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.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

the answer is na (sodium)