Final answer:
To calculate the total heat absorbed when heating water from 25 °C to 120 °C vapor, calculate the heat for heating liquid water to 100 °C, vaporizing it at 100 °C, and heating the steam up to 120 °C. Add the heat calculated for each step to find the total. The specific heat and enthalpy of vaporization values for water are required.
Step-by-step explanation:
To calculate the heat (q) involved in heating water from a 25 °C liquid to a 120 °C vapor, three steps should be considered:
- Heating the liquid water from 25 °C to 100 °C using the specific heat of liquid water.
- Vaporizing the water at 100 °C to steam using the enthalpy of vaporization.
- Heating the steam from 100 °C to 120 °C using the specific heat of water vapor.
Using the equation q = m × c × ΔT, where m is the mass of water, c is the specific heat capacity, and ΔT is the change in temperature, calculate the heat for each of the three steps. First, calculate q for the increase in temperature from 25 °C to 100 °C for liquid water. Then calculate the heat required for phase change at 100 °C. Finally, calculate q for the temperature increase from 100 °C to 120 °C for vaporized water.
To find the total heat absorbed, sum the heat for each step. Note that the latent heat of vaporization and the specific heats for liquid water and steam must be in consistent units (usually joules or kilojoules).