A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the boiling temperature, assuming that 70.0% of the available energy is absorbed by the water

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁) Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t Equation 2

make t  the subject of the equation

t = Q/P Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.