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Problem 4.10
[Difficulty: 2]
Given:
Data on heating and cooling a copper block
Find:
Final system temperature
Solution:
Basic equation
QW
−
∆
E
=
Assumptions:
1) Stationary system
∆
E =
∆
U
2) No work W = 0
3) Adiabatic Q = 0
Then for the system (water and copper)
∆
U0
=
or
M
copper
c
copper
⋅
T
copper
⋅
M
w
c
w
⋅
T
w
⋅
+
M
copper
c
copper
⋅
M
w
c
w
⋅
+
()
T
f
⋅
=
(1)
where T
f
is the final temperature of the water (
w
) and copper (
copper
)
The given data is
M
copper
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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