Answer:
a) annual thermal energy delivered to the house is 6.3 × 10⁹ J
b) corresponding electricity consumption is 3.15 × 10⁹ J
c) volume is needed if there are no losses from storage is 20.895 m³
d) E = 335 MJ
Explanation:
a)
Given that 1 ton refrigeration is 3.5 kw
so Annual thermal energy delivered to the house is
Energy = power × time
E = 3.5 × ( 500×3600)
we converted 500 hrs to seconds
E = 6.3 × 10⁶ kJ
E = 6.3 × 10⁹ J
therefore annual thermal energy delivered to the house is 6.3 × 10⁹ J
b)
corresponding electricity consumption {Cop = 2}
Electricity consumption = Refrigeration effect / work input
∴ EC = 6.3 × 10⁹J / 2
EC = 3.15 × 10⁹ J
∴ corresponding electricity consumption is 3.15 × 10⁹ J
c)
we know that latent heat of ice is 335 kj/kg = 335 × 10³ J/kg
now let m represent the mass of ice needed for required refrigeration
E = mL
6.3 × 10⁹ j = m × (335 × 10³J/kg)
m = 6.3 × 10⁹ J / 335 × 10³J/kg
m = 18805.97 kg
Given that density of ice = 0.9ton/m³ = 900 kg/m³
NOW
Volume of ice needed V = mass / density
v = 18805.97 kg / 900 kg/m³
v = 20.895 m³
volume is needed if there are no losses from storage is 20.895 m³
d)
cooling energy of 1 ton ( m = 1000 kg) ice
we know L = 335 kJ
E = mL
E = 1000 × 335KJ
E = 335 MJ
