Thermodynamics

Exercise L.V.3 — Thermodynamics & Heat

Clear, step-by-step solutions in HTML format

Q1 — Linear expansion

Calculate the increase in length of a rod 100 mm long if the temperature rises by 200 °C. (α = 4.5 × 10^-6/°C)

Formula: ΔL = L · α · ΔT

Given: L = 100 mm, α = 4.5 × 10^-6/°C, ΔT = 200 °C

Calculation:

ΔL = 100 × (4.5 × 10^-6) × 200 = 0.09 mm

Increase in length = 0.09 mm

Q2 — Energy removed cooling steam to liquid

How much energy is removed when 10 g of water is cooled from steam at 113 °C to liquid at 53 °C?

Given: m = 10 g = 0.01 kg
L_v = 2.26 × 10⁶ J/kg, C_water = 4.26 × 10³ J/(kg·°C), C_steam = 2.0 × 10³ J/(kg·°C)

1. Cool steam from 113 °C to 100 °C:
   Q₁ = m·C_steam·(113−100) = 0.01·2000·13 = 260 J
2. Condense steam at 100 °C to liquid at 100 °C:
   Q₂ = m·L_v = 0.01·2.26×10⁶ = 22,600 J
3. Cool liquid water from 100 °C to 53 °C:
   Q₃ = m·C_water·(100−53) = 0.01·4260·47 = 2,009 J

Total: Q = Q₁ + Q₂ + Q₃ = 260 + 22600 + 2009 = 24,869 J ≈ 24.9 kJ

Q3 — Mass of steam condensed

Steam at 100 °C was passed onto 200 g of cold water at 5 °C until its temperature rises to 100 °C. Calculate the mass of steam that condensed.

Given: m_w = 200 g = 0.2 kg, C_w = 4200 J/(kg·°C), L_v = 2.26×10⁶ J/kg

Heat gained by water: Q_w = m_w·C_w·(100−5) = 0.2·4200·95 = 79,800 J

Heat released by condensed steam: Q_s = m_s·L_v

Equate: m_s·2.26×10⁶ = 79,800 → m_s = 79,800 / 2.26×10⁶ = 0.0353 kg = 35.3 g

Q4 — Ideal gas law (final temperature)

A gas has initial volume 15 L, pressure 2 atm and temperature 310 K. After piston action the volume decreases to 10 L and the pressure becomes 3.5 atm. Find final temperature (use ideal gas law).

Use P1V1/T1 = P2V2/T2

T₂ = (P₂ V₂ T₁)/(P₁ V₁) = (3.5·10·310)/(2·15) = 361.7 K

Final temperature ≈ 362 K

Q5 — First law of thermodynamics (two parts)

a)

Given: Work done on gas W = 135 J (on the gas → +135 J), internal energy increase ΔU = +445 J.

First law (sign convention used here): ΔU = Q + W. So Q = ΔU − W = 445 − 135 = 310 J.

b)

Given: Initial internal energy U_i = 27 J, heat added Q = 383 J, system does work W = 67 J.

ΔU = Q − W = 383 − 67 = 316 J → U_f = U_i + ΔU = 27 + 316 = 343 J.

Prepared for Exercise L.V.3 — If you want the file exported or adjusted (fonts, sizes, orPdf-friendly layout), tell me and I will update it.