Answer to Question #179966 in Molecular Physics | Thermodynamics for Johannes Steven

Question

Air is contained in a vertical piston-cylinder assembly fitted with an electrical resistor. The atmospheric pressure is 100 kPa and point has a mass of 50 kg and a face area of 0.1 m². Electric current passes through the resistor, and the volume of of air slowly increases by 0.045 m³. The mass of the air is 0.3 kg, and its specific energy increases by 42.2 kJ/kg. Assume the assembly ( including the piston) is insulated and neglect the friction between the cylinder and piston, g = 9.8 m/s². Determine the heat transfer from the resistor to air for a system consisting of (a) the balloon, (b) the air and the piston

Answer 1

Assumptions:

1. Two closed systems are under consideration, as shown in schematic.

2. The only heat transfer is from the resistor to the air. AKE = APE=0 (for air)

3. The internal energy of the piston is not affected by the heat transfer.

a) Taking the air as the system,

(∆KE + ∆PE + ∆U) = Q - W

Q=W+ ∆U_air

For this system work is done at the bottom of the piston. The work done by the system is:

$W = \int^{V_2}_{V_1}pdV = p (V_2-V_1)$ ------- (at constant pressure)

The pressure acting on the air can be found from:

$PA_{piston }= m_{pistong}g+ P_{atm}A_{piston}$

$P =\dfrac{m_{piston}g}{A_{piston}}+ P_{atm}$

$P = \dfrac{50× 9.81}{0.1} + 100kPa$

P = 104.91 kPa

Thus, the work is

W = (104.91 kPa)(0.045m³) = 4.721kJ

With ∆U_air = m_air ∆u_air the heat transfer is;

Q =W + m_air ∆u_air = 4.721 kJ + (0.3 kg)(42.2 kJ/kg) = 17.38 kJ

b) system consisting the air and the piston. The first law becomes:

(∆KE + ∆PE + ∆U)_air + (∆KE + ∆PE + ∆U)_piston = Q - W,

where (∆KE = ∆PE)_air = 0 and (∆KE = ∆U)_piston = 0.

Thus, it simplifies to;

(∆U)_air + (∆PE)_piston = Q - W

For this system, work is done at the top of the piston and pressure is the atmospheric pressure. The work becomes

W = P_atm∆V = (100 kPa)(0.045m') = 4.5 kJ

The elevation change required to evaluate the potential energy change of the piston can be found from the volume change:

∆z = ∆V/A_piston = 0.045 m³/0.1 m² = 0.45 m

(∆PE =m_pistong∆z = (50 kg)(9.81 m/s')(0.45 m) = 220.73 J = 0.221 kJ

Q = W + (∆PE)_piston + m_air∆u_air

Q=4.5 kJ + 0.221 kJ + (0.3 kg)(42.2 kJ/kg) = 17.38 kJ

Note that the heat transfer is identical in both systems.