Respuesta :
Explanation:
The given data is as follows.
Volume of water = 0.25 [tex]m^{3}[/tex]
Density of water = 1000 [tex]kg/m^{3}[/tex]
Therefore,  mass of water = Density × Volume
            = [tex]1000 kg/m^{3} \times 0.25 m^{3}[/tex]
            = 250 kg Â
Initial Temperature of water ([tex]T_{1}[/tex]) = [tex]20^{o}C[/tex]
Final temperature of water = [tex]140^{o}C[/tex]
Heat of vaporization of water ([tex]dH_{v}[/tex]) at [tex]140^{o}C[/tex] Â is 2133 kJ/kg
Specific heat capacity of water = 4.184 kJ/kg/K
As 25% of water got evaporated at its boiling point ([tex]140^{o}C[/tex]) 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)
              = [tex]133.3 \times 10^{3}[/tex] 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 = [tex]\frac{133.3 \times 10^{3}kJ}{3600 s}[/tex] = 37 kJ/s or kW
The power rating of electric heating element is 37 kW.
Hence, heat required to raise the temperature from [tex]20^{o}C[/tex] to [tex]140^{o}C[/tex] of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)
           = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)
           = 125520 kJ Â
Time required = Heat required / Power rating
            = [tex]\frac{125520}{37}[/tex]
            = 3392 sec
Time required to raise the temperature from [tex]20^{o}C[/tex] to [tex]140^{o}C[/tex] of 0.25 [tex]m^{3}[/tex] water is calculated as follows.
          [tex]\frac{3392 sec}{60 sec/min}[/tex]
           = 56 min
Thus, we can conclude that the time required to raise the temperature is 56 min.
