C_C (T_C – T)

Or, m (3/4) C_B (T – 12) = m (4/5) C_B (28 – T)

Or, (3/4) (T – 12) = (4/5) (28 – T)

Or, 15 T – 180 = 448 – 16 T

Or, T = 628/31 = 20.3⁰C.

4. Four 2 cm * 2 cm * 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. Density of ice = 900 kg/m³, density of the drink = 1000 kg/m³, specific heat capacity of the drink = 4200 J/kg-K, latent heat of fusion of ice = 3.4 * 10⁵ J/kg.

Sol:

Given: Density of ice, ρ_i = 900 kg/m³; specific heat capacity of ice, C_i = 2108 J/kg-K; density of the drink, ρ_d = 1000 kg/m³; specific heat capacity of the drink, C_d = 4200 J/kg-K; latent heat of fusion of ice, L = 3.4 * 10⁵ J/kg; volume of ice, V_i = 4 * 2³ = 32 cm³ = 32 * 10^-6 m³; volume of drink, V_d = 200 ml = 2 * 10^-4 m³.

Mass of the ice, m_i = ρ_i V_i = 900 * 32 * 10^-6 = 288 * 10^-4 kg

Mass of the drink, m_d = ρ_d V_d = 1000 * 2 * 10^-4 = 0.2 kg

Heat required to melt ice completely is

→ Q_m = m_iL = 288 * 10^-4 * 3.4 * 10⁵ = 9792 J.

Heat release when drink comes from 10⁰ C to 0⁰ C is

→ Q_d = m_dC_dΔT = 0.2 * 4200 * (10 – 0) = 8400 J.

Since, Q_d is greater than Q_m. So complete ice will not melt.

(a) The temperature of the drink when thermal equilibrium is attained is 0⁰ C.

(b) Let the amount ice melt be m.

→ Latent heat of melted ice = Heat loss by the drink

Or, mL = 8400

Or, 3.4 * 10⁵ * m = 8400

Or, m = 8400/ (3.4 * 10⁵) = 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 = 2.27 * 10⁶ J/kg.

Sol:

Given: mass of water, m_w = 10 kg; C_w = 4200 J/kg-°C; latent heat of vaporization, L = 2.27 * 10⁶ J/kg; rate at which water comes, m’ = 0.2 g/s = 0.2 * 10^-3 kg/s; ΔT = 5⁰ C. let time = t.

According to the question,

→ Heat loss by the water = latent heat of vaporization of water

Or, m_w * C_w * ΔT = m’ * t * L

Or, 10 * 4200 * 5 = 0.2 * 10^-3 * 2.27 * 10⁶ * t

Or, t = 462.55 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 = 3.36 * 10⁵ J/kg.

Sol:

Given: density of iron, ρ_iron = 8000 kg/m³; specific heat capacity, C_iron = 470 J/kg-K; the latent heat, L = 3.36 * 10⁵ J/kg; density of ice, ρ_ice = 900 kg/m³; final temp of iron, T_f,iron = 0⁰ C; initial temperature of the iron cube, T_i,iron.

According to the question,

→ Volume of iron cube = volume of melt ice = V

Now, V ρ_iron C_iron (T_i,iron – T_f,iron) = V ρ_ice L

Or, 8000 * 470 * (T_i,iron – 0) = 900 * 3.36 * 10⁵

Or, T_i,iron = 80⁰ C.

Therefore, initial temperature of the iron cube is 80⁰ 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 the thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 * 10⁵ J/kg and latent heat of vaporization of water = 2.26 * 10⁶ J/kg.

Sol:

Given: mass of ice, m_i = 1 kg; mass of steam, m_s = 1 kg; Latent heat of fusion of ice, L_f = 3.36 * 10⁵ J/kg; latent heat of vaporization of water, L_v = 2.26 * 10⁶ J/kg.

Heat released in condensation, Q_c = m*L_v = 2.26 * 10⁶ J.

Heat required to melt and take water to 100⁰ C is

Q = m * L_f + m *C *ΔT = 3.36 * 10⁵ + 1* 4200 *100 = 7.56 * 10⁵ J.

Mass of steam condensed to melt and take water to 100⁰ C is

= Q/L_v = 7.56 * 10⁵/2.26 * 10⁶ = 0.335 kg = 335 g.

Therefore, mass of water in equilibrium = 1 kg + 0.335 kg = 1.335 kg.

And, mass of steam in equilibrium = 1 kg – 0.335 kg = 665 kg.

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.

Sol:

Given: mass of water, m = 20 kg; T_i = 10⁰C = 283 K; T_f = 35⁰C = 308 K; power of heater, P = 1000 W; efficiency of heater, η = 80% = 0.8; Specific heat capacity, C = 4200 J/kg-K.

Heat required to increase temperature, Q = mC (T_f – T_i)

Or, Q = 20 * 4200 * (308 – 283) = 2100 kJ

Efficiency of heater, η = output/input = 0.8

Or, output = 0.8 * input        [input = 1000 * time]

Or, output = 0.8 * 1000 * t = 800 t

Heat required, Q = output

Or, 2100 * 10³ = 800 * t

Or, t = 2625 s = 43.75 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 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.

Sol:

Given: initial temp of water, T_i = 20⁰ C; final temp of water, T_f = 5⁰ C; volume of water, ρ = 0.5 m³; density, ρ = 1000 kg/m³; specific heat capacity, C_w = 4.2 kJ/kg-K; Lift mass, m = 10 kg.

Heat loss from water, Q = ρV C_w (T_f – T_i)

Or, Q = 1000 * 0.5 * 4.2 * |(5 – 20)| = 31500 kJ

This heat energy is used to lift a 10 kg mass vertically.

So, Potential energy = heat energy

Or, mgh = 31500 kJ

Or, height, h = 31500/ (10 * 10) km = 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.

Sol:

Given: mass of bullet, m = 20 g = 0.02 kg; initial speed of bullet, u = 40 m/s; final speed of bullet, v = 0.

KE_initial = ½ mu² = ½ * 0.02 * 40² = 16 J

KE_final = ½ mv² = 0

By the conservation of energy,

→ Change in internal energy = change in kinetic energy

Or, Δu = | ½ mv² – ½ mu² | = |0 – 16 | = 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?

Sol:

Given: mass of man, m = 50 kg; speed, v = 18 km/h = 5 m/s; change in temp of water, Δ T = 10⁰ C; specific heat capacity, C_w = 4200 J/kg-K.

Let the mass of water be m_w.

The kinetic energy of the man, KE = ½ mv² = ½ * 50 * 5² = 625 J.

According to the question,

→ m_w C_w Δ T = KE

Or, m_w * 4200 * 10 = 625

Or, m_w = 0.015 kg = 15 g.

So, mass of water, m_w = 15 g.

12. A brick weighing 4 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.

Sol:

Given: mass of brick, m = 4 kg; height, h = 1 m + 2 m = 3 m.

Potential Energy = mgh = 4 * 10 * 3 = 120 J.

According to the question,

→ Thermal energy = 0.8 * Potential Energy

Or, Thermal energy = 0.8 * 120 = 96 J = 96/4.2 cal. = 23 cal.

13. 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.

Sol:

Given: mass of van, m = 1500 kg; initial speed of van, u = 54 km/h = 15 m/s; final speed of van, v = 0; time taken, t = 10 s.

We know,

Acceleration, a = (v – u)/t = (0 – 15)/10 = – 1.5 m/s².

Force, F = ma = 1500 * (– 1.5) = – 2250 N.

And, distance travelled, s = (v² – u²)/2a = 75 m.

Therefore, the mechanical energy lost = F * s = 2250 * 75 = 168750 J

All the mechanical energy lost appears as thermal energy in the brake mechanism.

So, thermal energy = 168750 J

And the average rate of production of thermal energy = 168750/10 = 16875 J/s = 4017 cal/s. [Since, 1 cal = 4.2 J]

Alternative process,

By the conservation of energy principle,

 → Thermal energy = change in kinetic energy

Or, Thermal energy = ½ mu² = ½ * 1500 * 15² = 168750 J

And the average rate of production of thermal energy = 168750/10 = 16875 J/s = 4017 cal/s. [Since, 1 cal = 4.2 J]

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.

Sol:

Given: mass, m = 100 g = 0.1 kg; initial speed, u = 10 m/s; final speed, v = 5 m/s.

By the conservation of energy principle,

 → Thermal energy = change in kinetic energy

Or, Thermal energy = ½ mu² – ½mv²

Or, Thermal energy = ½ * 0.1 * 10² – ½ * 0.1 * 5²

Or, Thermal energy = 3.75 J

So, the thermal energy developed in the process is 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.

Sol:

Given: mass of block 1, m₁ = 10 kg; mass of block 2, m₂ = 20 kg; velocity of block 1, v₁ = 10 m/s; velocity of block 2, v₂ = 20 m/s.

For perfectly inelastic collision, two bodies move together after collision and let their common velocity be V.

By the conservation of linear momentum,

→ V (m₁ + m₂) = m₁v₁ + m₂v₂

Or, V (10 + 20) = 10 * 10 + 20 * 20

Or, V = 500/30 = 50/3 m/s.

By the conservation of energy principle,

→ Thermal energy = Loss in kinetic energy

Or, Thermal energy = KE_i – KE_f

Or, Thermal energy = [½ m₁ (v₁)² + ½ m₂ (v₂)²] – ½ (m₁ + m₂) V²

Or, Thermal energy = [½* 10*(10)² + ½ * 20*(20)²] – ½ (10 + 20) (50/3)² 

Or, Thermal energy = 4500 – 4166.66 = 333.33 J.

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

Sol:

Given: Heat capacity of the ball, C_b = 800 J/K; change in height, Δh = (2 – 1.5) = 0.5 m; thermal energy = 40% of the mechanical energy.

By the conservation of energy principle,

→ Mechanical energy = change in potential energy

Or, Mechanical energy = mg Δh

And, thermal energy = 40% of the mechanical energy

Or, mC_b ΔT = 0.40 * mg Δh

Or, ΔT = (0.40 * 10 * 0.5)/800 = 0.0025⁰ C = 2.5 * 10^{-3 0}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 down through 60 cm. Specific heat capacity of copper = 420 J/kg-K.

Sol:

Given: Heat capacity of the ball, C_b = 420 J/kg-K; change in height, Δh = 0.6 sin 37 = 0.36 m; thermal energy = mechanical energy.

By the conservation of energy principle,

→ Mechanical energy = change in potential energy

Or, Mechanical energy = mg Δh

And, thermal energy = mechanical energy

Or, mC_b ΔT = mg Δh

Or, ΔT = (10 * 0.36)/420 = 0.0086⁰ C = 8.6 * 10^{-3 0}C.

18. A metal block of density 6000 kg/m³ 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 are negligible.

Sol:

Given: density of block, ρ_b = 6000 kg/m³; mass of block, m_b = 1.2 kg, spring constant, k = 200 N/m; mass of water, m_w = 250 g = 0.25 kg; Specific heat capacity of the block, C_b = 250 J/kg-K; Specific heat capacity of the water, C_w = 250 J/kg-K.

When spring is broken, block falls from 40 cm = 0.4 m.

Loss in energy = m_bgh = 1.2 * 9.8 * 0.4 = 4.707 J.

This heat will rise temp of block as well as water and also temp of both will be same because they are in equilibrium.

→ m_b C_b ΔT + m_w C_w ΔT = 4.707

Or, (1.2 * 250 + 0.25 * 4200) ΔT = 4.707

Or, ΔT = 0.003 ⁰C.

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