Question

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an insulated enclosure. Determine the final equilibrium temperature and the total entropy change for this process. The specific heat of aluminum at 400 K is cp = 0.949 kJ/kg·K. The specific heat of iron at room temperature is cp = 0.45 kJ/kg·K.

Answer 1

Answer:

The equilibrium temperature is 109.7°C

The total entropy change for the process is 0.23 kJ/K

Explanation:

The parameters given are;

The initial temperature of aluminium, T₁ₐ = 140°C = 413.15 K

Mass, m₁, of aluminium block = 28 kg

The specific heat of aluminium, c₁, at 400 K = 0.949 kJ/(kg·K)

The initial temperature of iron, T₁ₙ = 60°C = 333.15 K

Mass, m₂, of iron block = 36 kg

The specific heat of iron, c₂, at room temperature = 0.45 kJ/(kg·K)

The heat lost by the aluminium = Heat gained by the iron

The heat lost by the aluminium = m₁ × c₁ × (T₁ₐ - Tₓ)

The heat lost by the aluminium = 28 × 0.949 × (413.15 - Tₓ)

The heat gained by the iron =  m₂ × c₂ × (Tₓ - T₁ₙ)

The heat gained by the iron =  36 × 0.45 × (Tₓ - 333.15)

Hence;

28 × 0.949 × (413.15 - Tₓ) = 36 × 0.45 × (Tₓ - 333.15)

26.572 × (413.15 - Tₓ) = 16.2 × (Tₓ - 333.15)

∴ (26.572 + 16.2) × Tₓ = 26.572 × 413.15 + 16.2 × 333.15

42.772·Tₓ = 16375.2518

Tₓ = 382.85 K = 109.7°C

The total entropy change of the process is given by the following relation;

$\Delta S = \Delta S_{Fe} + \Delta S_{Al}$

$=m_{1}c_{1}ln\dfrac{T_{x}}{T_{1n}}+m_{2}c_{2}ln\dfrac{T_{x}}{T_{1a}}$

$=28 \times 0.949 \times ln\dfrac{382.85}{413.15}+36 \times 0.45 \times ln\dfrac{382.85}{333.15} = 0.23 \, \dfrac{kJ}{K}$