When the T ambient is low, we can see frost formation when the surface of window or surface is less than 0 degree C. But how to calculate the energy released during that frost formation at the window pane? Will it just simple be Q = m x L? But the formation is directly fromm gas to solid.
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$\begingroup$ What specifically is $L$ measuring here? $\endgroup$– JMacCommented Nov 12, 2019 at 13:29
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$\begingroup$ L will be the specific latent heat of ice [J/kg] during frost formation. No? I don't comprehend it completely though. $\endgroup$– ZeeshanCommented Nov 12, 2019 at 15:14
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2 Answers
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Hess's Law https://en.wikipedia.org/wiki/Hess%27s_law, states that the energy change in a reaction (physical or chemical) depends only on the starting and end points, and not the "route" or individual steps one chooses for the reaction.
So:
- Find the energy supplied to warm a gram of water vapor from room temperature to $100$ degrees Celsius. The specific heat of water vapor may vary significantly over this temperature range.
- Find the energy released when the gram of $100$ degree C water vapor condenses to boiling hot liquid water;
- Find the energy released when the gram of $100$ degree liquid water cools to $0$ degrees Celsius;
- Find the energy released when a gram of $0$ degree Celsius water freezes:
- Find the energy released when a gram of ice cools from $0$ degrees Celsius to the temperature of the frost
The sum of these steps, with appropriate positive and negative signs, will give the energy change for the transition from room temperature water vapor to sub-zero ice. It is irrelevant that boiling hot liquid water in fact never exists in the process under consideration.
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No, the latent heat is only responsible for a small amount of energy exchanged between water and the surface. What you primarily have to take into consideration are the enthalpies for the transformation from liquid water to ice (and also from vapor to liquid if water is solved in air initially).
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$\begingroup$ The latent heat is the heat required to undergo the phase change. Your answer seems to be implying that it isn't. $\endgroup$– JMacCommented Nov 12, 2019 at 21:11