Final answer:
The total heat required to convert 100 g of ice at -100°C to water at its boiling point involves calculating the heat needed to warm the ice to 0°C, melt the ice, and then heat the water to 100°C. The total heat is the sum of the heat for each of these processes, taking into account the specific heat capacities and the heat of fusion.
Step-by-step explanation:
Calculating the Heat Required for Phase and Temperature Change
To calculate the total heat required to convert 100 g of ice at -100°C to water at its boiling point, we'll need to consider several steps involving specific heat capacities and heat of fusion and vaporization. Here's the breakdown of the calculation:
- Heat to warm the ice from -100°C to 0°C.
- Heat to melt the ice at 0°C to water.
- Heat to raise the temperature of water from 0°C to 100°C.
The first step is to use the specific heat capacity of ice (which is approximately 2.06 J/g°C) to calculate the energy required to raise the temperature of 100 g of ice from -100°C to 0°C. Then, use the heat of fusion (which is approximately 334 J/g) to calculate the energy needed to melt the ice. Lastly, use the specific heat capacity of water (about 4.18 J/g°C) to determine the energy required to bring the water from 0°C to 100°C. The total heat required is the sum of the heat for each of these processes.
The calculations are as follows:
- Q1 = m * Cice * ∆T (heating the ice)
- Q2 = m * Lf (melting the ice)
- Q3 = m * Cwater * ∆T (heating the water)
Where Q is the heat energy, m is the mass, Cice is the specific heat capacity of ice, Cwater is the specific heat capacity of water, ∆T is the change in temperature, and Lf is the latent heat of fusion. Sum Q1, Q2, and Q3 to find the total heat required.