A 0.8-m³ rigid tank contains carbon dioxide (CO₂) gas at 250 K and 100 kPa. A 500-W electric resistance heater placed in the tank is now turned on and kept on for 40 min after which the pressure of CO₂ is measured to be 175 kPa. Assuming the surroundings to be at 300 K and using constant specific heats, determine (a) the final temperature of CO₂, (b) the net amount of heat transfer from the tank, and (c) the entropy generation during this process.

Explanation:

Given:

Power, p = 500 W

Time, t = 40 min

= 2400 s

Volume, V = 0.8 m^3

Temperature, T = 250 K

Pressure, P = 100 kPa

Temperature of surroundings, Ts = 300 K

Using ideal gas equation,

PV = nRT

n = (100 × 10^3 × 0.8)/(8.3145 × 250)

= 38.49 mole

Mass = number of moles × molar mass

Molar mass = 12 + (16 × 2)

= 44 g/mol

Mass = 44 × 38.49

= 1693.43 g

= 1.693 kg.

A.

P2 = 175 kPa

Using pressure law,

P1/T1 = P2/T2

T2 = (175 × 250)/100

= 437.5 K

B.

Cvco2 = 0.706 kJ/kg.K

Total energy, U = Qin - Qout

Qout = (p × t) - (m × cv × delta T)

= (500 × 2400) - (1.693 × 0.706 × (437.5 - 250))

= 1200 kJ - 224.11 kJ

= 975.889 kJ

= 975.9 kJ

C.

Cpco2 = 0.895 kJ/kg.K

Gas constant, Rc = R/molar mass of CO2

= 8.3145/44

= 0.189

Using the formula ,

Entropy, S = (m × (Cpco2 × ln(T2/T1) - Rc × ln(P2/P1)) + Qout/Ts

Inputting values,

= (1.693 × (0.895 × ln(437.5/250) - 0.189 × ln(175/100)) + 975.9/300

= 3.922 kJ/K.

it breaks and dirupture other materials and other sencews from the bottom

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.