A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that final temperature is 97 "C. The specific heats are Cp 1.005 kJ (kg. K) and C-0.718 kJ / (kg. K). Determine the total work done in kJ:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = - mCvΔT

⇒Q = - 1*0.718 * (97-15)

⇒Q = - 58.716 kJ