Question

One kg of ice at -5°C is exposed to the atmosphere which is at 25°C. The ice melts and comes into thermal equilibrium with the atomosphere. Determine the entropy increase of the universe. Assume the Cp of ice is 2-093 kJ/kg K and the latent heat of fusion of ice is 333-3 kJ/kg.​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:Entropy change of universe = 0.094 kJ/k

Explanation:

Given:

Mass of ice (m) = 1 kg

Initial temperature of ice (T1) = -5°C

Temperature of atmosphere (T0) = 20°C

Cp of ice = 2.093 kJ/kgK

Latent heat of fusion of ice = 333.3 kJ/kg

Now,

Heat absorbed by ice = Heat absorbed in solid phase + latent heat + Heat absorbed in liquid phase

∴ Q = [mCp(ΔT)]ice + m × Latent heat of fusion of ice + [mCp(ΔT)]water

∴ Q = 1 × 2.093 × [0 – (-5)] + 1 × 333.3 + 1 × 4.187 × (20 - 0)

∴ Q = 427.5 kJ

Now,

Entropy change of ice in liquid phase,

∴ (Δs)1 = 0.0389 kJ/k

Now,

Entropy change as ice melts to water at 0°C,

Now,

Entropy change of water from 0°C to 20°C

∴ (Δs)3 = 0.296 kJ/k  

Now,

Total entropy change of ice = Δs1 + Δs2 + Δs3.

∴ (Δs)ice = 0.0389 + 1.22 + 0.296

∴ (Δs)ice = 1.554 kJ/k

Now,

Entropy change of atmosphere = -1.46 kJ/k    

Hence,

Entropy change of universe = Δsice + Δsatmosphere

Entropy change of universe = 1.554 – 1.46

∴ Entropy change of universe = 0.094 kJ/k