A small immersion heater is rated at 255 W . The specific heat of water is 4186 J/kg⋅C°.
Estimate how long it will take to heat a cup of soup (assume this is 250 ml of water) from 17 °C to 62 °C. Ignore the heat loss to the surrounding environment.

0.2093 s

Explanation:

First, we the energy needed to heat the cup of soup

Q = cm(t₂-t₁) Equation 1

Where c = specific heat capacity of water, m = mass of water, t₂ = Final temperature, t₁ = initial temperature.

Given: c = 4186 J/kg.°C, m = density of water×volume of water,

Where density of water = 1 kg/m³, volume of water = 250/1000000 = 0.00025 m³,

Therefore, m = 1×0.00025  = 0.00025 kg, t₂ = 62°C, t₁ = 17 °C

Substitute into equation 1

Q = 4186(0.00025)(62-17)

Q = 47.0925 J.

Secondly we look for the time using

W = Q/t Equation 2

Where W = power rating of the heater.

make t the subject of the equation

t = Q/W Equation 3

Given: W = 225 W

Substitute into equation 4

t = 47.0925/225

t = 0.2093 s