Revisit SP6 Reanalyze that problem assuming the rangetowater

Revisit SP6. Reanalyze that problem, 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

For SP6, the efficiency ratio from the heat output from the stove to the water to be boiled was (1175.625 kJ / 1500 kJ) after ten minutes of heating, or 78.4% efficiency.

Solution

Latent heat of evaporation of water = 2264.76 KJ/kg

For 3.75 L of water, latent heat requirement = 3.75*2264.76 = 8492.85 KJ

So total time require to evaporate 3.75 L of water = (8492.85/1175.625)*10 min = 72.24 min = 1 hr 12 min 14.469 sec