lilsis76
  • lilsis76
An 8.5g Ice Cube is placed into 255g water. Calculate temperature change in water upon the complete melting of the ice. (Assume all the energy required to melt the ice comes from the water. )
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
aaronq
  • aaronq
okay for this you have to use this equation "q=m*C*\(\Delta T\)", and, q=\(\Delta H_{fusion}*n\). remember that this last one can be n or m, depending on what the units for \(\Delta H_{fusion}\) are. ------------------------------------------------------- Also, remember that \(\Delta T=T_f-T_i\) where, \(T_f\)=final temperature and \(T_i\)=initial temperature although this won't be used for this specific question ------------------------------------------------------- So, what you wanna do is to determine what you're ultimately looking for. The question asks for "Calculate temperature change in water", which is \(\Delta T\). So, first you have to find the heat (q) required to melt the ice (which will come from the water). so convert the mass of ice to grams, then mass to moles. (look up a conversion chart on google if that helps). and use \(q_{ice}=\Delta H_{fusion}*n\) (\(\Delta H_{fusion}\) needs to be looked up) Then, since all the heat to melt the ice came from the water, we express it mathematically, like this: \(q_{ice}=-q_{water}\) so, \(q_{ice}=-q_{water}=m*C*\Delta T\) Here you're using the heat capacity (C) for water, the mass of the water (not including the ice) and you're solving for \(\Delta T\).
aaronq
  • aaronq
EDIT: sorry i had a mental relapse when i said "convert the mass of ice to grams, then mass to moles." It should only be "mass to moles".
lilsis76
  • lilsis76
haha okay im almost done reading what you put.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

aaronq
  • aaronq
okay.
lilsis76
  • lilsis76
okay im on the first step let me look in the book or heat required to melt the ice unless its 40.7 kj/mol
aaronq
  • aaronq
no thats to vaporize (turn into steam).
lilsis76
  • lilsis76
okay so.. in the book to melt a solid ice cube to water is....2) Ice melting to liquid = 6.02 kJ/mol
aaronq
  • aaronq
okay, so convert the mass of ice to moles
lilsis76
  • lilsis76
so i could do the 6.02 kj/mol // 1000 kj= .00602 mol
lilsis76
  • lilsis76
right?
aaronq
  • aaronq
Just follow what i said
aaronq
  • aaronq
n=m/M= 8.5 g /18g/mol =
lilsis76
  • lilsis76
what does the q mean again?
aaronq
  • aaronq
heat
lilsis76
  • lilsis76
okay .472 moles
aaronq
  • aaronq
okay, not find the heat required to melt that much ice q=\(\Delta H_{vap}*n\)
aaronq
  • aaronq
now find*
lilsis76
  • lilsis76
okay ... .472mol (6.02 kj/ 1 mol)=2.84
aaronq
  • aaronq
good stuff! don't forget the units 2.84 kJ now, the heat required to melt the ice comes from the water, so, \(q_{ice}=- \;q_{water}\) 2.84 kJ = -2.84 kJ use the second formula, -2.84 kJ = m*C*\(\Delta T\) you should convert 2.84 kJ to J (multiply by 1000).
lilsis76
  • lilsis76
-2840J
aaronq
  • aaronq
nice, now solve for \(\Delta T\), -2840 J = m*C* \(\Delta T\) C=4.184 J/\(C^{\circ}g\)
lilsis76
  • lilsis76
okay @aaronq i got -5608.56 moles
aaronq
  • aaronq
your answer should be in terms of degrees celsius because you're solving for the change in temperature, not moles.
aaronq
  • aaronq
so you have 255 g of water -2840 J = (255 g)*(4.184 J/C*g)*\(\Delta T\)
lilsis76
  • lilsis76
sorry, celcius
lilsis76
  • lilsis76
hey @aaronq I have to go. I will continue this tomorrow. Have a goodnite!
aaronq
  • aaronq
okay, see ya !
lilsis76
  • lilsis76
Hi @aaronq Sorry I had go go, my folks got back home, and I didnt want them to see me crying.
lilsis76
  • lilsis76
Im ready, refreshed and Im ready to go :)
aaronq
  • aaronq
So did you get that last part?
lilsis76
  • lilsis76
yes i did :) it is wait haha
lilsis76
  • lilsis76
(-2840 J) = (255g)(4.184 J/ C*g) (delta T) I rearranged to get the delta T(temperature change) to one side so it looks like this, right? \[\Delta T = -2840 J (255g)(4.184\frac{ J }{ C*g }\]
aaronq
  • aaronq
not quite, you're algebraically solving. A*B=C ->solving for A, A=\(\dfrac{C}{B}\) right?
lilsis76
  • lilsis76
yes, okay so then it would look like this, hold on
lilsis76
  • lilsis76
\[-2840 J / (255 g) (4.184 \frac{ J }{ C * g } )\]
aaronq
  • aaronq
thats' right. notice that the units "g" and "J" will cancel out.
lilsis76
  • lilsis76
okay they cancel
lilsis76
  • lilsis76
i got then
lilsis76
  • lilsis76
-2.66 Celcius
aaronq
  • aaronq
okay good stuff.
lilsis76
  • lilsis76
is this correct? then Im guessing we have to do another calculation, right? @aaronq
aaronq
  • aaronq
no that's it, the question is asking you for the change in temperature. The change in temperature will be -2.66 celsius, since you're not given the temp of the water, you can't say what it's final temp will be.
lilsis76
  • lilsis76
oh thats true..................... YOURE SO AWESOME! :D
aaronq
  • aaronq
haha thanks ! chemistry is my strong point
lilsis76
  • lilsis76
it really is haha okay, I have another Problem, I will close this one. thank you very mucH!
aaronq
  • aaronq
no problem! okay
lilsis76
  • lilsis76
when I try messaging you it says only accepts people that are fanned, what does that mean? I was thinking about sending you a message if and when I need help later on.
lilsis76
  • lilsis76
@aaronq
aaronq
  • aaronq
yeah, sorry about that. I switched my settings so only people i'm a fan of will be able to msg me because i used to get A LOT of messages of people asking for help, it got pretty annoying. But i've fanned you, now you can msg me! lol
lilsis76
  • lilsis76
okay, thank you, i was like WHA?! 0.o haha
lilsis76
  • lilsis76
WAIT!!! i just saw the rest of the quesiton to the problem haha we found the temperature change. it says calculate temperature change in water upon the complete melthing of the ice. Assume all the energy required to melt the ice comes from the water. So then that -2.66 C is still the answer right? I just want to double chekc.
aaronq
  • aaronq
yep, we calculated the heat needed to melt the ice, which would be extracted from the liquid water, and we found the change of temperature of the water.
lilsis76
  • lilsis76
okay, haha I was gonna freak when I saw that i was like wait aminute did i forget something haha.
lilsis76
  • lilsis76
thank you Aaronq. Im going to close this problem, i have to clean up the house to give myself a quick break.
aaronq
  • aaronq
okay! clearing up your head with some other activity is a good idea, see ya
lilsis76
  • lilsis76
thank you Aaronq. Im going to close this problem, i have to clean up the house to give myself a quick break.

Looking for something else?

Not the answer you are looking for? Search for more explanations.