How much electricity is used to boil 600 g of water if the kettle has a power of 1500 W? The water boiled for 3 minutes and 9 seconds. Water density is 1000 kg/m3, specific heat of water is 4200 J/(kg· oC).

The electrical energy consumed in boiling the water is 0.0788 kWh

Explanation:

Given;

mass of the water, m = 600 g = 0.6 kg

power rating of the kettle, P = 1500 W = 1.5 kW

specific heat capacity of water, c = 4,200 J/kg⁰C

density of water, = 1000 kg/m³

time taken to boil the water, t = 3 mins + 9 s = (3 x 60s) + 9 s = 180 s + 9 s = 189 s =

The electrical energy consumed in boiling the water is calculated as;

E = P x t

E = 1.5 kW  x 0.0525 hr

E = 0.0788 kWh

Therefore, the electrical energy consumed in boiling the water is 0.0788 kWh

the area between the graph and the t axis

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day