How much heat is added to 10.0 g of ice at -200°C to convert it to steam at 120.00°C?
1 Answer
34561.5 Joules
Explanation:
The standard heating/cooling curve goes through four phases. All temperature changes in a single phase (solid, liquid or gas) depend on the specific heat (oC/g) of the compound in that phase, and the inter-phase transition energies.
So, in this case we have to heat solid water from -200oC to 0oC, then change phase from solid to liquid at 0oC, followed by heating liquid water from 0oC to 100oC. Then the water must change phase again from liquid to gas, and then finally heat the gas phase from 100oC to 120oC.
Solid Specific Heat: 2.05 J/g-oK
Liquid Specific Heat: 4.178 J/g-oK
Vapor Specific Heat: 1.89 J/g-oK
Heat of Fusion: 333.55 J/g
Heat of Vaporization: 2257 J/g
10.0g * (200oC) * 2.05 J/g-oC + 10.0g * 333.55 J/g + 10.0g * (100oC) * 4.178 J/g-oC + 10.0g * 2257 J/g + 10.0g * (20oC) * 1.89 J/g-oC
10.0j * ((200oC * 2.05 J/g-oC) + 333.55 J/g + (100oC * 4.178 J/g-oC) + 2257 J/g + (20oC * 1.89 J/g-oC))
10.0g * (410J/g + 333.55 J/g + 417.8 J/g + 2257 J/g + 37.8 J/g)
= 34561.5 Joules
A good example of this type of problem is explained at the Khan Academy here:
https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/states-of-matter/v/specific-heat-heat-of-fusion-and-vaporization
