Final answer:
To calculate the total heat absorbed by water to become steam, we consider the heat required to raise the temperature to the boiling point, the heat of vaporization, and the heat needed to raise the temperature of steam to the final temperature.
Step-by-step explanation:
The student is asking about the amount of heat absorbed when water is heated from a certain temperature to steam at another temperature. This process involves two main phases: the heating of liquid water to its boiling point and the phase change from water to steam. To calculate the overall heat absorbed, we must account for the heat required to raise the temperature of water to its boiling point and then the heat required for the phase change.
The specific heat of water is 4.184 J/(g°C), which is the amount of heat needed to raise 1 gram of water by 1°C. At the boiling point, water undergoes a phase change from liquid to steam, which requires a heat of vaporization of 539 cal/g (or 2256 J/g, converting calories to joules by multiplying by 4.184 J/cal). Beyond the boiling point, the water, now in gaseous form, continues to absorb heat at a specific heat for steam which is 2.02 J/(g°C).
To calculate the total heat absorbed, the process is as follows:
- Heat required to raise temperature from 60.0°C to 100.0°C (boiling point of water):
q=m × c × ΔT
where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. - Heat required for the phase change from water at 100.0°C to steam at 100.0°C:
q = m × ΔHvap
where ΔHvap is the heat of vaporization. - Heat required to raise the temperature of steam from 100.0°C to 140.0°C:
q = m × c × ΔT
where c is the specific heat of steam.
By summing all these heat values, we get the total heat absorbed by the 2 g of water to become steam at 140.0°C.