Question

A student conducts an experiment boiling water in a 3.96L pot. The Diameter of the base of the pot is 7-3/4” and the height is 5-1/8”.

2 Liters of water @ 23°C takes 17 minutes before it reaches a boil. It has a height in the pot of 2-3/4”. The process took 1 hour and 13 minutes to boil dry. How much energy was required to do this?

Answer 1

Total energy required = Energy required to turn water at 23⁰ celsius to water at 100⁰ celsius + water at 100⁰ celsius to steam at 100⁰ celsius

Energy required to turn water at 23⁰ celsius to water at 100⁰ celsius = mass*(specific heat)*(temperature change)
but mass = density * volume = 1g/cc * 2lit = 1g/cc * 2000 cc    (since 1 lit = 1000cc)
mass = 2 kg
specifc heat of water = 4.187 kJ/(Kg-⁰C)
temperature change = 100-23 = 77 ⁰C

Thus, Energy required to turn water at 23⁰ celsius to water at 100⁰ celsius = 2kg*4.187 kJ/(Kg-⁰C)*77⁰C = 644.798 kJ

water at 100⁰ celsius to steam at 100⁰ celsius = mass * latent heat of evaporation
latent heat of evaporation of water = 2260 kJ/kg

Thus, water at 100⁰ celsius to steam at 100⁰ celsius = 2kg*2260kJ/kg = 4520 kJ

total energy required for boiling = 4520 + 644.798 = 5164.798 kJ = 5.164 MJ