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E D t 34% BE Tue Feb 7 1:46 AM Q E Safari File Edit View History Bookmarks Window Help Homework: HW3 of Ch18 CA https www.fl pitphysics.co Reade Course/ViewProblem? unit tem 2422613&enrollmentID; 236317 WebAdvisor Main Menu Fresno City ...ege Canvas Fresno State...board Learn Google Scholar plagiarism Online Text Correction Homework: HW3 of Ch18 Physics question I Chegg.com A 44-kg block of ice originally at 263 K Contact US Standard Exercise Cooling Copper A 8 kg cube of copper (cc 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.6 kg of water (cwater 4186 J/kg-K) with an initial temperature of 293 K. Standard Exercise Ice and Silver 1) What is the final temperature of the water-and-cube system? 344 Submi Tipler6 18 P.030. You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. Standard Exercise Tipler6 18. P.031. 2) If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent Standard Exercise heat of vaporization 2.26 x 106 J/kg) will be left after the water stops boiling? Tipler 18.P.042 Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question 3) Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need? Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this questionExplanation / Answer
1) let T is the final temperature.
use, Heat lost by the copper = heat gained by the water
8.1*386*(750 - T) = 5.6*4186*(T - 293)
on solving the above equation we get
T = 347
2) In this case the final temperature will be 373 k
Heat lost by copper, Q = 8.1*386*(1350 - 373)
= 3.054*10^6 J
Heat required to raise the temperature from 293 to 373
Q' = 5.6*4186*(373 - 293)
= 1.875*10^6 J
mass of water that vaporize, m = (Q - Q')/Lv
= (3.054 - 1.875)*10^6/(2.26*10^6)
= 0.522 kg
liquid water left out = 5.6 - 0.522
= 5.08 kg
3) let m is the mass of water that we need.
use, Heat lost by the copper = heat gained by the water
8.1*386*(750 - 373) = m*4186*(373 - 293) + m*2.26*10^6
on solving the above equation we get
m = 0.454 kg
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