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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