# Consider an appliance which consumes 5.2 kWh of energy per day. a) What is the average power...

## Question:

Consider an appliance which consumes 5.2 kWh of energy per day.

a) What is the average power consumption, in watts, of the appliance per day?

b) How much energy, in joules, does this appliance consume in a year?

## Power Consumption:

The energy consumed by an appliance in the given amount of time shows the power consumption of the appliance. If the rated power is more for an appliance, then the energy consumption will be more for the same time period.

Given Data

Power consumption {eq}(P) = 5.2 \ kWh \ day \\ {/eq}

(a)

The energy conversion into Joule

{eq}1\ kWh = (10^{3} \ W) \times (3600 \ s) \\ 1\ kwh = (10^{3} \ J/s) \times (3600 \ s) \\ 1\ kWh = 36*10^{5} \ J {/eq}

Now, the total energy consumption in one day

{eq}E = 5.2 \kWh \\ E = (5.2)(36*10^{5}) \ J \\ E = 187.2*10^{5} \ J {/eq}

now, the time in seconds would be

{eq}(t) = 24 \ hr \\ t = 24*3600 \ s {/eq}

Now, the average power would be

{eq}P_{av} = \dfrac{E}{t} \\ P_{av} = \dfrac{187.2*10^{5} \ J }{24*3600\ s} \\ P_{av} = 216.67 \ W {/eq}

(b)

Now, the seconds in a year would be

{eq}T = 1 \ year \\ T = 365 \ days \\ T = (365 \ days) (24 \ hr/days) (3600 \ s/hr ) \\ T = 3153600 \ s {/eq}

Now, the total energy would be

{eq}(E_{t}) = P_{av}*T \\ E_{t} = (216.67)(3153600) \\ E_{t} = 6.832*10^{9} \ J {/eq}