Question

An electric heater used to boil small amounts of water consists of a 15-Ω coil that is immersed directly in the water. It operates from a 60-V socket. How much time is required for this heater to raise the temperature of 0.80 kg of water from 20° C to the normal boiling point?

Answer 1

Answer:

t = 1120 seconds

Explanation:

Law of Conservation of Energy: It states that energy can neither be created or destroyed but it can be transformed from one form to another

The heater converts electric energy to heat energy.

According to the law of conservation of Energy,

Heat supplied by the heater = heat gained by the water.

Heat supplied by the heater = (V²/R)t, Where V = voltage, R = Resistance, t = Time.

Heat gained by the water = cm(θ₂ -θ₁)

Where c = specific heat capacity of water, m = mass of water, θ₂= final temperature of water, θ₁= initial temperature of water.

∴ (V²/R)t =  cm(θ₂ -θ₁)

Making  t the subject of the equation,

t =  cm(θ₂ -θ₁) ×R/V²

Where c= 4200 Jkg/K, m=0.80 kg, R = 15 Ω, V = 60 V, θ₁= 20°C, θ₂ = the normal boiling point of water = 100°C

∴ t = (4200×0.8)(100 -20)×15/60²

t = (3360×80×15)/3600

t = 4032000/3600 = 1120 seconds

∴  The time required by the heater to raise the temperature of water to normal boiling point is 1120 seconds.