Thermodynamics Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1Water flow: 3.0 L min⁻¹ = 3000 g min⁻¹; ΔT = 77 − 27 = 50 °C; c = 4.2 J g⁻¹ °C⁻¹.
2. 2Heat needed per minute: ΔQ = mcΔT = 3000 × 4.2 × 50 = 6.3 × 10⁵ J min⁻¹.
3. 3Fuel rate = ΔQ/heat of combustion = 6.3 × 10⁵/(4.0 × 10⁴).
4. 4= 15.75 g min⁻¹ ≈ 16 g min⁻¹.

Step-by-step solution

1. 1Moles: n = m/M = 20/28 = 0.714 mol.
2. 2C_p = (7/2)R = (7/2) × 8.3 = 29.05 J mol⁻¹ K⁻¹ (diatomic gas).
3. 3ΔQ = nC_pΔT = 0.714 × 29.05 × 45.
4. 4ΔQ = 933.4 J ≈ 9.33 × 10² J.

Step-by-step solution

1. 1(a) Heat flows until both bodies reach equilibrium; the equilibrium temperature equals the arithmetic mean only when the two bodies have equal thermal capacities.
2. 2(b) A high specific heat means the liquid can absorb a large amount of heat for a small temperature rise, so it keeps plant parts from overheating.
3. 3(c) Friction heats the air inside the tyre during driving; at roughly constant volume the pressure follows the temperature rise (p ∝ T).
4. 4(d) The nearby sea has a large specific heat, so it warms and cools slowly, moderating the harbour's temperature; the desert has no such water body and swings between extremes.

Step-by-step solution

1. 1Insulated walls and piston mean the compression is adiabatic (Q = 0).
2. 2For hydrogen (diatomic), γ = 7/5 = 1.4.
3. 3Adiabatic relation: P₁V₁^γ = P₂V₂^γ with V₂ = V₁/2.
4. 4P₂/P₁ = (V₁/V₂)^γ = 2^1.4.
5. 5P₂/P₁ = 2.639.

Step-by-step solution

1. 1Adiabatic path: ΔQ = 0, and work 22.3 J is done on the system, so ΔU = +22.3 J.
2. 2ΔU depends only on the endpoints, so the second path has the same ΔU = 22.3 J.
3. 3Second path: ΔQ = 9.35 cal = 9.35 × 4.19 = 39.18 J.
4. 4First law, work by system: ΔW = ΔQ − ΔU = 39.18 − 22.3.
5. 5ΔW = 16.88 J ≈ 16.9 J (work done by the system).

Step-by-step solution

1. 1(a) The gas rushes to fill double the volume; for an ideal gas at constant temperature, pressure halves. Final pressure in each cylinder = 0.5 atm.
2. 2(b) There is nothing to push against (true free expansion into vacuum), so no work is done and ΔU = 0.
3. 3(c) With ΔU = 0, the temperature does not change: ΔT = 0.
4. 4(d) No — free expansion passes through non-equilibrium intermediate states, which do not obey the gas equation and do not lie on the P-V-T surface.

Step-by-step solution

1. 1First law (power form): dU/dt = dQ/dt − dW/dt.
2. 2dQ/dt = 100 J s⁻¹, dW/dt = 75 J s⁻¹.
3. 3dU/dt = 100 − 75 = 25 J s⁻¹ = 25 W.

Step-by-step solution

1. 1From the figure, D is at (600 N m⁻², 2.0 m³), E at (300 N m⁻², 5.0 m³) and F at (300 N m⁻², 2.0 m³).
2. 2Work done along D → E → F equals the area of triangle DEF.
3. 3Base DF = change in pressure = 600 − 300 = 300 N m⁻²; height FE = 5.0 − 2.0 = 3.0 m³.
4. 4Area = ½ × 300 × 3.0 = 450 J.