How much heat must be removed when 25 grams of steam at 100°C Is condensed to form liquid water at 100°C?
Well, to not steamroll you with a technical answer, let's just say that a lot of heat needs to be removed. I mean, we're going from steam to liquid water here, so it's definitely going to be pretty steamy. But on the bright side, you won't need to summon Elsa to do it - just some good old-fashioned cooling down should do the trick!
To calculate the amount of heat that must be removed when steam is condensed to form liquid water, we need to use the formula for heat absorption or release during a phase change:
Q = m * ΔH
Where:
Q = heat absorbed or released
m = mass of the substance undergoing the phase change
ΔH = enthalpy of the phase change (latent heat)
In this case, the steam is being condensed, so the enthalpy of vaporization (latent heat of steam) would be used. The latent heat of vaporization for steam is approximately 2260 J/g.
Given:
Mass of steam (m) = 25 grams
Plugging the values into the formula:
Q = 25 g * 2260 J/g
Calculating:
Q = 56,500 J
Therefore, 56,500 Joules of heat must be removed when 25 grams of steam at 100°C is condensed to form liquid water at 100°C.
To solve this problem, we need to understand the concept of heat transfer during a phase change. When steam condenses to form liquid water, it undergoes a phase change from a gas to a liquid. During this phase change, there is a release of heat energy.
Now, we can calculate the amount of heat released during the condensation process using the formula:
Q = m * ΔH
where Q represents the heat energy, m is the mass of the substance, and ΔH is the heat of vaporization, which is the amount of energy required to change 1 gram of a substance from a gas to a liquid at its boiling point.
For water, the heat of vaporization is approximately 2260 joules per gram (J/g).
Given:
Mass of steam (m) = 25 grams
Heat of vaporization (ΔH) = 2260 J/g
Now, we can calculate the heat energy released during the condensation process:
Q = 25 g * 2260 J/g
Q = 56500 J
Therefore, when 25 grams of steam at 100°C is condensed to liquid water at 100°C, approximately 56500 joules of heat energy must be removed.