A four element electric stove has an 23 cm diameter burner t

\"A four element electric stove has an ~23 cm diameter burner that provides 2.5 kW of power. A pan is filled with 4 quarts (~3.75 l) of water initially at 25 C. The water boils in 10 minutes. What fraction of the burner output power is transferred to the water and what fraction is lost? Report your answer in %. [Recall from CHEM 115 that the specific heat for water is 4.18 kJ/kg-K.]\"

Reanalyze that problem above, assuming the range-to-water heat transfer efficiency is the same. Estimate the time it would take for the entire 3.75 l of water to evaporate. Report your answer in the form hr:min:sec.sec.

Solution

Burner Capacity : 2.5 kW

It can provide energy in 10 minutes = 2.5x10x60 = 1500 kJ.

Energy requirement for 3.75 l of water to boil = 3.75x1x4.18x(100-25) = 1175.625 kJ.

Fraction of burner output power is transferred to the water = 100 x 1175.625/1500 = 78.375 %.

Fraction of burner output power is lost = 100 - 78.375 = 21.625 %.

Energy requirement to evaporate 3.75 l of water = 1175.625 + 3.75 x 2257 = 9639.375 kJ.

Here 1 kg of water needs 2257 kJ to convert steam from water at 100C.

Time requirement = 9639.375/2.5 = 3855.75 seconds. = 1 hr 04 min 15.75 sec or 01:04:15.75