Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Water and Ice mixture, Find original temperature of water
Reply to thread
Message
[QUOTE="Moon_tm, post: 5443531, member: 591026"] [h2]Homework Statement [/h2] An [I]8 cm3 [/I]ice cube (temperature = [I]0.00 °C[/I]) is dropped into a glass with [I]3 dL [/I]juice which results in the ice cube being melted. By doing this, the juice drops to a temperature of [I]3.00 °C[/I]. What was the temperature of the juice before the ice cube was added? (Treat juice as water in this assignment. The density for ice is [I]920 kg/m3[/I]. Assume no energy is lost to the surroundings). ice m[SUB]1[/SUB]=(920)(8E-6) = 7.36E-3 T[SUB]1[/SUB]=0 C juice=water m[SUB]2[/SUB]=(1000)(0.3) = 0.3kg T[SUB]0water[/SUB]=? mixture T[SUB]final[/SUB]=3°C[h2]Homework Equations[/h2] [/B] Q=mcΔT H=mL[SUB]f[/SUB][h2]The Attempt at a Solution[/h2] Energy gained by ice = Energy lost by water (Ice → Water[SUB]0C[/SUB]) + (Water[SUB]0C[/SUB] → Water[SUB]3C[/SUB]) = Water[SUB]initial[/SUB] → Water[SUB]3C[/SUB] m[SUB]1[/SUB]L[SUB]f[/SUB]+m[SUB]1[/SUB]c(T[SUB]3[/SUB]-T[SUB]0[/SUB]) = m[SUB]2[/SUB]c(T[SUB]initial[/SUB]-T[SUB]3[/SUB]) m[SUB]1[/SUB][L[SUB]f[/SUB]+c(T[SUB]3[/SUB]-T[SUB]0[/SUB])]/m[SUB]2[/SUB]c=T[SUB]initial[/SUB]-T[SUB]3[/SUB] [B]I am not sure whether in the case of water ΔT is (T[SUB]final[/SUB]-T[SUB]initial[/SUB]) or vice versa.[/B] I get either a negative temperature, [I]which is bs, [/I]or after swapping them around [B]T[SUB]initial[/SUB]= 5°C[/B] [I]which I do not quite believe, because who would want to cool down water by 2°C?! [/I] Please help, I am on the verge of questioning my sanity after spending more than an hour on this... [I](One year ago it would've been no problem for me, but after a year of little to no physics my brain went ..dumb :D)[/I] [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Water and Ice mixture, Find original temperature of water
Back
Top