QUICK CHEMISTRY QUESTION [physical chemistry]?

Question

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
Water 4.18
Ice 2.09
Steam 2.01

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?

Answer 1

Mass of Ice = 2751 grams
Moles of Ice = 152.83
Heat = 201.24 kJ
Heat = 920.04 kJ
Heat = 1149.92 kJ
Heat = 6218.65 kJ
Heat = 165.89 kJ

For total, add them up.

It will require more energy.

Answer 2

PV=nRT n=0.050 mol and m=6.9g CCl3F 13.0-6.9=6.1