Final answer:
To find the heat required to warm 10.0 g of ice from -10.0°C to steam at 110°C, multiple stages of heating and phase changes must be considered with their respective specific heats and latent heats. The total heat absorbed is the sum of the heats required for each stage, and this sum in joules can be converted to kilojoules for the final answer.
Step-by-step explanation:
To calculate the amount of heat required to warm 10.0 g of ice from -10.0°C to steam at 110°C, we need to consider the various stages of heat transfer, including the warming of ice, melting of ice, heating the resulting water, vaporizing the water, and then heating the steam. Each of these stages has a different specific heat value or requires heat of fusion/vaporization that we must account for. Using the given heat capacities, we will calculate the heat for each of these steps:
- Heating the ice from -10.0°C to 0°C: q1 = m x Cice x ΔT
- Melting the ice at 0°C (heat of fusion): q2 = m x ΔHfus
- Heating the water from 0°C to 100°C: q3 = m x Cwater x ΔT
- Vaporizing the water at 100°C (heat of vaporization): q4 = m x ΔHvap
- Heating the steam from 100°C to 110°C: q5 = m x Csteam x ΔT
Where m is the mass of the ice (10 g), Cice is the specific heat capacity of ice (2.09 J/g°C), Cwater is the specific heat capacity of water (usually about 4.18 J/g°C), Csteam is the specific heat capacity of steam (2.01 J/g°C), ΔHfus is the heat of fusion for ice (about 334 J/g), and ΔHvap is the heat of vaporization for water (about 2260 J/g). Note that the values for ΔHfus and ΔHvap are not given in the question but are commonly known. The sum of q1 through q5 will give us the total energy required in joules, which we can then convert to kilojoules (kJ) by dividing by 1000.
Calculations must be performed with the appropriate values for heat capacities and latent heats of fusion and vaporization, which are not provided in the question prompt but are typically known constants.