Answer:
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
