M = 1.5 kg of water t = 5 min U = 220V I = 5A find the temperature for which 1.5 kg of water can

M = 1.5 kg of water t = 5 min U = 220V I = 5A find the temperature for which 1.5 kg of water can be heated on an electric stove.

m = 1.5 kg.

t = 5 min = 300 s.

U = 220 V.

I = 5 A.

С = 4200 J / kg * ° C.

ΔТ -?

When the electric current passes through the spiral, the amount of heat Q is released in it, which is determined by the Joule-Lenz law: Q = I * U * t, where I is the current in the spiral, U is the current voltage, t is the time of passage of the current.

The amount of heat required for heating water Q is expressed by the formula: Q = C * m * ΔТ, where C is the specific heat capacity of water, m is the mass of heated water, ΔТ is the change in water temperature.

I * U * t = C * m * ΔТ.

ΔТ = I * U * t / C * m.

ΔТ = 5 A * 220 V * 300 s / 4200 J / kg * ° C * 1.5 kg = 52.4 ° C.

Answer: the temperature of the water on the tile will increase by ΔТ = 52.4 ° C.