Answer:
a. E = 900000 J = 900 KJ
b. ΔT = 8.18 °C
c. Cost = $ 7.2
Explanation:
a.
The energy can be given by:
[tex]E = Pt[/tex]
where,
E = Energy = ?
P = Power = 500 W
t = time = (0.5 h)(3600 s/1 h) = 1800 s
Therefore,
[tex]E = (500\ W)(1800\ s)[/tex]
E = 900000 J = 900 KJ
b.
The change in temperature of room air is given by the following formula:
[tex]0.5E = mC\Delta T\\[/tex] (since 50% of energy is used to heat air)
where,
m = mass of air = 50 kg
C = specific heat of air = 1.1 KJ/kg.°C
ΔT = Change in temperature of air = ?
Therefore,
[tex](0.5)(900\ KJ) = (50\ kg)(1.1\ KJ/Kg.^oC)\Delta T[/tex]
ΔT = 8.18 °C
c.
First, we will calculate the total energy consumed:
[tex]E = (0.5\ KW)(6\ h/d)(30\ d)\\E = 90\ KWh[/tex]
Now, for the cost:
[tex]Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(90\ KWh)[/tex]
Cost = $ 7.2
