A piston-cylinder device initially contains 0.5m3 of nitrogen gas at 400 kPa and 27 °C. Anelectric heater within the device is turned on and is allowed to pass a current of 2 A for 5min from a 120V source. Nitrogen expands at constant pressure, and a heat loss of 2800 Joccurs during the process. Determine the final temperature of nitrogen.Assume nitrogen to be an ideal gas with M = 28 g/mol and cp=1.039 kJ/kg.K

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Respuesta :

arjunrv arjunrv · Answer 1 · 2020-03-18T10:39:59+01:00

The final temperature of nitrogen is 56.968° C.

Explanation:

Given data:

v = 0.5 [tex]m^{3}[/tex]

P = 400 k Pa

T = 27+273 = 300 K

[tex]C_{p}[/tex] = 1.039 KJ/Kg K

[tex]Q_{out}[/tex] = 2800

J = 2.8 KJ

M = 28 g/mol

Solution:

Electric work ([tex]W_{e}[/tex]) = VI Δt

[tex]W_{e}[/tex] = 120V × 2A × (5×60s)(1/1000)

[tex]W_{e}[/tex] = 72 KJ

m = [tex]\frac{PV}{RT}[/tex]

=[tex]\frac{(400)(0.5)}{(0.297)(300)}[/tex]

= 2.245 Kg

Equation for energy balance is

[tex]E_{in}[/tex] - [tex]E_{out}[/tex] = ΔE system

[tex](W_{e} )in[/tex] - [tex]Q_{out}[/tex] - [tex](W_{e} )out[/tex] = ΔU

[tex](W_{e} )in[/tex] - [tex]Q_{out}[/tex] = ΔH

= m ([tex]h_{2}[/tex] - [tex]h_{1}[/tex])

= m [tex]C_{p}[/tex]([tex]T_{2}[/tex] - [tex]T_{1}[/tex])

(72 - 2.8) KJ = 2.245 × 1.039 ([tex]T_{2}[/tex] - 27°C)

69.2 = 2.3325 ( [tex]T_{2}[/tex] - 27°C)

( [tex]T_{2}[/tex] - 27°C) = 29.968

[tex]T_{2}[/tex] = 56.968° C