A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa and 400°C. It is cooled to 40°C and 300 kPa. Calculate the heat transfer for the process.

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m (u2 - u1) = Q - W

Q is heat transfer

and in process P = A + N that is linear spring

so

W = ∫PdV

= 0.5 (P1+P2) (V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 (500 + 300) (0.3944 - 0.5087)

W = - 45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv (T2-T1) + W

Q = 2 * 0.83 (40 - 400) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ