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H D 4 35% EE Tue Feb 7 1:44 AM Q BE Safari File Edit View History Bookmarks Window Help Homework: HW3 of Ch18 https www.fl pitphysics.co Course/ViewProblem? unit tem 2422614&enrollment; ID 236317 Reade iiii WebAdvisor Main Menu Gmail Fresno City ...ege Canvas Fresno State...board Learn Google Scholar plagiarism Online Text Correction Homework: HW3 of Ch18 7-kg Cube of Copper (cCu 386 J/kg-K 4.4-kg block o nally at 263 K Contact US Unit 3: Homework Problems Homework: HW3 Of Ch18 Print Assignment View Deadline: 100% until Monday, February 6 at 11:59 PM Standard Exercise Boiling Water Ice and Silver Standard Exercise Cooling Copper A 4-4-kg block of ice originally at 263 K is placed in thermal contact with a 14.9-kg block of silver (CAE 233 J/kg-K) which is initially at 1027 K. The H20 silver system is insulated so no heat flows into or out of it Standard Exercise Ice and Silver 1) At what temperature will the system achieve equilibrium? 383 K Submit Tipler6 18 P.030. You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Standard Exercise Tipler6 18. P.031. 2) What will be the phase of the H20 at equilibrium? Standard Exercise Tipler 18.P.042 Solid (ice) Liquid (water) Gas (vapor) Submit You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question.Explanation / Answer
a) 0 = (mi)(ci)(273.15 - 263) + (mi)(Lf) + (ms)(cs)(T - 1027) + (mi)(cw)(T - 273.15)
mi = mass of ice = 4.4-kg
ci = specific heat ice = 2100 J/kg-K
273.15 when ice melts. We now have water!
Lf = heat of fusion = 334000 J/Kg
ms = mass of silver = 15.6-kg
cs = specific heat silver = 233 J/kg-K
cw = specific heat water = 4190 J/kg-K
sub all these values into the eqution and rearrange for T.
0= (4.4*2100)10.15 + (4.4*334000) +(15.6*233)(T - 1027) +(4.4*4190)(T - 273.15)
0=1563386 + 3634.8T - 3732939.6 + 18436T - 5035793
T = 7205346.6 / 22070.8
T= 326.46 K
b) Phase of H2O at equilibrium is liquid(water)
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