Explanation:
The given data is as follows.
Volume of water = 0.25 
Density of water = 1000 
Therefore, mass of water = Density × Volume
= 
= 250 kg
Initial Temperature of water (
) = 
Final temperature of water = 
Heat of vaporization of water (
) at
is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point (
) in 60 min.
Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg
Heat required to evaporate = Amount of water evapotaed × Heat of vaporization
= 62.5 (kg) × 2133 (kJ/kg)
=
kJ
All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec
Therefore, heat supplied per unit time = Heat required/time =
= 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from
to
of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)
= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)
= 125520 kJ
Time required = Heat required / Power rating
=
= 3392 sec
Time required to raise the temperature from
to
of 0.25
water is calculated as follows.

= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.