Question

What will be the final temperature of the liquid water resulting from mixing 10 grams of steam at 130 degrees C with 40 g of ice at -10 degrees C?

Answer 1

I thought of breaking it down into steps (like do a phase change, then temp, then phase change etc) and for some reason... I don't think that's gonna work.

Answer 2

Here is the the speicfic heat etc info:
C of H20 (solid): 2.09 J/gC,
C of H20 (liquid): 4.18
C of H20 (gas): 2.03

Heat fusion of H20 (solid): 333 J/g
Heat vaporization of H20 (liquid): 2260 J/g

Answer 3

When we combine the two, the heat energy from the steam will first heat the ice to the melting point, then we'll have a phase change which will absorb more energy, then remaining energy from the steam will heat the resulting water, and if there's enough energy it might even cause another phase change into steam and heat the steam.

What are the formulas we have to work with?

Answer 4

q=mc(tf-ti) where tf = final temp and ti= initial temp and then you use the q=m x (heat fusion or vapor) for the actual points of phse change (solid-liquid, liquid-gas)

Answer 5

so i should go from 130 C to 100 C, then phase change, then go from 100 to 0, then phase change, then go from 0 to -10?

Answer 6

Okay, to heat the ice to the melting point, we need to supply
\[Q_{warm ice} = m_{ice}c_{H20 solid}\Delta{t} =(40g)(2.09 J/gC)(0 C -(-10C)) = 836J \]
To melt the ice, we need to supply
\[Q_{melt ice} = m_{ice}H_{fusion} = (40g)(333 J/g) = 13320J\]

Answer 7

But we can't just go marching through in this fashion because we don't know how much heat we have to supply yet, and we don't know if some of the steam will condense into water. We need to write a big equation that shows the heat gained by the ice = the heat lost by the steam, both sides being a function of the final temperature.

Answer 8

ohh, ok

Answer 9

look at what i found lol!!!
http://www.onlineconversion.com/mixing_water.htm

Answer 10

but my choices are a. 80.4 C b. 72.6 C c. 54.3 C d. 46.1 C e. 63.3 C

Answer 11

so the closest answer is e.. hm..

Answer 12

Before we do that, just to get a better feel for what's going to happen, let's do the corresponding calculation to see how much energy would be given up if we condensed all the steam into water:
\[Q_{coolsteam} = m_{steam}c_{H20gas}\Delta{t} = (10g)(2.03J/gC)(100C-130C) = 609J\]
\[Q_{condensesteam} = m_{steam}H_{H20vapor} = (10g)(2260J/g) = 22600J\]

Answer 13

I figured there must be such a calculator around, but in the Nietzschean spirit of "that which does not destroy us only makes us stronger" decided it would be better to work through the problem :-)

Answer 14

Ill upload my work in just a few secs :)

Answer 15

We do know now (thanks to our "tip") that all the steam condenses to water!

I just got t = 63.36

Answer 16

whoa, ur awesome!!! umm i like was following something and i ended up getting 117.27 C which is obviously NOT the answer lol

Answer 17

After melting the ice and condensing the steam, there's 22600+609-(13320+836) = 9053J of heat energy left over.

9053J = heat by warming water - heat lost by cooling steam
\[9053J = m_{ice}c_{H20liquid}\Delta{t_ice} - m_{steam}c_{H20liquid}\Delta{t_steam}\]
\[9053J = (40g)(4.18 J/gC)(t_{mix} - 0C) - (10g)(4.18J/gC)(100C-t_{mix})\]
\[9053J = 167.2t_{mix}-4180J+41.8t_{mix}\]\[13233J = 209t_{mix}\]\[t_{mix} = 63.3C\]

(earlier report of 63.36 was due to data entry mistake)

Answer 18

My description in the first statement there isn't so great :-)

Answer 19

and I forgot some {}'s to make the subscripts on delta t look nice. too much blood in my caffeine stream this morning :-)

Answer 20

yay, thank you!!! I am going to go back and look at all the work and see how u did it :D

Answer 21

I wonder why that calculator gets the slightly different answer? Different values for the constants, perhaps?

Answer 22

Yes thats what i assumed, and why is it [heat by warming water - heat lost by cooling steam] and not the other way around?

Answer 23

9053 J is the net heat: raising the water to the final temperature takes so much, and cooling the steam to the final temperature adds, because it's subtracting a negative number....

Answer 24

Oh, Ok. Thanks!

Answer 25

Here's how I should have written out:

\[Heat_{gained} = Heat_{lost}\]
\[m_{ice}C_{ice}(0C-t_{ice})+m_{ice}H_{fusion}+m_{ice}C_{water}(t_{f}-0C) =\]\[m_{stm}C_{stm}(t_{stm}-100C)+m_{stm}H_{vapor}+m_{stm}C_{water}(100C-t_{f})\]

Answer 26

Thank you!! I got it lol :D