In an electrically heated boiler, water is boiled at 140°c by a 90 cm long, 8 mm diameter horizontal heating element immersed in water. the heating element is made of mechanically polished stainless steel. the boiler initially contains 0.25 m of water at 20°c. once boiling starts, it is observed that 25% of the water in the boiler evaporated in 60 min. i. determine the power rating of the electric heating element immersed in water and the surface temperature of the heating element. [8 marks] ii. using the above heater, estimate the time required to raise the temperature of 0.25 m2 of cold water from 20°c to 140°c

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.

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