An electric kettle heats water with a volume of 1.5 liters from 10 to 100 degrees in 10 minutes.

An electric kettle heats water with a volume of 1.5 liters from 10 to 100 degrees in 10 minutes. What is the power consumed by the kettle if the kettle’s efficiency is 75%?

Here is given in the SI system:
V = 1.5 l = 0.0015 m³
τ = 10 min = 600 s
η = 75% = 0.75
1. The amount of heat spent on heating the body is equal to the product of the specific heat capacity of the substance, body weight and the difference between the final and initial temperatures.
Q = c * m * (t2-t1), where c is the specific heat capacity of the substance, m is the mass of the substance, t2 and t1 are the final and initial temperatures, respectively.
Specific heat of water c = 4200 J / kg * K, according to the reference book.
2. Power required to heat water:
P = Q / τ, Q – where the amount of heat, τ – heating time.
3. The power consumed by the kettle is found through the efficiency:
η = P / W
W = P / η
4. Substitute into this expression the power required for heating:
W = P / η = Q / τ * η = c * m * (t2-t1) / (τ * η).
5. Let’s write down the expression to determine the mass by specific gravity (density):
m = ρ * V, where ρ is the density of the substance, V is the volume of the substance.
The density of water according to the reference book is 1000 kg / m³.
6. Substitute this formula into the expression from item 4.
W = c * ρ * V * (t2-t1) / (τ * η) = (4200 * 1000 * 0.0015 * (100-10) / (600 * 0.75)) = 1260 W

Answer: power 1260 W.