What is the final equilibrium temperature after you mix 10 grams of ice at 0 degrees C with 50 grams of water at 90 degrees C? I have posted my proffesor's solution I understand it for the most part but I'm confused on why in her solution she does the mass of ice times the specific heat of WATER times change in temp. Why does she use the specific heat of water and not ice?? I understand the rest of the problem but that is tripping me up
What is the final equilibrium temperature after you mix 10 grams of ice at 0 degrees C with 50 grams of water at 90 degrees C? I have posted my proffesor's solution I understand it for the most part but I'm confused on why in her solution she does the mass of ice times the specific heat of WATER times change in temp. Why does she use the specific heat of water and not ice?? I understand the rest of the problem but that is tripping me up
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What is the final equilibrium temperature after you mix 10 grams of ice at 0 degrees C with 50 grams of water at 90 degrees C? I have posted my proffesor's solution I understand it for the most part but I'm confused on why in her solution she does the mass of ice times the specific heat of WATER times change in temp. Why does she use the specific heat of water and not ice?? I understand the rest of the problem but that is tripping me up

Transcribed Image Text:udy
8350J = m
3 EQ-0
(334x10³₂)
Qice = mice Lfice + Mic Cw (Tf -0°C)
трее
Qunter = M₁ Cm (Tf -90°c)
I
mi 4f +M; C₂(Te -0°C) +mw Cw (Tf -90°C) = 0
Mi 4f + mi CwTf +mw Cw Tf - M₁²w (90°c) = 0
Tf (mic₁ + m₂ cw) = M₁ Cw (90° ( ) - milf
Te
m = 0.025 kg or
= mwcw 90°C - milf
(mitmw) Си
4 H = KATH-T² = 0.024 0.5
H = 1104 I
Q Search
72-(-20°C)
1x10'³ m
-·05kg 4190 90°℃-.01 (3346²)
(Obky) 4190
Tf
prt sc
F10
home
F11
end
25 grams
61.7°C
F12
Insert
6:14 AM
12/14/2022
delete
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