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.

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 26 (Laws of Thermodynamics)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 26 - Laws Of Thermodynamics:

EXERCISE

1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal expansion, (a) how much heat is transferred to the liquid—vessel system? (b) How much work has been done on this system? (c) How much is the increase in internal energy of the system?

Sol:

System: water + copper

Given: insulated; closed system; t_i = 15 ⁰c; t_e = 17 ⁰c; mass of copper, m_cu = 100 g = 0.1 kg; mass of water, m_w = 200 g = 0.2 kg; c_cu = 420 J/kg-K and c_w = 4200 J/kg-K.

(a) Due to insulated boundary, heat transfer to the system is zero.

(b) From the 1^st law,

We know, Q = ΔU + W = U₂ – U₁ + W

Or, W = U₁ – U₂              [Q = 0]

Or, W = (U₁ – U₂)_cu + (U₁ – U₂)_w

Or, W = m_cu c_cu (t_i – t_e) + m_w c_w (t_i – t_e)

Or, W = 0.1 * 420 (15 – 17) + 0.2 * 4200 (15 – 17)

Or, W = – 1764 J.

– Ve sign means work done on the system.

(c) The increase in internal energy of the system is ΔU = W = 1764 J.

2. Figure (26-E1) shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) how much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Sol:

System: liquid + wheel + container

Given: adiabatic system; heat capacity of liquid, C_L = 4200 J/K, mass of block, m = 12 kg, change of height of block, Δh = 70 cm = 0.7 m.

(a) Heat is given to the liquid is zero, because system is adiabatic.

Q = 0

(b) Work is done on the liquid, W = change of potential energy of block

Or, W = mg (Δ h) = 12 * 10 * 0.7 = 84 J.

(c) 1^st law, Q = ΔU + W

Or, 0 = C_L (Δ t) – 84 [W = – 84 work done on the system]

Or, Δ t = 84/4200 = 0.02 ⁰c.

3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

Sol:

Given: mass of block = 100 kg; u = 2 m/s; μ = 0.2; v = 0.

1^st law: Q = Δu + W

In this case Q = 0

Or, – Δu = W = change in K.E.

Or, Δu = – (½ mv² – ½ mu²)

Or, Δu = ½ * 100 * 2² = 200 J.

(a) The change in the internal energy of the block-belt system is 200 J.

(b) The increase in the kinetic energy of the block is 200 J.

4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.

Sol:

Given: Q = 100 J; ΔV = 0.

1^st Law: Q = ΔU + W = ΔU + pΔV

Or, 100 = ΔU + 0 → ΔU = 100 J

So, the change in internal energy of a gas is 100 J.

5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Sol:

Given: Q = 0; v₁ = 200 cc = 2 * 10^-4 m³; v₂ = 50 cc = 0.5 * 10^-4; p₁ = 10 kPa; p₂ = 50 kPa.

(a) Work done during the process 1-2,

W = p₁ (v₂ – v₁) + ½ * (p₂ – p₁)(v₂ – v₁)

Or, W = 10 * (0.5 – 2)*10^-4 + ½ *40 *(0.5 – 2) * 10^-4

Or, W = – 0.0015 – 0.0030 = – 0.0045 kJ

Or, W = – 4.5 J

So, the work done by the gas is – 4.5 J.

(b) 1^st law: Q = ΔU + W

Or, 0 = ΔU + W

Or, ΔU = – W = 4.5 J.

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