Answer:
The amount that is "vented" out by "the fans" is $0.50 for 10 hours.
Option: a
Explanation:
"Energy discharged by air in every hour" can be determined by,
[tex]\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}[/tex]
Q = heat energy (Joules, J)
m = mass of a substance (kg)
c = specific heat (units J/kg∙K)
[tex]\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}[/tex]
[tex]\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}[/tex]
[tex]\text { Density of air } \rho=1.20 \times 200 \times 2.4[/tex]
[tex]\text { Density of air } \rho=576 \mathrm{kg}[/tex]
∆T = 10 hours
[tex]\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}[/tex]
Q = 576 × 1.00 × 10
Q = 5760 kJ/hours
W = 1.6 kwh
We know that, “Coefficient of performance” (COP)
[tex]\mathrm{Cop}=\frac{Q}{w}[/tex]
[tex]\mathrm{W}=\frac{Q}{\mathrm{cop}}[/tex]
Given that, COP = 3.2
[tex]\mathrm{W}=\frac{1.6}{3.2}[/tex]
W = 0.5 kwh
The unit cost of electricity is $0.10/kWh
The unit cost of electricity is $0.10/kWh
Unit electricity cost for 10 hours = 0.5 × 10 × 0.1$
Unit electricity cost for 10 hours = $0.5
The amount that is "vented out" by "the fans" is $0.50 for 10 hours.
