How much heat do you get out of steam when it goes from vapor (steam) to liquid water. Isn't that the heat of vaporization? I think it's something like 2257 J/g but you need to confirm that. If you have 10 g steam then the heat available to melt the ice is
10 g x 2257 J/g = 22570 J.
Now you know that ice melts and we must add heat to do that. How much heat? It takes about 334 J/g (but you need to confirm that by looking up the heat of fusion or enthalpy of fusion.
So mass ice x 334 j/g = 22570 J and solve for mass ice. Subtracting that mass from 80 g will tell you how much ice is left if you started with 80 g. That multiplied by the heat of fusion will tell you how much more heat is needed to melt the remaining ice. Then see if there is that much heat left in water at 100 to go to zero degrees? Post your work if you need further assistance. Check my thinking, too. I don't THINK I omitted any steps.
The question is:
A small block of ice at 0oC is subjected to 10 g of 100oC steam and melts completely. Show that the mass of the block of ice can be no more than 80 g.
There were no equations in my chapter to use for this question except Q ~ change in T. I can't figure out how this would relate to the problem, so it is either not the equation I need or I'm just braindead. I don't want the answer, but I need to know what I'm trying to do here. Thanks for any help Mary :-)
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Ok. I multiplied 10g x 2,256J/g (that was what was in the text) to get 22,560J.
I then set up the equation to find the maximum mass of the ice that could be melted with 22,560J:
m(334J/g) = 22,560J
m = 67.54g
According to my calculations, then, the ice couldn't have a mass of any more than 67.54g - not 80g, as the problem states.
Am I missing something here? Thanks!
I then set up the equation to find the maximum mass of the ice that could be melted with 22,560J:
m(334J/g) = 22,560J
m = 67.54g
According to my calculations, then, the ice couldn't have a mass of any more than 67.54g - not 80g, as the problem states.
Am I missing something here? Thanks!
Yes, I think you are missing something.
There is still heat in the steam which has now turned to water. It's temperature is 100 C so it can melt more ice. How much more?
What mass of ice can that heat melt?
mass ice x 334 = mass water x specific heat water x delta T. (The 100 C water COULD go from 100 to zero if enough ice is present.)
mass ice x 334 = 10 x 4.18 x 100 or
mass ice = 12.5 g
So 67.5 + 12.5 = 80. Thus, 80 g is the maximum amount of ice that could be melted by 10 g steam at 100 C. If MORE ice were present, there would be ice remaining after removal of all of the heat from the 100 C steam.
There is still heat in the steam which has now turned to water. It's temperature is 100 C so it can melt more ice. How much more?
What mass of ice can that heat melt?
mass ice x 334 = mass water x specific heat water x delta T. (The 100 C water COULD go from 100 to zero if enough ice is present.)
mass ice x 334 = 10 x 4.18 x 100 or
mass ice = 12.5 g
So 67.5 + 12.5 = 80. Thus, 80 g is the maximum amount of ice that could be melted by 10 g steam at 100 C. If MORE ice were present, there would be ice remaining after removal of all of the heat from the 100 C steam.
Thank you sooo much! I get it! I would never have understood all of the steps on my own. The text my class is using is great about explaining concepts, but terrible about connecting those concepts to the problems they have at the end of the chapter. I am so frustrated. This chapter, although I understood all of the concepts, is the first one where I have been unable to solve any of the problems on my own. I really appreciate your time - Thanks! Mary
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