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A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move. Now air at 500 kPa and 70°C is allowed to enter the cylinder from a supply line until the volume increases by 50%. Using constant specific heats at room temperature, determine (a) the final temperature, (b) the amount of mass that has entered, and (d) the work done; and (d) entropy generation

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Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered  0.5742 Kg.

(c) The work done is 56.52 kJ.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

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The Final Temperature; amount of mass that has entered; work done; entropy generation are respectively; 315.25 K; 0.5742 Kg 56.52 kJ;  0.0398 kJ/kgK.

What is the Work done and Entropy?

A) The pressure is gotten from;

P = mRT/V

We are given;

mass; m₁ = 1.3 kg

Temperature; T₁ = 30°C = 303 K

Volume; V₁ = 0.4 m³

R is ideal gas constant = 0.287 kJ/kg.K

Thus;

P = (1.3 * 0.287 * 303)/0.4

P = 282.62 kPa

B) Mass that entered is gotten from;

Δm = (p * ΔV)/RT

Δm = (282.62 * 0.5 * 0.4)/(0.287 * 303)

Δm = 0.5742 kg

C) The work done is gotten from;

W = p * ΔV

W = 282.62 * (0.5 * 0.4)

W = 56.52 kJ

D) Final temperature is;

T₂ = pV₂/(m₂R)

T₂ = 282.62 * (1.5 * 0.4)/((1.3 + 0.5742)*0.287)

T₂ = 315.25 K

Entropy generated is gotten from the formula;

s_gen = c_v(In T₂/T₁) + R(In α₁/α₂)

s_gen = 0.718(In (315.25/303) + 0.287(In (0.6/(1.3 + 0.5742))/(0.4/1.3))

s_gen = 0.0398 kJ/Kg.k

Read more about entropy at; https://brainly.com/question/16014998

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