Question

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

Answer 1

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore,  mass of water = Density × Volume

                       = $1000 kg/m^{3} \times 0.25 m^{3}$

                       = 250 kg  

Initial Temperature of water ($T_{1}$) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ($dH_{v}$) at $140^{o}C$  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ($140^{o}C$) 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)

                           = $133.3 \times 10^{3}$ 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 = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ 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

                       = $\frac{125520}{37}$

                       = 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

                    $\frac{3392 sec}{60 sec/min}$

                     = 56 min

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