Step-by-step explanation:
The given data is as follows.
Volume of water = 0.25
![m^(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/on1zxagzm25f0eany388hhun268dwjz20o.png)
Density of water = 1000
![kg/m^(3)](https://img.qammunity.org/2020/formulas/physics/high-school/lirf47ojrjk11hgq6hwp0oq73cln4ag79q.png)
Therefore, mass of water = Density × Volume
=
![1000 kg/m^(3) * 0.25 m^(3)](https://img.qammunity.org/2020/formulas/chemistry/college/ghd7ugsl0oge2lsg9xz8og1g1pyoc5t96d.png)
= 250 kg
Initial Temperature of water (
) =
![20^(o)C](https://img.qammunity.org/2020/formulas/chemistry/college/27wi7gyiix2e69f68lel9js8ini01jp0nl.png)
Final temperature of water =
![140^(o)C](https://img.qammunity.org/2020/formulas/chemistry/college/ab4p8e6xulj73iog6xmk9rc1ggzgr226lr.png)
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.
![(3392 sec)/(60 sec/min)](https://img.qammunity.org/2020/formulas/chemistry/college/pp09kofh1v5r7ckguleuijexne681c86uj.png)
= 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.