A 115-g sample of steam at 100 °C is emitted from a volcano. It condenses, cools, and falls as snow at 0.0 °C. How many kilojoules were released?
Can some explain how to do these kind of questions?
q1 for condensation at 100 C = mass steam x heat condensation = ?
q2 for water @ 100 C to water @ 0 C = mass H2O x specific heat H2O x (Tfinal-Tinitial) = ?
q3 to freezing H2O = mass H2O x heat fusion H2O @ 0 C = ?
Add q1 + q2 + q3 for total. q1, q2, & q3 are negative quantities and total will be negative.
How to do these in general follows two formulas as follows:
When you go from one phase to another (melting/freezing or boiling/condensation) it is
q for changing from solid to liquid or liquid to solid = mass of whatever x heat fusion if melting or heat solidification if freezing. Remember that heat fusion is the negative of heat solidification and heat vaporization is the negative of heat of condensation.
When you stay in the same phase,(changing temperature withing liquid phase or solid phase or vapor phase) it is
q = mass of whatever x specific heat in that phase x (Tfinal-Tinitial)
Those two equations will do it all. You may have four or five values for q all the way from q1 to q5. For example, from ice @ -20 C to steam @ 135 C goes through these steps.
q1 = within the solid phase from -20 C to 0 C.
q2 = phase change for solid ice @ 0 C to liquid @ 0 C.
q3 = within the liquid phase for water from 0 C to 100 C.
q4 = phase change for liquid @ 100 C to vapor (steam) @ 100 C.
q5 = within the vapor phase @ 100 C to vapor @ 135 C.
But get all of that with the two equations at the appropriate time, calculate each segment, then add all of them together for the total. Hope this helps.
I did all that writing for nothing? I poured my soul into that. But that's my problem. To answer your question about q1, no, what you suggest is not correct. Here is what I wrote?
q1 for condensation at 100 C = mass steam x heat condensation = ?
Mass 115 g from the problem.
The problem says steam so you're supposed to know that that is the boiling point of water; therefore, you are going from the vapor phase to the liquid phase @ 100 C and that tells you to use the formula q1 = mass x heat condensation = ? So I plugged in q1 = what we want. Mass steam is 115 g from the problem. I assumed you have or have access to the heat of condensation. I looked that up on google and came up with 2260 J/g for H2O So plug in and the completed formula is q1 = 115 g x -2260 J/g*C = -259900 J which is the same as -259.9 kJ. Those negative signs tell you that steam is giving away heat (not absorbing it). Let's go to q2.
q2 is to get us from liquid water @ 100 C to liquid water @ 0 C. That formula is
q2 = mass water x specific heat H2O x (Tfinal-Tinitial). Plug in the numbers.
q2 = 115 g x 4.184 J/g*C x (0 - 100) = -48,116 J or -48.2 kJ to round to three significant figures.
I won't go through q3; it's just like q1 except the heat solidification (at freezing) is not the same as the heat condensation (at boiling).
That should clear up q1, q2, and q3. Let me know if you have further questions. Show your work and be especially clear about what you don't understand. Giving examples of what you're have trouble with may help.
First, make sure you confirm those final numbers for q1 and q2. I have been known to punch in the wrong numbers and/or to make a typo.
Second, I punched in the numbers for q3 and didn't get your answer.
I have 115 x -334 J/g*C = 38410 J or 38.4 kJ. Can't figure out how you came up with 51.77 kJ.
Glad to help. I agree that summer classes are a bit fast.
yes.
I'm sorry, I still don't understand how to do this.
So for q1, it's 100 = 115 * 0?
I don't understand how to q2 or q3.
Sorry I made you write all of that.. It's my first time taking Chemistry and it's an online summer course, so they're just throwing chapters for us to read too quickly for my brain to process everything.
I think I'm understanding it a little more, hopefully.
So q3 is 115g * -334 J/g°C = -51770 J = -51.77 kJ...?
Then you said to add them all up, so..
-259.9 -48.2 -51.77 = -359.87 kJ ?
Thank you so much for taking the time to explain everything to me, I really appreciate it.
Oops, I accidentally entered 155 * -334 in my calculator instead of 115
So the new total would be -346.41.
So the answer is -346kJ..?
Thank you very much!
A 135 g - g sample of steam at 100 ∘C is emitted from a volcano. It condenses, cools, and falls as snow at 0 ∘C . (For water, 80. cal (334 J ) is needed to melt 1 g of ice or must be removed to freeze 1 g of water and 540 cal (2260 J ) is needed to convert 1 g of water to vapor at 100 ∘C .)
How many kJ of heat were released?