Will my heating process take more energy or less?
Substance - WATER
Specific Latent Heat of Fusion (kJ/mol) - 6.02
Melting Pt (°C) - 0
Specific Latent Heat of Vaporization (kJ/mol) - 40.69
Boiling Pt (°C) - 100
Substance (s) J/g x °C
A pot sits on a stove and holds 3.00 L of ice at –35 °C. The stove is turned on and begins to release heat at a steady rate to the pan of ice. Using the above info, calculate the following:
Mass of ice (density of 0.917 g/mL)
Moles of ice in pot
Amount of heat required to change the ice to 0.0 °C
A of H reqrd to liquefy ice at 0.0 °C
A of H reqrd to raise the temp of the water to 100 °C
A of H reqrd to vaporize water at 100 °C
A of H reqrd to raise the temp of the vapor to 130 °C
Total heat reqd to change the ice from its initial to 130 °C
The pot absorbs heat from the stove. Air molecules dissipate some of it before its transferred to the water. Will the heating process take more or less energy?
Given the setting you are descibing, it will take more heat than is required to get the ice to melt, then to raise the temperature of the cold water to boiling point. At 100 degrees Celsius, the liquid water achieves boiling point, where the liquid water molecules overcome their attraction for one another and become steam, or water in gaseous form. All the while this is going on, heat is transferred to the pot the ice is in, so some of the heat is lost in conduction to the pot. The air surrounding the ice receives some of the heat as it comes off the water, as it is absorbing heat on its way to the boiling point. So the answer to your general "lead" question is it will take more heat energy than is required to get the ice to the liquid state, then to the gaseous state, because much of the heat is being lost to the surrounding atmosphere.