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Hydrogen in rigid vessel: heat, entropy change

Short answer

At constant volume, Q = m cv ΔT = 0.4×10.183×150 = 611 kJ. ΔS_H₂ = m cv ln(T₂/T₁) = 1.652 kJ/K. Reservoir entropy change = −Q/T_R = −1.018 kJ/K. Entropy produced S_gen = 0.633 kJ/K.

A rigid vessel consist of 0.4 kg of hydrogen initially at 200 kPa and 27°C, heat is transferred to the system from a reservoir at 600K until its temperature reaches at 450 K. Determine heat transfer, the change in entropy of hydrogen and the amount of entropy produced. (Take cv = 10.183 J/KgK)

Reading of the data. cv = 10.183 is given in J/kg·K, but for hydrogen cv ≈ 10.183 kJ/kg·K, so it is used in kJ/kg·K.

Given: m = 0.4 kg, P₁ = 200 kPa, T₁ = 27 + 273 = 300 K, T₂ = 450 K, reservoir at T_R = 600 K, rigid vessel (V = constant, W = 0).

Heat transfer

First law with W = 0: Q = ΔU = m cv (T₂ − T₁)

Q = 0.4 × 10.183 × (450 − 300) = 611.0 kJ (into the hydrogen)

Entropy change of hydrogen

Constant volume: ΔS_H₂ = m cv ln(T₂/T₁)

ΔS_H₂ = 0.4 × 10.183 × ln(450/300) = 4.0732 × 0.40547 = 1.6515 kJ/K

(The final pressure is P₂ = P₁T₂/T₁ = 300 kPa; it is not needed for ΔS here.)

Entropy change of the reservoir

The reservoir loses Q at constant 600 K:

ΔS_res = −Q/T_R = −611.0/600 = -1.0183 kJ/K

Entropy produced

S_gen = ΔS_H₂ + ΔS_res = 1.6515 − 1.0183

Answer: Q = 611 kJ, ΔS_H₂ = 1.652 kJ/K, S_gen = 0.633 kJ/K. S_gen > 0, so heat transfer across the finite temperature difference is irreversible.

✓ Answer checked by Claude Opus 5.5 (high effort)

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