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 se
1 answer:
Answer:
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
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Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
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The energy of the incident light is calculated as follows;
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Ф = E - K.E
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Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;