The initial water temperature is 20 degrees, the efficiency of the kettle is 75%, the cost of 1 kW of electricity

The initial water temperature is 20 degrees, the efficiency of the kettle is 75%, the cost of 1 kW of electricity is 3.01 rubles. How much does it cost to boil half a liter? Specific heat 4200

t1 = 20 ° C.

t2 = 100 ° C.

V = 0.5 l = 0.5 * 10-3 m3.

ρ = 1000 kg / m3.

Efficiency = 75%.

С = 4200 J / kg * ° C.

C1 = 3.01 rubles.

C -?

Let us write the formula for determining the efficiency: efficiency = Ap * 100% / Az, where Ap is the useful work that goes to heating water to the boiling point, Az is the work expended.

We express the useful work An by the formula: An = C * m * (t2 – t1).

We express the mass of heated water m by the formula: m = ρ * V.

An = C * ρ * V * (t2 – t1).

Efficiency = C * ρ * V * (t2 – t1) * 100% / Az.

Az = C * ρ * V * (t2 – t1) * 100% / efficiency.

Az = 4200 J / kg * ° C * 1000 kg / m3 * 0.5 * 10-3 m3 * (100 ° C – 20 ° C) * 100% / 75% = 224000 J.

Nz = Az / 1 hour.

Nz = 224000 J / 3600 s = 62.2 W.

Ts = Nz * Ts1.

C = 62.2 * 3.01 / 1000 = 0.187 rubles.

Answer: half a liter of water is worth boiling C = 0.187 rubles.