How did they get kg at the end please help mi u mass oidlan g iceHeat Transfer rate due to conduetivity,dQdtdQ is change in energydt b timeKA AT%3DdQ- m;(bdtk = 0.033 W/mkAT is temperature dfferencek io conductivity og styrfoomA i surface ariat u thikness.%3DThr = 3600 seconds4 is ladent heat of fusion of icemi n mass oficdmelted.A =0.016 mAT = 35 - 0 = 35t = 2x10-mL=33600 J/KStep2b)= KA AT dt0.033 X0.0l6 X 35 X 360 02x10-3 x 336 00mimi - 0.99 kgmow ofmassice melted .

Answered: Image /qna-images/answer/16643487-fbe2-4c53-962e-ff7268608dae.jpg

Answered: Step2 b) .*. mi = KA AT dt 0.033 XO.016…

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Step2 b) .*. mi = KA AT dt 0.033 XO.016 X 35 X 360 0 2x10-3 x 336 00 こ mi - 0.99 kg mass of ice melted .

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

How did they get kg at the end please help

mi u mass oidlan g ice
Heat Transfer rate due to conduetivity,
dQ
dt
dQ is change in energy
dt b time
KA AT
%3D
dQ- m;(b
dt
k = 0.033 W/mk
AT is temperature dfference
k io conductivity og styrfoom
A i surface aria
t u thikness.
%3D
Thr = 3600 seconds
4 is ladent heat of fusion of ice
mi n mass oficdmelted.
A =
0.016 m
AT = 35 - 0 = 35
t = 2x10-m
L=33600 J/K
Step2
b)
= KA AT dt
0.033 X0.0l6 X 35 X 360 0
2x10-3 x 336 00
mi
mi - 0.99 kg
mow of
mass
ice melted .

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