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39. We compute with Celsius temperature, which poses no difficulty for the J/kg·K
values of specific heat capacity (see Table 183) since a change of Kelvin temperature is
numerically equal to the corresponding change on the Celsius scale. If the equilibrium
temperature is
T
f
then the energy absorbed as heat by the ice is
Q
I
= L
F
m
I
+ c
w
m
I
(
T
f
– 0°C),
while the energy transferred as heat from the water is
Q
w
= c
w
m
w
(
T
f
–
T
i
). The system is
insulated, so
Q
w
+ Q
I
= 0, and we solve for
T
f
:
.
()
ww
i
FI
f
IC
w
cmT Lm
T
mm
c
−
=
+
(a) Now
T
i
= 90°C so
3
(4190J / kg C )(0.500kg)(90 C) (333 10 J / kg)(0.500kg)
5.3 C.
(0.500kg
0.500kg)(4190J / kg C )
f
T
⋅°
° −
×
==
°
+⋅
°
(b) Since no ice has remained at
5.3
f
TC
=°
, we have
0
f
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 Spring '08
 Any
 Physics, Heat

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